6 Vector Calculus 向量微积分

本页译自 OpenStax《Calculus Volume 3》第 6 章 Vector Calculus。公式经本地 MathJax 渲染，自定义宏已注入。

Chapter Outline 章节概览

• 6.1 Vector Fields
• 6.2 Line Integrals
• 6.3 Conservative Vector Fields
• 6.4 Green’s Theorem
• 6.5 Divergence and Curl
• 6.6 Surface Integrals
• 6.7 Stokes’ Theorem
• 6.8 The Divergence Theorem

• 6.1 向量场
• 6.2 线积分
• 6.3 保守向量场
• 6.4 格林定理
• 6.5 散度与旋度
• 6.6 曲面积分
• 6.7 斯托克斯定理
• 6.8 散度定理

6.1 Vector Fields 6.1 向量场

• 6.1.1 Recognize a vector field in a plane or in space.
• 6.1.2 Sketch a vector field from a given equation.
• 6.1.3 Identify a conservative field and its associated potential function.

• 6.1.1 识别平面或空间中的向量场。
• 6.1.2 根据给定方程描绘向量场。
• 6.1.3 识别保守场及其对应的势函数。

Vector fields are an important tool for describing many physical concepts, such as gravitation and electromagnetism, which affect the behavior of objects over a large region of a plane or of space. They are also useful for dealing with large-scale behavior such as atmospheric storms or deep-sea ocean currents. In this section, we examine the basic definitions and graphs of vector fields so we can study them in more detail in the rest of this chapter.

Examples of Vector Fields 向量场示例

How can we model the gravitational force exerted by multiple astronomical objects? How can we model the velocity of water particles on the surface of a river? Figure 6.2 gives visual representations of such phenomena.

Figure 6.2(a) shows a gravitational field exerted by two astronomical objects, such as a star and a planet or a planet and a moon. At any point in the figure, the vector associated with a point gives the net gravitational force exerted by the two objects on an object of unit mass. The vectors of largest magnitude in the figure are the vectors closest to the larger object. The larger object has greater mass, so it exerts a gravitational force of greater magnitude than the smaller object.

Figure 6.2(b) shows the velocity of a river at points on its surface. The vector associated with a given point on the river’s surface gives the velocity of the water at that point. Since the vectors to the left of the figure are small in magnitude, the water is flowing slowly on that part of the surface. As the water moves from left to right, it encounters some rapids around a rock. The speed of the water increases, and a whirlpool occurs in part of the rapids.

Each figure illustrates an example of a vector field. Intuitively, a vector field is a map of vectors. In this section, we study vector fields in $\mathbb{R}^{2}$ and $\mathbb{R}^{3}.$

A vector field $\mathbf{\text{F}}$ in $\mathbb{R}^{2}$ is an assignment of a two-dimensional vector $\mathbf{\text{F}}\left( {x,y} \right)$ to each point $\left( {x,y} \right)$ of a subset *D* of $\mathbb{R}^{2}.$ The subset *D* is the domain of the vector field.

A vector field F in $\mathbb{R}^{3}$ is an assignment of a three-dimensional vector $\mathbf{\text{F}}\left( {x,y,z} \right)$ to each point $\left( {x,y,z} \right)$ of a subset *D* of $\mathbb{R}^{3}.$ The subset *D* is the domain of the vector field.

Vector Fields in $\mathbb{R}^{2}$ $\mathbb{R}^{2}$ 中的向量场

A vector field in $\mathbb{R}^{2}$ can be represented in either of two equivalent ways. The first way is to use a vector with components that are two-variable functions:

$$\mathbf{\text{F}}(x,y) = \left\langle {P(x,y),Q(x,y)} \right\rangle.$$ (6.1)

The second way is to use the standard unit vectors:

$$\mathbf{\text{F}}(x,y) = P(x,y)\mathbf{\text{i}} + Q(x,y)\mathbf{\text{j}}.$$ (6.2)

A vector field is said to be *continuous* if its component functions are continuous.

Finding a Vector Associated with a Given Point 求给定点处关联的向量

Let $\mathbf{\text{F}}(x,y) = (2y^{2} + x - 4)\mathbf{\text{i}} + \text{cos}(x)\mathbf{\text{j}}$ be a vector field in $\mathbb{R}^{2}.$ Note that this is an example of a continuous vector field since both component functions are continuous. What vector is associated with point $\left( {0,-1} \right)?$

Solution 解

Substitute the point values for *x* and *y*:

$$\begin{array}{cl}{\mathbf{\text{F}}(0,-1)} & {= (2{(-1)}^{2} + 0 - 4)\mathbf{\text{i}} + \text{cos}(0)\mathbf{\text{j}}} \\ & {= -2\mathbf{\text{i}} + \mathbf{\text{j}}.}\end{array}$$

Let $\mathbf{\text{G}}\left( {x,y} \right) = x^{2}y\mathbf{\text{i}} - \left( {x + y} \right)\mathbf{\text{j}}$ be a vector field in $\mathbb{R}^{2}.$ What vector is associated with the point $\left( {-2,3} \right)?$

Drawing a Vector Field 描绘向量场

We can now represent a vector field in terms of its components of functions or unit vectors, but representing it visually by sketching it is more complex because the domain of a vector field is in $\mathbb{R}^{2},$ as is the range. Therefore the “graph” of a vector field in $\mathbb{R}^{2}$ lives in four-dimensional space. Since we cannot represent four-dimensional space visually, we instead draw vector fields in $\mathbb{R}^{2}$ in a plane itself. To do this, draw the vector associated with a given point at the point in a plane. For example, suppose the vector associated with point $\left( {4,-1} \right)$ is $\left\langle {3,1} \right\rangle.$ Then, we would draw vector $\left\langle {3,1} \right\rangle$ at point $\left( {4,-1} \right).$

We should plot enough vectors to see the general shape, but not so many that the sketch becomes a jumbled mess. If we were to plot the image vector at each point in the region, it would fill the region completely and is useless. Instead, we can choose points at the intersections of grid lines and plot a sample of several vectors from each quadrant of a rectangular coordinate system in $\mathbb{R}^{2}.$

There are two types of vector fields in $\mathbb{R}^{2}$ on which this chapter focuses: radial fields and rotational fields. Radial fields model certain gravitational fields and energy source fields, and rotational fields model the movement of a fluid in a vortex. In a radial field, all vectors either point directly toward or directly away from the origin. Furthermore, the magnitude of any vector depends only on its distance from the origin. In a radial field, the vector located at point $\left( {x,y} \right)$ is perpendicular to the circle centered at the origin that contains point $\left( {x,y} \right),$ and all other vectors on this circle have the same magnitude.

Drawing a Radial Vector Field 描绘径向向量场

Sketch the vector field $\mathbf{\text{F}}(x,y) = \frac{x}{2}\mspace{2mu}\mathbf{\text{i}} + \frac{y}{2}\mspace{2mu}\mathbf{\text{j}}.$

Solution 解

To sketch this vector field, choose a sample of points from each quadrant and compute the corresponding vector. The following table gives a representative sample of points in a plane and the corresponding vectors.

| | | | | | |
|-------------------------|-------------------------------------------------|-------------------------|-----------------------------------------|--------------------------|-------------------------------------------------------------|
| $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ | $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ | $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ |
| $\left( {1,0} \right)$ | $\left\langle {\frac{1}{2},0} \right\rangle$ | $\left( {2,0} \right)$ | $\left\langle {1,0} \right\rangle$ | $\left( {1,1} \right)$ | $\left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle$ |
| $\left( {0,1} \right)$ | $\left\langle {0,\frac{1}{2}} \right\rangle$ | $\left( {0,2} \right)$ | $\left\langle {0,1} \right\rangle$ | $\left( {-1,1} \right)$ | $\left\langle {- \frac{1}{2},\frac{1}{2}} \right\rangle$ |
| $\left( {-1,0} \right)$ | $\left\langle {- \frac{1}{2},0} \right\rangle$ | $\left( {-2,0} \right)$ | $\left\langle {-1,0} \right\rangle$ | $\left( {-1,-1} \right)$ | $\left\langle {- \frac{1}{2}, - \frac{1}{2}} \right\rangle$ |
| $\left( {0,-1} \right)$ | $\left\langle {0, - \frac{1}{2}} \right\rangle$ | $\left( {0,-2} \right)$ | $\left\langle {0,-1} \right\rangle$ | $\left( {1,-1} \right)$ | $\left\langle {\frac{1}{2}, - \frac{1}{2}} \right\rangle$ |

Figure 6.3(a) shows the vector field. To see that each vector is perpendicular to the corresponding circle, Figure 6.3(b) shows circles overlain on the vector field.

Draw the radial field $\mathbf{\text{F}}(x,y) = - \frac{x}{3}\mspace{2mu}\mathbf{\text{i}} - \frac{y}{3}\mspace{2mu}\mathbf{\text{j}}.$

In contrast to radial fields, in a rotational field, the vector at point $\left( {x,y} \right)$ is tangent (not perpendicular) to a circle with radius $r = \sqrt{x^{2} + y^{2}}.$ In a standard rotational field, all vectors point either in a clockwise direction or in a counterclockwise direction, and the magnitude of a vector depends only on its distance from the origin. Both of the following examples are clockwise rotational fields, and we see from their visual representations that the vectors appear to rotate around the origin.

Chapter Opener: Drawing a Rotational Vector Field 章首图：描绘旋转向量场

Sketch the vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {y,\text{−}x} \right\rangle.$

Solution 解

Create a table (see the one that follows) using a representative sample of points in a plane and their corresponding vectors. Figure 6.6 shows the resulting vector field.

| | | | | | |
|-------------------------|-----------------------------------------|-------------------------|-----------------------------------------|--------------------------|-----------------------------------------|
| $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ | $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ | $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ |
| $\left( {1,0} \right)$ | $\left\langle {0,-1} \right\rangle$ | $\left( {2,0} \right)$ | $\left\langle {0,-2} \right\rangle$ | $\left( {1,1} \right)$ | $\left\langle {1,-1} \right\rangle$ |
| $\left( {0,1} \right)$ | $\left\langle {1,0} \right\rangle$ | $\left( {0,2} \right)$ | $\left\langle {2,0} \right\rangle$ | $\left( {-1,1} \right)$ | $\left\langle {1,1} \right\rangle$ |
| $\left( {-1,0} \right)$ | $\left\langle {0,1} \right\rangle$ | $\left( {-2,0} \right)$ | $\left\langle {0,2} \right\rangle$ | $\left( {-1,-1} \right)$ | $\left\langle {-1,1} \right\rangle$ |
| $\left( {0,-1} \right)$ | $\left\langle {-1,0} \right\rangle$ | $\left( {0,-2} \right)$ | $\left\langle {-2,0} \right\rangle$ | $\left( {1,-1} \right)$ | $\left\langle {-1,-1} \right\rangle$ |

Analysis 分析

Note that vector $\mathbf{\text{F}}\left( {a,b} \right) = \left\langle {b,\text{−}a} \right\rangle$ points clockwise and is perpendicular to radial vector $\left\langle {a,b} \right\rangle.$ (We can verify this assertion by computing the dot product of the two vectors: $\left\langle {a,b} \right\rangle \cdot \left\langle {\text{−}b,a} \right\rangle = \text{−}ab + ab = 0.)$ Furthermore, vector $\left\langle {b,\text{−}a} \right\rangle$ has length $r = \sqrt{a^{2} + b^{2}}.$ Thus, we have a complete description of this rotational vector field: the vector associated with point $\left( {a,b} \right)$ is the vector with length *r* tangent to the circle with radius *r*, and it points in the clockwise direction.

Sketches such as that in Figure 6.6 are often used to analyze major storm systems, including hurricanes and cyclones. In the northern hemisphere, storms rotate counterclockwise; in the southern hemisphere, storms rotate clockwise. (This is an effect caused by Earth’s rotation about its axis and is called the Coriolis Effect.)

Sketching a Vector Field 描绘向量场

Sketch vector field $\mathbf{\text{F}}(x,y) = \frac{y}{x^{2} + y^{2}}\mspace{2mu}\mathbf{\text{i}} - \frac{x}{x^{2} + y^{2}}\mspace{2mu}\mathbf{\text{j}}.$

Solution 解

To visualize this vector field, first note that the dot product $\mathbf{\text{F}}(a,b) \cdot (a\mathbf{\text{i}} + b\mathbf{\text{j}})$ is zero for any point $\left( {a,b} \right).$ Therefore, each vector is tangent to the circle on which it is located. Also, as $\left( {a,b} \right)\rightarrow\left( {0,0} \right),$ the magnitude of $\mathbf{\text{F}}\left( {a,b} \right)$ goes to infinity. To see this, note that

$${\left| \left| {\mathbf{\text{F}}\left( {a,b} \right)} \right| \right| = \sqrt{\frac{a^{2} + b^{2}}{\left( {a^{2} + b^{2}} \right)^{2}}} = \sqrt{\frac{1}{a^{2} + b^{2}}}}.$$

Since $\frac{1}{a^{2} + b^{2}}\rightarrow\infty$ as $\left( {a,b} \right)\rightarrow\left( {0,0} \right),$ then $\left| \left| {\mathbf{\text{F}}\left( {a,b} \right)} \right| \right|\rightarrow\infty$ as $\left( {a,b} \right)\rightarrow\left( {0,0} \right).$ This vector field looks similar to the vector field in Example 6.3, but in this case the magnitudes of the vectors close to the origin are large. The table below shows a sample of points and the corresponding vectors, and Figure 6.6 shows the vector field. Note that this vector field models the whirlpool motion of the river in Figure 6.2(b). The domain of this vector field is all of $\mathbb{R}^{2}$ except for point $\left( {0,0} \right).$

| | | | | | |
|-------------------------|-----------------------------------------|-------------------------|-------------------------------------------------|--------------------------|-------------------------------------------------------------|
| $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ | $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ | $\left( {x,y} \right)$ | $\mathbf{\text{F}}\left( {x,y} \right)$ |
| $\left( {1,0} \right)$ | $\left\langle {0,-1} \right\rangle$ | $\left( {2,0} \right)$ | $\left\langle {0, - \frac{1}{2}} \right\rangle$ | $\left( {1,1} \right)$ | $\left\langle {\frac{1}{2}, - \frac{1}{2}} \right\rangle$ |
| $\left( {0,1} \right)$ | $\left\langle {1,0} \right\rangle$ | $\left( {0,2} \right)$ | $\left\langle {\frac{1}{2},0} \right\rangle$ | $\left( {-1,1} \right)$ | $\left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle$ |
| $\left( {-1,0} \right)$ | $\left\langle {0,1} \right\rangle$ | $\left( {-2,0} \right)$ | $\left\langle {0,\frac{1}{2}} \right\rangle$ | $\left( {-1,-1} \right)$ | $\left\langle {- \frac{1}{2},\frac{1}{2}} \right\rangle$ |
| $\left( {0,-1} \right)$ | $\left\langle {-1,0} \right\rangle$ | $\left( {0,-2} \right)$ | $\left\langle {- \frac{1}{2},0} \right\rangle$ | $\left( {1,-1} \right)$ | $\left\langle {- \frac{1}{2}, - \frac{1}{2}} \right\rangle$ |

Sketch vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {-2y,2x} \right\rangle.$ Is the vector field radial, rotational, or neither?

Velocity Field of a Fluid 流体的速度场

Suppose that $\mathbf{\text{v}}\left( {x,y} \right) = - \frac{2y}{x^{2} + y^{2}}\mspace{2mu}\mathbf{\text{i}} + \frac{2x}{x^{2} + y^{2}}\mspace{2mu}\mathbf{\text{j}}$ is the velocity field of a fluid. How fast is the fluid moving at point $\left( {1,-1} \right)?$ (Assume the units of speed are meters per second.)

Solution 解

To find the velocity of the fluid at point $\left( {1,-1} \right),$ substitute the point into v:

$$\mathbf{\text{v}}\left( {1,-1} \right) = - \frac{2(-1)}{1 + 1}\mspace{2mu}\mathbf{\text{i}} + \frac{2(1)}{1 + 1}\mspace{2mu}\mathbf{\text{j}} = \mathbf{\text{i}} + \mathbf{\text{j}}.$$

The speed of the fluid at $\left( {1,-1} \right)$ is the magnitude of this vector. Therefore, the speed is $\left| \middle| \left. \mathbf{\text{i}} + \mathbf{\text{j}} \right| \right| = \sqrt{2}$ m/sec.

Vector field $v(x,y) = \left\langle {4|x|,1} \right\rangle$ models the velocity of water on the surface of a river. What is the speed of the water at point $\left( {2,3} \right)?$ Use meters per second as the units.

We have examined vector fields that contain vectors of various magnitudes, but just as we have unit vectors, we can also have a unit vector field. A vector field F is a unit vector field if the magnitude of each vector in the field is 1. In a unit vector field, the only relevant information is the direction of each vector.

A Unit Vector Field 一个单位向量场

Show that the vector field $\mathbf{\text{G}}\left( {x,y} \right) = \left\langle {\frac{y}{\sqrt{x^{2} + y^{2}}}, - \frac{x}{\sqrt{x^{2} + y^{2}}}} \right\rangle$ is a unit vector field.

Solution 解

To show that G is a unit field, we must show that the magnitude of each vector is 1. Note that

$$\begin{array}{cl} \sqrt{\left( \frac{y}{\sqrt{x^{2} + y^{2}}} \right)^{2} + \left( {- \frac{x}{\sqrt{x^{2} + y^{2}}}} \right)^{2}} & {= \sqrt{\frac{y^{2}}{x^{2} + y^{2}} + \frac{x^{2}}{x^{2} + y^{2}}}} \\ & {= \sqrt{\frac{x^{2} + y^{2}}{x^{2} + y^{2}}}} \\ & {= 1.} \end{array}$$

Therefore, G is a unit vector field.

Is vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {\text{−}y,x} \right\rangle$ a unit vector field?

Why are unit vector fields important? Suppose we are studying the flow of a fluid, and we care only about the direction in which the fluid is flowing at a given point. In this case, the speed of the fluid (which is the magnitude of the corresponding velocity vector) is irrelevant, because all we care about is the direction of each vector. Therefore, the unit vector field associated with velocity is the field we would study.

If $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ is a vector field, then the corresponding unit vector field is $\left\langle {\frac{P}{\left| \left| \mathbf{\text{F}} \right| \right|},\frac{Q}{\left| \left| \mathbf{\text{F}} \right| \right|},\frac{R}{\left| \left| \mathbf{\text{F}} \right| \right|}} \right\rangle.$ Notice that if $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {y,\text{−}x} \right\rangle$ is the vector field from Example 6.3, then the magnitude of F is $\sqrt{x^{2} + y^{2}},$ and therefore the corresponding unit vector field is the field G from the previous example.

If F is a vector field, then the process of dividing F by its magnitude to form unit vector field $\mathbf{\text{F}}\text{/}\left| \left| \mathbf{\text{F}} \right| \right|$ is called *normalizing* the field F.

Vector Fields in $\mathbb{R}^{3}$ $\mathbb{R}^{3}$ 中的向量场

We have seen several examples of vector fields in $\mathbb{R}^{2};$ let’s now turn our attention to vector fields in $\mathbb{R}^{3}.$ These vector fields can be used to model gravitational or electromagnetic fields, and they can also be used to model fluid flow or heat flow in three dimensions. A two-dimensional vector field can really only model the movement of water on a two-dimensional slice of a river (such as the river’s surface). Since a river flows through three spatial dimensions, to model the flow of the entire depth of the river, we need a vector field in three dimensions.

The extra dimension of a three-dimensional field can make vector fields in $\mathbb{R}^{3}$ more difficult to visualize, but the idea is the same. To visualize a vector field in $\mathbb{R}^{3},$ plot enough vectors to show the overall shape. We can use a similar method to visualizing a vector field in $\mathbb{R}^{2}$ by choosing points in each octant.

Just as with vector fields in $\mathbb{R}^{2},$ we can represent vector fields in $\mathbb{R}^{3}$ with component functions. We simply need an extra component function for the extra dimension. We write either

$$\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {P\left( {x,y,z} \right),Q\left( {x,y,z} \right),R\left( {x,y,z} \right)} \right\rangle$$ (6.3)

or

$$\mathbf{\text{F}}\left( {x,y,z} \right) = P\left( {x,y,z} \right)\mathbf{\text{i}} + Q\left( {x,y,z} \right)\mathbf{\text{j}} + R\left( {x,y,z} \right)\mathbf{\text{k}}.$$ (6.4)

Sketching a Vector Field in Three Dimensions 绘制三维中的向量场

Describe vector field $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {1,1,z} \right\rangle.$

Solution 解

For this vector field, the *x* and *y* components are constant, so every point in $\mathbb{R}^{3}$ has an associated vector with *x* and *y* components equal to one. To visualize F, we first consider what the field looks like in the *xy*-plane. In the *xy*-plane, $z = 0.$ Hence, each point of the form $\left( {a,b,0} \right)$ has vector $\left\langle {1,1,0} \right\rangle$ associated with it. For points not in the *xy*-plane but slightly above it, the associated vector has a small but positive *z* component, and therefore the associated vector points slightly upward. For points that are far above the *xy*-plane, the *z* component is large, so the vector is almost vertical. Figure 6.7 shows this vector field.

Sketch vector field $\mathbf{\text{G}}\left( {x,y,z} \right) = \left\langle {2,\frac{z}{2},1} \right\rangle.$

In the next example, we explore one of the classic cases of a three-dimensional vector field: a gravitational field.

Describing a Gravitational Vector Field 描述引力向量场

Newton’s law of gravitation states that $\mathbf{\text{F}} = \text{−}G\mspace{2mu}\frac{m_{1}m_{2}}{r^{2}}\mathbf{\hat{r}},$ where *G* is the universal gravitational constant. It describes the gravitational field exerted by an object (object 1) of mass $m_{1}$ located at the origin on another object (object 2) of mass $m_{2}$ located at point $\left( {x,y,z} \right).$ Field F denotes the gravitational force that object 1 exerts on object 2, *r* is the distance between the two objects, and $\mathbf{\hat{r}}$ indicates the unit vector from the first object to the second. The minus sign shows that the gravitational force attracts toward the origin; that is, the force of object 1 is attractive. Sketch the vector field associated with this equation.

Solution 解

Since object 1 is located at the origin, the distance between the objects is given by $r = \sqrt{x^{2} + y^{2} + z^{2}}.$ The unit vector from object 1 to object 2 is $\mathbf{\hat{r}} = \frac{\left\langle {x,y,z} \right\rangle}{\left| \left| \left\langle {x,y,z} \right\rangle \right| \right|},$ and hence $\mathbf{\hat{r}} = \left\langle {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right\rangle.$ Therefore, gravitational vector field F exerted by object 1 on object 2 is

$$\mathbf{\text{F}} = \text{−}Gm_{1}m_{2}\left\langle {\frac{x}{r^{3}},\frac{y}{r^{3}},\frac{z}{r^{3}}} \right\rangle.$$

This is an example of a radial vector field in $\mathbb{R}^{3}.$

Figure 6.8 shows what this gravitational field looks like for a large mass at the origin. Note that the magnitudes of the vectors increase as the vectors get closer to the origin.

The mass of asteroid 1 is 750,000 kg and the mass of asteroid 2 is 130,000 kg. Assume asteroid 1 is located at the origin, and asteroid 2 is located at $\left( {15,-5,10} \right),$ measured in units of 10 to the eighth power kilometers. Given that the universal gravitational constant is $G = 6.67384\ \times \ 10^{-11}{\ \text{m}}^{3}\text{kg}^{-1}\text{s}^{-2},$ find the gravitational force vector that asteroid 1 exerts on asteroid 2.

Gradient Fields 梯度场

In this section, we study a special kind of vector field called a gradient field or a conservative field. These vector fields are extremely important in physics because they can be used to model physical systems in which energy is conserved. Gravitational fields and electric fields associated with a static charge are examples of gradient fields.

Recall that if $f$ is a (scalar) function of *x* and *y*, then the gradient of $f$ is

$$\text{grad}\ f = \text{∇}f = f_{x}(x,y)\mathbf{\text{i}} + f_{y}(x,y)\mathbf{\text{j}}.$$

We can see from the form in which the gradient is written that $\text{∇}f$ is a vector field in $\mathbb{R}^{2}.$ Similarly, if $f$ is a function of *x*, *y*, and *z*, then the gradient of $f$ is

$$\text{grad}\ f = \text{∇}f = f_{x}(x,y,z)\mathbf{\text{i}} + f_{y}(x,y,z)\mathbf{\text{j}} + f_{z}(x,y,z)\mathbf{\text{k}}.$$

The gradient of a three-variable function is a vector field in $\mathbb{R}^{3}.$

A gradient field is a vector field that can be written as the gradient of a function, and we have the following definition.

A vector field $\mathbf{\text{F}}$ in $\mathbb{R}^{2}$ or in $\mathbb{R}^{3}$ is a gradient field if there exists a scalar function $f$ such that $\text{∇}f = \mathbf{\text{F}}.$

Sketching a Gradient Vector Field 绘制梯度向量场

Use technology to plot the gradient vector field of $f\left( {x,y} \right) = x^{2}y^{2}.$

Solution 解

The gradient of $f$ is $\text{∇}f = \left\langle {2xy^{2},2x^{2}y} \right\rangle.$ To sketch the vector field, use a computer algebra system such as Mathematica. Figure 6.9 shows $\text{∇}f.$

Use technology to plot the gradient vector field of $f\left( {x,y} \right) = \text{sin}\ x\ \text{cos}\ y.$

Consider the function $f\left( {x,y} \right) = x^{2}y^{2}$ from Example 6.9. Figure 6.11 shows the level curves of this function overlaid on the function’s gradient vector field. The gradient vectors are perpendicular to the level curves, and the magnitudes of the vectors get larger as the level curves get closer together, because closely grouped level curves indicate the graph is steep, and the magnitude of the gradient vector is the largest value of the directional derivative. Therefore, you can see the local steepness of a graph by investigating the corresponding function’s gradient field.

As we learned earlier, a vector field $\mathbf{\text{F}}$ is a conservative vector field, or a gradient field if there exists a scalar function $f$ such that $\text{∇}f = \mathbf{\text{F}}.$ In this situation, $f$ is called a potential function for $\mathbf{\text{F}}.$ Conservative vector fields arise in many applications, particularly in physics. The reason such fields are called *conservative* is that they model forces of physical systems in which energy is conserved. We study conservative vector fields in more detail later in this chapter.

You might notice that, in some applications, a potential function $f$ for F is defined instead as a function such that $\text{−}\text{∇}f = \mathbf{\text{F}}.$ This is the case for certain contexts in physics, for example.

Verifying a Potential Function 验证势函数

Is $f\left( {x,y,z} \right) = x^{2}yz - \text{sin}\left( {xy} \right)$ a potential function for vector field

$$\mathbf{\text{F}}(x,y,z) = \left\langle {2xyz - y\ \text{cos}(xy),x^{2}z - x\ \text{cos}(xy),x^{2}y} \right\rangle?$$

Solution 解

We need to confirm whether $\text{∇}f = \mathbf{\text{F}}.$ We have

$$f_{x} = 2xyz - y\ \text{cos}\left( {xy} \right),f_{y} = x^{2}z - x\ \text{cos}\left( {xy} \right),\ \text{and}\ f_{z} = x^{2}y.$$

Therefore, $\text{∇}f = \mathbf{\text{F}}$ and $f$ is a potential function for $\mathbf{\text{F}}.$

Is $f(x,y,z) = x^{2}\text{cos}(yz) + y^{2}z^{2}$ a potential function for $\mathbf{\text{F}}(x,y,z) = \left\langle {2x\ \text{cos}(yz),\text{−}x^{2}z\ \text{sin}(yz) + 2yz^{2},y^{2}} \right\rangle?$

Verifying a Potential Function 验证势函数

The velocity of a fluid is modeled by field $\mathbf{\text{v}}\left( {x,y} \right) = \left\langle {xy,\frac{x^{2}}{2} - y} \right\rangle.$ Verify that $f\left( {x,y} \right) = \frac{x^{2}y}{2} - \frac{y^{2}}{2}$ is a potential function for v.

Solution 解

To show that $f$ is a potential function, we must show that $\text{∇}f = \mathbf{\text{v}}.$ Note that $f_{x} = xy$ and $f_{y} = \frac{x^{2}}{2} - y.$ Therefore, $\text{∇}f = \left\langle {xy,\frac{x^{2}}{2} - y} \right\rangle$ and $f$ is a potential function for v (Figure 6.11).

Verify that $f\left( {x,y} \right) = x^{3}y^{2} + 1$ is a potential function for velocity field $\textbf{v}(x,y) = \left\langle 3x^{2}y^{2},2x^{3}y \right\rangle.$

If F is a conservative vector field, then there is at least one potential function $f$ such that $\text{∇}f = \mathbf{\text{F}}.$ But, could there be more than one potential function? If so, is there any relationship between two potential functions for the same vector field? Before answering these questions, let’s recall some facts from single-variable calculus to guide our intuition. Recall that if $k(x)$ is an integrable function, then *k* has infinitely many antiderivatives. Furthermore, if *F* and *G* are both antiderivatives of *k*, then *F* and *G* differ only by a constant. That is, there is some number *C* such that $F(x) = G(x) + C.$

Now let $\mathbf{F}$ be a conservative vector field and let $f$ and *g* be potential functions for $\mathbf{F}$. Since the gradient is like a derivative, $\mathbf{F}$ being conservative means that $\mathbf{F}$ is “integrable” with “antiderivatives” $f$ and *g*. Therefore, if the analogy with single-variable calculus is valid, we expect there is some constant *C* such that $f(x) = g(x) + C.$ The next theorem says that this is indeed the case.

To state the next theorem with precision, we need to assume the domain of the vector field is connected and open. To be connected means if $P_{1}$ and $P_{2}$ are any two points in the domain, then you can walk from $P_{1}$ to $P_{2}$ along a path that stays entirely inside the domain.

Uniqueness of Potential Functions 势函数的唯一性

Let F be a conservative vector field on an open and connected domain and let $f$ and *g* be functions such that $\text{∇}f = \mathbf{\text{F}}$ and $\text{∇}g = \mathbf{\text{F}}.$ Then, there is a constant *C* such that $f = g + C.$

Proof 证明

Since $f$ and *g* are both potential functions for F, then $\text{∇}\left( {f - g} \right) = \text{∇}f - \text{∇}g = \mathbf{\text{F}} - \mathbf{\text{F}} = 0.$ Let $h = f - g,$ then we have $\text{∇}h = 0.$ We would like to show that *h* is a constant function.

Assume *h* is a function of *x* and *y* (the logic of this proof extends to any number of independent variables). Since $\text{∇}h = 0,$ we have $h_{x} = 0$ and $h_{y} = 0.$ The expression $h_{x} = 0$ implies that *h* is a constant function with respect to *x—*that is, $h\left( {x,y} \right) = k_{1}(y)$ for some function *k₁*. Similarly, $h_{y} = 0$ implies $h\left( {x,y} \right) = k_{2}(x)$ for some function *k₂*. Therefore, function *h* depends only on *y* and also depends only on *x*. Thus, $h\left( {x,y} \right) = C$ for some constant *C* on the connected domain of F. Note that we really do need connectedness at this point; if the domain of F came in two separate pieces, then *k* could be a constant *C₁* on one piece but could be a different constant *C₂* on the other piece. Since $f - g = h = C,$ we have that $f = g + C,$ as desired.

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Conservative vector fields also have a special property called the *cross-partial property*. This property helps test whether a given vector field is conservative.

The Cross-Partial Property of Conservative Vector Fields 保守向量场的交叉偏导性质

Let F be a vector field in two or three dimensions such that the component functions of F have continuous first-order partial derivatives on the domain of F.

If $\mathbf{\text{F}}(x,y) = \left\langle {P(x,y),Q(x,y)} \right\rangle$ is a conservative vector field in $\mathbb{R}^{2},$ then $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}.$ If $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {P\left( {x,y,z} \right),Q\left( {x,y,z} \right),R\left( {x,y,z} \right)} \right\rangle$ is a conservative vector field in $\mathbb{R}^{3},$ then

$$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x},\frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y},\ \text{and}\ \frac{\partial R}{\partial x} = \frac{\partial P}{\partial z}.$$

Proof 证明

Since F is conservative, there is a function $f(x,y)$ such that $\text{∇}f = \mathbf{\text{F}}.$ Therefore, by the definition of the gradient, $f_{x} = P$ and $f_{y} = Q.$ By Clairaut’s theorem, $f_{xy} = f_{yx},$ But, $f_{xy} = P_{y}$ and $f_{yx} = Q_{x},$ and thus $P_{y} = Q_{x}.$

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Clairaut’s theorem gives a fast proof of the cross-partial property of conservative vector fields in $\mathbb{R}^{3},$ just as it did for vector fields in $\mathbb{R}^{2}.$

The Cross-Partial Property of Conservative Vector Fields shows that most vector fields are not conservative. The cross-partial property is difficult to satisfy in general, so most vector fields won’t have equal cross-partials.

Showing a Vector Field Is Not Conservative 证明向量场不是保守场

Show that rotational vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {y,\text{−}x} \right\rangle$ is not conservative.

Solution 解

Let $P\left( {x,y} \right) = y\ \text{and}\ Q\left( {x,y} \right) = \text{−}x.$ If F is conservative, then the cross-partials would be equal—that is, $P_{y}$ would equal $Q_{x.}$ Therefore, to show that F is not conservative, check that $P_{y} \neq Q_{x}.$ Since $P_{y} = 1$ and $Q_{x} = -1,$ the vector field is not conservative.

Show that the vector field $\mathbf{\text{F}}\left( {x,y} \right) = y\mathbf{\text{i}} - x^{2}xy\mathbf{\text{j}}$ is not conservative.

Showing a Vector Field Is Not Conservative 证明向量场不是保守场

Is vector field $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {7,-2,x^{3}} \right\rangle$ conservative?

Solution 解

The integrand is $f\left( {x,y} \right) = 2.$ Figure 6.14 shows the graph of $f\left( {x,y} \right) = 2,$ curve *C*, and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width $\pi$ and length 2. Therefore, $\int_{C}{2ds = 2\pi.}$

To see that $\int_{C}{2ds = 2\pi}$ using the definition of line integral, we let $\mathbf{\text{r}}(t)$ be a parameterization of *C*. Then, $f\left( {\mathbf{\text{r}}\left( t_{i} \right)} \right) = 2$ for any number $t_{i}$ in the domain of r. Therefore,

$$\begin{matrix}{\int_{C}fds} & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}f\left( \textbf{r}\left( t_{i}^{*} \right) \right)\text{Δ}s_{i}} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}2\text{Δ}s_{i}} \\ & {= 2\underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\text{Δ}s_{i}} \\ & {= 2\left( \text{length of C} \right)} \\ & {= 2\pi.} \end{matrix}$$

Find the value of $\int_{C}{\left( {x + y} \right)ds,}$ where $C$ is the curve parameterized by $x = t,$ $y = t,$ $0 \leq t \leq 1.$

Note that in a scalar line integral, the integration is done with respect to arc length *s*, which can make a scalar line integral difficult to calculate. To make the calculations easier, we can translate $\int_{C}{fds}$ to an integral with a variable of integration that is *t*.

Let $\mathbf{\text{r}}(t) = \left\langle {x(t),y(t),z(t)} \right\rangle$ for $a \leq t \leq b$ be a parameterization of $C.$ Since we are assuming that $C$ is smooth, $\mathbf{r^{\prime}}(t) = \left\langle {x^{\prime}(t),y^{\prime}(t),z^{\prime}(t)} \right\rangle$ is continuous for all $t$ in $\left\lbrack {a,b} \right\rbrack.$ In particular, $x\text{'}(t),y\text{'}(t),$ and $z\text{'}(t)$ exist for all $t$ in $\left\lbrack {a,b} \right\rbrack.$ According to the arc length formula, we have

$$\text{length}\left( C_{i} \right) = \text{Δ}s_{i} = {\int_{t_{i - 1}}^{t_{i}}\left\| {\mathbf{r^{\prime}}\left. (t) \right\|} \right.}dt.$$

If width $\text{Δ}t_{i} = t_{i} - t_{i - 1}$ is small, then function $\int_{t_{i - 1}}^{t_{i}}{\left\| {\mathbf{r^{\prime}}(t)} \right\|\left. {dt \approx} \right\|\left. {r^{\prime}\left( t_{i}^{*} \right)} \right\|\text{Δ}t_{i},}$ $\left\| {\mathbf{r^{\prime}}(t)} \right\|$ is almost constant over the interval $\left\lbrack {t_{i - 1},t_{i}} \right\rbrack.$ Therefore,

$${\int_{t_{i - 1}}^{t_{i}}{\left\| {\mathbf{r^{\prime}}(t)} \right\| dt}} \approx \left\| {\mathbf{r^{\prime}}(t_{i}^{*})} \right\|\text{Δ}t_{i},$$

and we have

$${\sum\limits_{i = 1}^{n}{f(\mathbf{\text{r}}(t_{i}^{*}))\text{Δ}s_{i}}} = {\sum\limits_{i = 1}^{n}{f(\mathbf{\text{r}}(t_{i}^{*}))\left\| {\mathbf{r^{\prime}}(t_{i}^{*})} \right\|}}\text{Δ}t_{i}.$$ (6.6)

See Figure 6.15.

Note that

$$\underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}{f(\mathbf{\text{r}}(t_{i}^{*}))}}\left\| {\mathbf{r^{\prime}}(t_{i}^{*})} \right\|\text{Δ}t_{i} = {\int_{a}^{b}{f(\mathbf{\text{r}}(t))}}\left\| {\mathbf{r^{\prime}}(t)} \right\| dt.$$

In other words, as the widths of intervals $\lbrack t_{i - 1},t_{i}\rbrack$ shrink to zero, the sum ${\sum\limits_{i = 1}^{n}{f(\mathbf{\text{r}}(t_{i}^{*}))}}\left\| {\mathbf{r^{\prime}}(t_{i}^{*})} \right\|\text{Δ}t_{i}$ converges to the integral ${\int_{a}^{b}{f(\mathbf{\text{r}}(t))}}\left\| {\mathbf{r^{\prime}}(t)} \right\| dt.$ Therefore, we have the following theorem.

Evaluating a Scalar Line Integral 计算标量线积分

Let $f$ be a continuous function with a domain that includes the smooth curve $C$ with parameterization $\mathbf{\text{r}}(t),a \leq t \leq b.$ Then

$${\int_{C}{fds = {\int_{a}^{b}{f(\mathbf{\text{r}}(t))}}\left\| {\mathbf{r^{\prime}}(t)} \right\| dt}}.$$ (6.7)

Although we have labeled Equation 6.6 as an equation, it is more accurately considered an approximation because we can show that the left-hand side of Equation 6.6 approaches the right-hand side as $n\rightarrow\infty.$ In other words, letting the widths of the pieces shrink to zero makes the right-hand sum arbitrarily close to the left-hand sum. Since

$$\left\| {\mathbf{r^{\prime}}(t)} \right\| = \sqrt{\left( {x\prime(t)} \right)^{2} + \left( {y\prime(t)} \right)^{2} + \left( {z\prime(t)} \right)^{2},}$$

we obtain the following theorem, which we use to compute scalar line integrals.

Scalar Line Integral Calculation 标量线积分的计算

Let $f$ be a continuous function with a domain that includes the smooth curve *C* with parameterization $\mathbf{\text{r}}(t) = \left\langle {x(t),y(t),z(t)} \right\rangle,a \leq t \leq b.$ Then

$$\int_{C}{f\left( {x,y,z} \right)ds = {\int_{a}^{b}{f\left( {\mathbf{\text{r}}(t)} \right)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2} + \left( {z^{\prime}(t)} \right)^{2}}}}dt.}$$ (6.8)

Similarly,

$$\int_{C}{f\left( {x,y} \right)ds = {\int_{a}^{b}{f\left( {\mathbf{\text{r}}(t)} \right)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2}}}}dt}$$

if *C* is a planar curve and $f$ is a function of two variables.

Note that a consequence of this theorem is the equation $ds = \left\| {\mathbf{r^{\prime}}(t)} \right\| dt.$ In other words, the change in arc length can be viewed as a change in the *t* domain, scaled by the magnitude of vector $\mathbf{r^{\prime}}(t).$

Evaluating a Line Integral 计算一个线积分

Find the value of integral $\int_{C}{\left( {x^{2} + y^{2} + z} \right)ds,}$ where $C$ is part of the helix parameterized by $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t,t} \right\rangle,$ $0 \leq t \leq 2\pi.$

Solution 解

To compute a scalar line integral, we start by converting the variable of integration from arc length *s* to *t*. Then, we can use Equation 6.8 to compute the integral with respect to *t*. Note that $f\left( {\mathbf{\text{r}}(t)} \right) = \text{cos}^{2}t + \text{sin}^{2}t + t = 1 + t$ and

$$\begin{array}{cl}\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2} + \left( {z^{\prime}(t)} \right)^{2}} & {= \sqrt{\left( {\text{−}\text{sin}(t)} \right)^{2} + \text{cos}^{2}(t) + 1}} \\ & {= \sqrt{2}.} \end{array}$$

Therefore,

$$\int_{C}{\left( {x^{2} + y^{2} + z} \right)ds = {\int_{0}^{2\pi}{\left( {1 + t} \right)\sqrt{2}dt.}}}$$

Notice that Equation 6.8 translated the original difficult line integral into a manageable single-variable integral. Since

$$\begin{array}{cl}{{\int_{0}^{2\pi}\left( {1 + t} \right)}\sqrt{2}dt} & {= \left\lbrack {\sqrt{2}t + \frac{\sqrt{2}t^{2}}{2}} \right\rbrack_{0}^{2\pi}} \\ & {= 2\sqrt{2}\pi + 2\sqrt{2}\pi^{2},} \end{array}$$

we have

$${\int_{C}{\left( {x^{2} + y^{2} + z} \right)ds = 2\sqrt{2}\pi + 2\sqrt{2}\pi^{2}}}.$$

Evaluate $\int_{C}{\left( {x^{2} + y^{2} + z} \right)ds,}$ where *C* is the curve with parameterization $\textbf{r}(t) = \left\langle \text{sin}(3t),\text{cos}(3t)\text{,t} \right\rangle,0 \leq t \leq 2\pi.$

Independence of Parameterization 与参数化无关

Find the value of integral $\int_{C}{\left( {x^{2} + y^{2} + z} \right)ds,}$ where $C$ is part of the helix parameterized by $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\left( {2t} \right),\text{sin}\left( {2t} \right),2t} \right\rangle,0 \leq t \leq \pi.$ Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast.

Solution 解

As with the previous example, we use Equation 6.8 to compute the integral with respect to *t*. Note that $f\left( {\mathbf{\text{r}}(t)} \right) = \text{cos}^{2}\left( {2t} \right) + \text{sin}^{2}\left( {2t} \right) + 2t = 2t + 1$ and

$$\begin{matrix}\sqrt{\left( x^{'}(t) \right)^{2} + \left( y^{'}(t) \right)^{2} + \left( z'(t) \right)^{2}} & {= \sqrt{\left( {\text{−2sin}\left( {\text{2}t} \right)} \right)^{2} + \left( {2\text{cos}\left( {\text{2}t} \right)} \right)^{2} + 2^{2}}} \\ & \sqrt{4\text{sin}^{2}\left( {2t} \right) + 4\text{cos}^{2}\left( {2t} \right) + 4} \\ & {= 2\sqrt{2}} \end{matrix}$$

so we have

$$\begin{array}{cl}{\int_{C}{\left( {x^{2} + y^{2} + z} \right)ds}} & {= 2\sqrt{2}{\int_{0}^{\pi}{\left( {1 + 2t} \right)dt}}} \\ & {= 2\sqrt{2}\left\lbrack {t + t^{2}} \right\rbrack_{0}^{\pi}} \\ & {= 2\sqrt{2}\left( {\pi + \pi^{2}} \right).} \end{array}$$

Notice that this agrees with the answer in the previous example. Changing the parameterization did not change the value of the line integral. Scalar line integrals are independent of parameterization, as long as the curve is traversed exactly once by the parameterization.

Evaluate line integral ${\int_{C}{\left( {x^{2} + yz} \right)ds}},$ where $C$ is the line with parameterization $\mathbf{\text{r}}(t) = \left\langle {2t,5t,\text{−}t} \right\rangle,0 \leq t \leq 10.$ Reparameterize *C* with parameterization $\mathbf{\text{s}}(t) = \left\langle {4t,10t,-2t} \right\rangle,0 \leq t \leq 5,$ recalculate line integral ${\int_{C}{\left( {x^{2} + yz} \right)ds}},$ and notice that the change of parameterization had no effect on the value of the integral.

Now that we can evaluate line integrals, we can use them to calculate arc length. If $f\left( {x,y,z} \right) = 1,$ then

$$\begin{matrix}{\int_{C}f(x,y,z)ds} & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}f\left( P_{i}^{*} \right)\text{Δ}s_{i}} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\text{Δ}s_{i}} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\text{length}(C)} \\ & {= \text{length}(C).} \end{matrix}$$

Therefore, $\int_{C}{2ds = 2\pi}$ is the arc length of $C.$

Calculating Arc Length 计算弧长

A wire has a shape that can be modeled with the parameterization $\textbf{r}(t) = \left\langle \text{cos}\ t,\text{sin}\ t,\frac{2}{3}t^{3/2} \right\rangle,0 \leq t \leq 4\pi.$ Find the length of the wire.

Solution 解

The length of the wire is given by ${\int_{C}{1ds}},$ where *C* is the curve with parameterization r. Therefore,

$$\begin{array}{cl}\text{The length of the wire} & {= {\int_{C}{1ds}}} \\ & {= {\int_{0}^{4\pi}{\left\| {\mathbf{r^{\prime}}(t)} \right\| dt}}} \\ & {= {\int_{0}^{4\pi}{\sqrt{\left( {\text{−}\text{sin}\ t} \right)^{2} + \text{cos}^{2}t + t}dt}}} \\ & {= {\int_{0}^{4\pi}{\sqrt{1 + t}dt}}} \\ & {= \left\lbrack \frac{2\left( {1 + t} \right)^{3\text{/}2}}{3} \right\rbrack_{0}^{4\pi}} \\ & {= \frac{2}{3}\left( {\left( {1 + 4\pi} \right)^{3\text{/}2} - 1} \right).} \end{array}$$

Find the length of a wire with parameterization $\mathbf{\text{r}}(t) = \left\langle {3t + 1,4 - 2t,5 + 2t} \right\rangle,0 \leq t \leq 4.$

Vector Line Integrals 向量线积分

The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let

$$\mathbf{\text{F}}\left( {x,y,z} \right) = P\left( {x,y,z} \right)\mathbf{\text{i}} + Q\left( {x,y,z} \right)\mathbf{\text{j}} + R\left( {x,y,z} \right)\mathbf{\text{k}}$$

be a continuous vector field in $\mathbb{R}^{3}$ that represents a force on a particle, and let *C* be a smooth curve in $\mathbb{R}^{3}$ contained in the domain of $\mathbf{\text{F}}.$ How would we compute the work done by $\mathbf{\text{F}}$ in moving a particle along *C*?

To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve *C*; such a specified direction is called an orientation of a curve. The specified direction is the *positive* direction along *C*; the opposite direction is the *negative* direction along *C*. When *C* has been given an orientation, *C* is called an *oriented curve* (Figure 6.16). The work done on the particle depends on the direction along the curve in which the particle is moving.

A closed curve is one for which there exists a parameterization $\mathbf{\text{r}}(t),$ $a \leq t \leq b,$ such that $\mathbf{\text{r}}(a) = \mathbf{\text{r}}(b),$ and the curve is traversed exactly once. In other words, the parameterization is one-to-one on the domain $\left( {a,b} \right).$

Let $\mathbf{\text{r}}(t)$ be a parameterization of *C* for $a \leq t \leq b$ such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along *C*. Divide the parameter interval $\left\lbrack {a,b} \right\rbrack$ into *n* subintervals $\left\lbrack {t_{i - 1},t_{i}} \right\rbrack,0 \leq i \leq n,$ of equal width. Denote the endpoints of $\mathbf{\text{r}}\left( t_{0} \right),\mathbf{\text{r}}\left( t_{1} \right)\text{,…},\mathbf{\text{r}}\left( t_{n} \right)$ by $P_{0}\text{,…},P_{n}.$ Points *P_i* divide *C* into *n* pieces. Denote the length of the piece from *P_i−1* to *P_i* by $\text{Δ}s_{i}.$ For each *i*, choose a value $t_{i}^{*}$ in the subinterval $\left\lbrack {t_{i - 1},t_{i}} \right\rbrack.$ Then, the endpoint of $\mathbf{\text{r}}(t_{i}^{*})$ is a point in the piece of *C* between $P_{i - 1}$ and *P_i* (Figure 6.17). If $\text{Δ}s_{i}$ is small, then as the particle moves from $P_{i - 1}$ to $P_{i}$ along *C*, it moves approximately in the direction of $\mathbf{\text{T}}\left( P_{i} \right),$ the unit tangent vector at the endpoint of $\mathbf{\text{r}}(t_{i}^{*}).$ Let $P_{i}^{*}$ denote the endpoint of $\mathbf{\text{r}}(t_{i}^{*}).$ Then, the work done by the force vector field in moving the particle from $P_{i - 1}$ to *P_i* is $\mathbf{\text{F}}\left( P_{i}^{*} \right) \cdot \left( {\text{Δ}s_{i}\mathbf{\text{T}}\left( P_{i}^{*} \right)} \right),$ so the total work done along *C* is

$${\sum\limits_{i = 1}^{n}{\mathbf{\text{F}}\left( P_{i}^{*} \right) \cdot \left( {\text{Δ}s_{i}\mathbf{\text{T}}\left( P_{i}^{*} \right)} \right)}} = {\sum\limits_{i = 1}^{n}{\mathbf{\text{F}}\left( P_{i}^{*} \right) \cdot \mathbf{\text{T}}\left( P_{i}^{*} \right)}}\text{Δ}s_{i}.$$

Letting the arc length of the pieces of *C* get arbitrarily small by taking a limit as $n\rightarrow\infty$ gives us the work done by the field in moving the particle along *C*. Therefore, the work done by F in moving the particle in the positive direction along *C* is defined as

$$W = {\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds,}}$$

which gives us the concept of a vector line integral.

The vector line integral of vector field F along oriented smooth curve *C* is

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds =}}\underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}{\mathbf{\text{F}}\left( P_{i}^{*} \right)}} \cdot \mathbf{\text{T}}\left( P_{i}^{*} \right)\text{Δ}s_{i}$$

if that limit exists.

With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once by the parameterization, the value of the line integral is unchanged. With vector line integrals, the orientation of the curve does matter. If we think of the line integral as computing work, then this makes sense: if you hike up a mountain, then the gravitational force of Earth does negative work on you. If you walk down the mountain by the exact same path, then Earth’s gravitational force does positive work on you. In other words, reversing the path changes the work value from negative to positive in this case. Note that if *C* is an oriented curve, then we let −*C* represent the same curve but with opposite orientation.

As with scalar line integrals, it is easier to compute a vector line integral if we express it in terms of the parameterization function r and the variable *t*. To translate the integral $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}$ in terms of *t*, note that unit tangent vector T along *C* is given by $\mathbf{\text{T}} = \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}$ (assuming $\left\| {\mathbf{r^{\prime}}(t)} \right\| \neq 0).$ Since $ds = \left\| {\mathbf{r^{\prime}}(t)} \right\| dt,$ as we saw when discussing scalar line integrals, we have

$$\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds = \mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) \cdot \frac{\mathbf{r^{\prime}}(t)}{\left\| {\mathbf{r^{\prime}}(t)} \right\|}\left\| {\mathbf{r^{\prime}}(t)} \right\| dt = \mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) \cdot \mathbf{r^{\prime}}(t)dt.$$

Thus, we have the following formula for computing vector line integrals:

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}} = {\int_{a}^{b}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right)}} \cdot \mathbf{r^{\prime}}(t)dt.$$ (6.9)

Because of Equation 6.9, we often use the notation $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ for the line integral $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds.}$

If $\mathbf{\text{r}}(t) = \left\langle {x(t),y(t),z(t)} \right\rangle,$ then *d*r denotes vector differential ${\left\langle {x^{\prime}(t),y^{\prime}(t),z^{\prime}(t)} \right\rangle dt}.$

Evaluating a Vector Line Integral 计算一个向量线积分

Find the value of integral $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}$ where $C$ is the semicircle parameterized by $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle,$ $0 \leq t \leq \pi$ and $\mathbf{\text{F}} = \left\langle {\text{−}y,x} \right\rangle.$

Solution 解

We can use Equation 6.9 to convert the variable of integration from *r* to *t*. We then have

$$\mathbf{\text{F}}(\mathbf{\text{r}}(t)) = \left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle\ \text{and}\ \mathbf{r^{\prime}}(t) = \left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle.$$

Therefore,

$$\begin{array}{cl}{{\int_{C}\mathbf{\text{F}}} \cdot d\mathbf{\text{r}}} & {= {\int_{0}^{\pi}{\left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle \cdot \left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle dt}}} \\ & {= {\int_{0}^{\pi}{\text{sin}^{2}t}} + \text{cos}^{2}tdt} \\ & {= {\int_{0}^{\pi}{1dt}} = \pi.} \end{array}$$

See Figure 6.18.

Reversing Orientation 反向定向

Find the value of integral $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}$ where $C$ is the semicircle parameterized by $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ (t + \pi),\text{sin}\ t} \right\rangle,0 \leq t \leq \pi$ and $\mathbf{\text{F}} = \left\langle {\text{−}y,x} \right\rangle.$

Solution 解

Notice that this is the same problem as Example 6.18, except the orientation of the curve has been reversed. In this example, the parameterization starts at $\mathbf{\text{r}}(0) = \left\langle {–1,0} \right\rangle$ and ends at $\mathbf{\text{r}}(\pi) = \left\langle {1,0} \right\rangle.$ By Equation 6.9,

$$\begin{array}{cl} {{\int_{C}\mathbf{\text{F}}} \cdot d\mathbf{\text{r}}} & {= {\int_{0}^{\pi}{\left\langle {\text{−}\text{sin}\ t,\text{cos}\ (t + \pi)} \right\rangle \cdot \left\langle {\text{−}\text{sin}\ (t + \pi),\text{cos}\ t} \right\rangle dt}}} \\ & {= {\int_{0}^{\pi}{\left\langle {\text{−}\text{sin}\ t,\text{−}\text{cos}\ t} \right\rangle \cdot \left\langle {\text{sin}\ t,\text{cos}\ t} \right\rangle dt}}} \\ & {= {\int_{0}^{\pi}{\left( {\text{−}\text{sin}^{2}t - \text{cos}^{2}t} \right)dt}}} \\ & {= {\int_{0}^{\pi}{-1dt}}} \\ & {= \text{−}\pi.} \end{array}$$

Notice that this is the negative of the answer in Example 6.18. It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.

Let *C* be an oriented curve and let −*C* denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:

$${\int_{–C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} = \text{−}{\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}}}.$$

Let $\mathbf{\text{F}} = x\mathbf{\text{i}} + y\mathbf{\text{j}}$ be a vector field and let *C* be the curve with parameterization $\left\langle {t,t^{2}} \right\rangle$ for $0 \leq t \leq 2.$ Which is greater: $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}$ or ${\int_{\text{−}C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}}?$

Another standard notation for integral $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ is $\int_{C}{Pdx + Qdy + Rdz.}$ In this notation, *P*, *Q*, and *R* are functions, and we think of *d*r as vector $\left\langle {dx,dy,dz} \right\rangle.$ To justify this convention, recall that $d\mathbf{\text{r}} = \mathbf{\text{T}}ds = \mathbf{r^{\prime}}(t)dt = \left\langle {\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}} \right\rangle dt.$ Therefore,

$$\mathbf{\text{F}} \cdot d\mathbf{\text{r}} = \left\langle {P,Q,R} \right\rangle \cdot \left\langle {dx,dy,dz} \right\rangle = Pdx + Qdy + Rdz.$$

If $d\mathbf{\text{r}} = \left\langle {dx,dy,dz} \right\rangle,$ then $\frac{d\mathbf{\text{r}}}{dt} = \left\langle {\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}} \right\rangle,$ which implies that ${d\mathbf{\text{r}}} = \left\langle {\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}} \right\rangle dt.$ Therefore

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{C}{Pdx + Qdy + Rdz}}} \\ & {= {\int\left( {P\left( {\mathbf{\text{r}}(t)} \right)\frac{dx}{dt} + Q\left( {\mathbf{\text{r}}(t)} \right)\frac{dy}{dt} + R\left( {\mathbf{\text{r}}(t)} \right)\frac{dz}{dt}} \right)}dt.} \end{array}$$ (6.10)

Finding the Value of an Integral of the Form $\int_{C}{Pdx + Qdy + Rdz}$ 计算形如 $\int_{C}{Pdx + Qdy + Rdz}$ 的积分值

Find the value of integral $\int_{C}{zdx + xdy + ydz,}$ where *C* is the curve parameterized by $\mathbf{\text{r}}(t) = \left\langle {t^{2},\sqrt{t},t} \right\rangle,1 \leq t \leq 4.$

Solution 解

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of *t*. In this case, Equation 6.10 allows us to make this change:

$$\begin{array}{cl} {{\int_{C}z}dx + xdy + ydz} & {= {\int_{1}^{4}{\left( {t\left( {2t} \right) + t^{2}\left( \frac{1}{2\sqrt{t}} \right) + \sqrt{t}} \right)dt}}} \\ & {= {\int_{1}^{4}{\left( {2t^{2} + \frac{t^{3\text{/}2}}{2} + \sqrt{t}} \right)dt}}} \\ & {= \left\lbrack {\frac{2t^{3}}{3} + \frac{t^{5\text{/}2}}{5} + \frac{2t^{3\text{/}2}}{3}} \right\rbrack_{t = 1}^{t = 4}} \\ & {= \frac{793}{15}.} \end{array}$$

Find the value of $\int_{C}{4xdx + zdy + 4y^{2}dz,}$ where $C$ is the curve parameterized by $\mathbf{\text{r}}(t) = \left\langle {4\ \text{cos}\left( {2t} \right),2\ \text{sin}\left( {2t} \right),3} \right\rangle,0 \leq t \leq \frac{\pi}{4}.$

We have learned how to integrate smooth oriented curves. Now, suppose that *C* is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that *C* is a piecewise smooth curve. To be precise, curve *C* is piecewise smooth if *C* can be written as a union of *n* smooth curves $C_{1},C_{2}\text{,…},C_{n}$ such that the endpoint of $C_{i}$ is the starting point of $C_{i + 1}$ (Figure 6.19). When curves $C_{i}$ satisfy the condition that the endpoint of $C_{i}$ is the starting point of $C_{i + 1},$ we write their union as $C_{1} + C_{2} + \cdots + C_{n}.$

The next theorem summarizes several key properties of vector line integrals.

Properties of Vector Line Integrals 向量线积分的性质

Let F and G be continuous vector fields with domains that include the oriented smooth curve *C*. Then

1. ${\int_{C}{(\mathbf{\text{F}} + \mathbf{\text{G}}) \cdot d\mathbf{\text{r}}}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{C}{\mathbf{\text{G}} \cdot d\mathbf{\text{r}}}}$

2. ${\int_{C}{k\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = k{\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}}$ where *k* is a constant

3. ${\int_{\text{−}C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} =}}{\text{−}\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}$

4. Suppose instead that *C* is a piecewise smooth curve in the domains of F and G, where $C = C_{1} + C_{2} + \cdots + C_{n}$ and $C_{1},C_{2}\text{,…},C_{n}$ are smooth curves such that the endpoint of $C_{i}$ is the starting point of $C_{i + 1}.$ Then

$$\int_{C}\textbf{F} \cdot d\textbf{r} = \int_{C_{1}}\textbf{F} \cdot d\textbf{r} + \int_{C_{2}}\textbf{F} \cdot d\textbf{r} + \cdots + \int_{C_{n}}\textbf{F} \cdot d\textbf{r}.$$

Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along *C*, then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation $\int_{a}^{b}{f(x)dx = \text{−}{\int_{b}^{a}{f(x)dx.}}}$ Finally, if $\left\lbrack {a_{1},a_{2}} \right\rbrack,\left\lbrack {a_{2},a_{3}} \right\rbrack\text{,…},\left\lbrack {a_{n - 1},a_{n}} \right\rbrack$ are intervals, then

$${\int_{a_{1}}^{a_{n}}{f(x)dx =}}{\int_{a_{1}}^{a_{2}}{f(x)dx}} + {\int_{a_{2}}^{a_{3}}{f(x)dx + \cdots + {\int_{a_{n - 1}}^{a_{n}}{f(x)dx}}}},$$

which is analogous to property iv.

Using Properties to Compute a Vector Line Integral 利用性质计算向量线积分

Find the value of integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}}}ds,$ where *C* is the rectangle (oriented counterclockwise) in a plane with vertices $\left( {0,0} \right),\left( {2,0} \right),\left( {2,1} \right),\ \text{and}\ \left( {0,1} \right),$ and where $\mathbf{\text{F}} = \left\langle {x - 2y,y - x} \right\rangle$ (Figure 6.20).

Solution 解

Note that curve *C* is the union of its four sides, and each side is smooth. Therefore *C* is piecewise smooth. Let $C_{1}$ represent the side from $\left( {0,0} \right)$ to $\left( {2,0} \right),$ let $C_{2}$ represent the side from $\left( {2,0} \right)$ to $\left( {2,1} \right),$ let $C_{3}$ represent the side from $\left( {2,1} \right)$ to $\left( {0,1} \right),$ and let $C_{4}$ represent the side from $\left( {0,1} \right)$ to $\left( {0,0} \right)$ (Figure 6.20). Then,

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}d\mathbf{\text{r}}}} = {\int_{C_{1}}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}d\mathbf{\text{r}}}} + {\int_{C_{2}}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}d\mathbf{\text{r}}}} + {\int_{C_{3}}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}d\mathbf{\text{r}}}} + {\int_{C_{4}}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}d\mathbf{\text{r}}}}.$$

We want to compute each of the four integrals on the right-hand side using Equation 6.8. Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations (note that they traverse *C* counterclockwise):

$$\begin{array}{l} {C_{1}:\left\langle {t,0} \right\rangle,0 \leq t \leq 2} \\ {C_{2}:\left\langle {2,t} \right\rangle,0 \leq t \leq 1} \\ {C_{3}:\left\langle {2 - t,1} \right\rangle,0 \leq t \leq 2} \\ {C_{4}:\left\langle {0,1 - t} \right\rangle,0 \leq t \leq 1.} \end{array}$$

Therefore,

$$\begin{matrix} {\int_{C_{1}}\textbf{F} \cdot \textbf{T}d\textbf{r}} & {= \int_{0}^{2}\textbf{F}\left( \textbf{r}(t) \right) \cdot \mathbf{r}^{'}(t)dt} \\ & {= \int_{0}^{2}\left\langle t - 2(0),0 - t \right\rangle \cdot \left\langle 1,0 \right\rangle dt = \int_{0}^{2}tdt} \\ & {= \left\lbrack \frac{t^{2}}{2} \right\rbrack_{0}^{2} = 2.} \end{matrix}$$

Notice that the value of this integral is positive, which should not be surprising. As we move along curve *C₁* from left to right, our movement flows in the general direction of the vector field itself. At any point along *C₁*, the tangent vector to the curve and the corresponding vector in the field form an angle that is less than 90°. Therefore, the tangent vector and the force vector have a positive dot product all along *C₁*, and the line integral will have positive value.

The calculations for the three other line integrals are done similarly:

$$\begin{array}{cl} {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{0}^{1}\left\langle {2 - 2t,t - 2} \right\rangle} \cdot \left\langle {0,1} \right\rangle dt} \\ & {= {\int_{0}^{1}{\left( {t - 2} \right)dt}}} \\ & {= \left\lbrack {\frac{t^{2}}{2} - 2t} \right\rbrack_{0}^{1} = - \frac{3}{2},} \end{array}$$ $$\begin{array}{cl} {\int_{C_{3}}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}} & {= {\int_{0}^{2}\left\langle {(2 - t) - 2,1 - (2 - t)} \right\rangle} \cdot \left\langle {-1,0} \right\rangle dt} \\ & {= {\int_{0}^{2}{tdt}} = 2,} \end{array}$$

and

$$\begin{array}{cl} {\int_{C_{4}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{0}^{1}\left\langle {-2(1 - t),1 - t} \right\rangle} \cdot \left\langle {0,-1} \right\rangle dt} \\ & {= {\int_{0}^{1}{\left( {t - 1} \right)dt}}} \\ & {= \left\lbrack {\frac{t^{2}}{2} - t} \right\rbrack_{0}^{1} = - \frac{1}{2}.} \end{array}$$

Thus, we have ${\int_{C}\mathbf{\text{F}}} \cdot d\mathbf{\text{r}} = 2.$

Calculate line integral ${\int_{C}\mathbf{\text{F}}} \cdot d\mathbf{\text{r}},$ where F is vector field $\left\langle {y^{2},2xy + 1} \right\rangle$ and *C* is a triangle with vertices $\left( {0,0} \right),$ $\left( {4,0} \right),$ and $\left( {0,5} \right),$ oriented counterclockwise.

Applications of Line Integrals 线积分的应用

Scalar line integrals have many applications. They can be used to calculate the length or mass of a wire, the surface area of a sheet of a given height, or the electric potential of a charged wire given a linear charge density. Vector line integrals are extremely useful in physics. They can be used to calculate the work done on a particle as it moves through a force field, or the flow rate of a fluid across a curve. Here, we calculate the mass of a wire using a scalar line integral and the work done by a force using a vector line integral.

Suppose that a piece of wire is modeled by curve *C* in space. The mass per unit length (the linear density) of the wire is a continuous function $\rho\left( {x,y,z} \right).$ We can calculate the total mass of the wire using the scalar line integral $\int_{C}{\rho\left( {x,y,z} \right)ds.}$ The reason is that mass is density multiplied by length, and therefore the density of a small piece of the wire can be approximated by $\rho\left( {x*,y*,z*} \right)\text{Δ}s$ for some point $\left( {x*,y*,z*} \right)$ in the piece. Letting the length of the pieces shrink to zero with a limit yields the line integral $\int_{C}{\rho\left( {x,y,z} \right)ds.}$

Calculating the Mass of a Wire 计算金属丝的质量

Calculate the mass of a spring in the shape of a curve parameterized by $\left\langle {t,2\ \text{cos}\ t,2\ \text{sin}\ t} \right\rangle,$ $0 \leq t \leq \frac{\pi}{2},$ with a density function given by $\rho(x,y,z) = e^{x} + yz$ kg/m (Figure 6.21).

Solution 解

To calculate the mass of the spring, we must find the value of the scalar line integral ${\int_{C}{\left( {e^{x} + yz} \right)ds}},$ where *C* is the given helix. To calculate this integral, we write it in terms of *t* using Equation 6.8:

$$\begin{array}{cl} {{\int_{C}e^{x}} + yzds} & {= {\int_{0}^{\pi\text{/}2}{\left( {\left( {e^{t} + 4\ \text{cos}\ t\ \text{sin}\ t} \right)\sqrt{1 + \left( {-2\ \text{cos}\ t} \right)^{2} + \left( {2\ \text{sin}\ t} \right)^{2}}} \right)dt}}} \\ & {= {\int_{0}^{\pi\text{/}2}{\left( {\left( {e^{t} + 4\ \text{cos}\ t\ \text{sin}\ t} \right)\sqrt{5}} \right)dt}}} \\ & {= \sqrt{5}\left\lbrack {e^{t} + 2\ \text{sin}^{2}t} \right\rbrack_{t = 0}^{t = \pi\text{/}2}} \\ & {= \sqrt{5}\left( {e^{\pi\text{/}2} + 1} \right).} \end{array}$$

Therefore, the mass is $\sqrt{5}\left( {e^{\pi\text{/}2} + 1} \right)$ kg.

Calculate the mass of a spring in the shape of a helix parameterized by $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t,t} \right\rangle,0 \leq t \leq 6\pi,$ with a density function given by $\rho\left( {x,y,z} \right) = x + y + z$ kg/m.

When we first defined vector line integrals, we used the concept of work to motivate the definition. Therefore, it is not surprising that calculating the work done by a vector field representing a force is a standard use of vector line integrals. Recall that if an object moves along curve *C* in force field F, then the work required to move the object is given by ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$

Calculating Work 计算功

How much work is required to move an object in vector force field $\mathbf{\text{F}} = \left\langle {yz,xy,xz} \right\rangle$ along path $\mathbf{\text{r}}(t) = \left\langle {t^{2},t,t^{4}} \right\rangle,$ $0 \leq t \leq 1?$ See Figure 6.22.

Solution 解

Let *C* denote the given path. We need to find the value of ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$ To do this, we use Equation 6.9:

$$\begin{array}{cl} {{\int_{C}\mathbf{\text{F}}} \cdot d\mathbf{\text{r}}} & {= {\int_{0}^{1}{\left( {\left\langle {t^{5},t^{3},t^{6}} \right\rangle \cdot \left\langle {2t,1,4t^{3}} \right\rangle} \right)dt}}} \\ & {= {\int_{0}^{1}{\left( {2t^{6} + t^{3} + 4t^{9}} \right)dt}}} \\ & {= \left\lbrack {\frac{2t^{7}}{7} + \frac{t^{4}}{4} + \frac{2t^{10}}{5}} \right\rbrack_{t = 0}^{t = 1} = \frac{131}{140}.} \end{array}$$

Flux and Circulation 通量与环流量

We close this section by discussing two key concepts related to line integrals: flux across a plane curve and circulation along a plane curve. Flux is used in applications to calculate fluid flow across a curve, and the concept of circulation is important for characterizing conservative gradient fields in terms of line integrals. Both these concepts are used heavily throughout the rest of this chapter. The idea of flux is especially important for Green’s theorem, and in higher dimensions for Stokes’ theorem and the divergence theorem.

Let *C* be a plane curve and let F be a vector field in the plane. Imagine *C* is a membrane across which fluid flows, but *C* does not impede the flow of the fluid. In other words, *C* is an idealized membrane invisible to the fluid. Suppose F represents the velocity field of the fluid. How could we quantify the rate at which the fluid is crossing *C*?

Recall that the line integral of F along *C* is $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}$—in other words, the line integral is the dot product of the vector field with the unit tangential vector with respect to arc length. If we replace the unit tangential vector with unit normal vector $\mathbf{\text{N}}(t)$ and instead compute integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}},$ we determine the flux across *C*. To be precise, the definition of integral $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}$ is the same as integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}},$ except the T in the Riemann sum is replaced with N. Therefore, the flux across *C* is defined as

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds =}}\underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}{\mathbf{\text{F}}\left( P_{i}^{*} \right)}} \cdot \mathbf{\text{N}}\left( P_{i}^{*} \right)\text{Δ}s_{i},$$

where $P_{i}^{*}$ and $\text{Δ}s_{i}$ are defined as they were for integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}}.$ Therefore, a flux integral is an integral that is *perpendicular* to a vector line integral, because N and T are perpendicular vectors.

If F is a velocity field of a fluid and *C* is a curve that represents a membrane, then the flux of F across *C* is the quantity of fluid flowing across *C* per unit time, or the rate of flow.

More formally, let *C* be a plane curve parameterized by $\mathbf{\text{r}}(t) = \left\langle {x(t),y(t)} \right\rangle,$ $a \leq t \leq b.$ Let $\mathbf{\text{n}}(t) = \left\langle {y^{\prime}(t),\text{−}x^{\prime}(t)} \right\rangle$ be the vector that is normal to *C* at the endpoint of $\mathbf{\text{r}}(t)$ and points to the right as we traverse *C* in the positive direction (Figure 6.23). Then, $\mathbf{\text{N}}(t) = \frac{\mathbf{\text{n}}(t)}{\left\| {\mathbf{\text{n}}(t)} \right\|}$ is the unit normal vector to *C* at the endpoint of $\mathbf{\text{r}}(t)$ that points to the right as we traverse *C*.

The flux of F across *C* is line integral ${\int_{C}{\mathbf{\text{F}} \cdot \frac{\mathbf{\text{n}}(t)}{\left\| {\mathbf{\text{n}}(t)} \right\|}}}ds.$

We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see Equation 6.9).

Calculating Flux across a Curve 计算穿过曲线的通量

Let F be a vector field and let *C* be a smooth curve with parameterization $\mathbf{\text{r}}(t) = \left\langle {x(t),y(t)} \right\rangle,a \leq t \leq b.$ Let $\mathbf{\text{n}}(t) = \left\langle {y^{\prime}(t),\text{−}x^{\prime}(t)} \right\rangle.$ The flux of F across *C* is

$$\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds = {\int_{a}^{b}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) \cdot \mathbf{\text{n}}(t)dt}}}$$ (6.11)

Proof 证明

The proof of Equation 6.11 is similar to the proof of Equation 6.8. Before deriving the formula, note that $\left\| {\mathbf{\text{n}}(t)} \right\| = \left\| \left\langle {y\prime(t),\text{−}x\prime(t)} \right\rangle \right\| = \sqrt{\left( {y\prime(t)} \right)^{2} + \left( {x\prime(t)} \right)^{2}} = \left\| {\mathbf{r^{\prime}}(t)} \right\|.$ Therefore,

$$\begin{array}{cl} {{\int_{C}\mathbf{\text{F}}} \cdot \mathbf{\text{N}}ds} & {= {\int_{C}{\mathbf{\text{F}} \cdot \frac{\mathbf{\text{n}}(t)}{\left\| {\mathbf{\text{n}}(t)} \right\|}ds}}} \\ & {= {\int_{a}^{b}{\mathbf{\text{F}} \cdot \frac{\mathbf{\text{n}}(t)}{\left\| {\mathbf{\text{n}}(t)} \right\|}\left\| {\mathbf{r^{\prime}}(t)} \right\| dt}}} \\ & {= {\int_{a}^{b}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) \cdot \mathbf{\text{n}}(t)dt}}.} \end{array}$$

□

Flux across a Curve 穿过曲线的通量

Calculate the flux of $\mathbf{\text{F}} = \left\langle {2x,2y} \right\rangle$ across a unit circle oriented counterclockwise (Figure 6.24).

Solution 解

To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle,$ $0 \leq t \leq 2\pi.$ The normal vector to a unit circle is $\left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle.$ Therefore, the flux is

$$\begin{array}{cl} {{\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}}}ds} & {= {\int_{0}^{2\pi}{\left\langle {2\ \text{cos}\ t,2\ \text{sin}\ t} \right\rangle \cdot \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle}}\ dt} \\ & {= {\int_{0}^{2\pi}\left( {2\ \text{cos}^{2}t + 2\ \text{sin}^{2}t} \right)}\ dt = 2{\int_{0}^{2\pi}\left( {\text{cos}^{2}t + \text{sin}^{2}t} \right)}\ dt} \\ & {= 2{\int_{0}^{2\pi}{dt}} = 4\pi.} \end{array}$$

Calculate the flux of $\mathbf{\text{F}} = \left\langle {x + y,2y} \right\rangle$ across the line segment from $(0,0)$ to $(2,3),$ where the curve is oriented from left to right.

Let $\mathbf{\text{F}}(x,y) = \left\langle {P(x,y),Q(x,y)} \right\rangle$ be a two-dimensional vector field. Recall that integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}}}ds$ is sometimes written as ${\int_{C}{Pdx + Qdy}}.$ Analogously, flux $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}$ is sometimes written in the notation $\int_{C}{\text{−}Qdx + Pdy,}$ because the unit normal vector N is perpendicular to the unit tangent T. Rotating the vector $d\mathbf{\text{r}} = \left\langle {dx,dy} \right\rangle$ by 90° results in vector $\left\langle {dy,\text{−}dx} \right\rangle.$ Therefore, the line integral in Example 6.21 can be written as ${\int_{C}{-2ydx + 2xdy}}.$

Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field F along an oriented closed curve is called the circulation of F along *C*. Circulation line integrals have their own notation: ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}}.$ The circle on the integral symbol denotes that *C* is “circular” in that it has no endpoints. Example 6.18 shows a calculation of circulation.

To see where the term *circulation* comes from and what it measures, let v represent the velocity field of a fluid and let *C* be an oriented closed curve. At a particular point *P*, the closer the direction of v(*P*) is to the direction of T(*P*), the larger the value of the dot product $\mathbf{\text{v}}(P) \cdot \mathbf{\text{T}}(P).$ The maximum value of $\mathbf{\text{v}}(P) \cdot \mathbf{\text{T}}(P)$ occurs when the two vectors are pointing in the exact same direction; the minimum value of $\mathbf{\text{v}}(P) \cdot \mathbf{\text{T}}(P)$ occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation $\int_{C}{\mathbf{\text{v}} \cdot \mathbf{\text{T}}ds}$ measures the tendency of the fluid to move in the direction of *C*.

Calculating Circulation 计算环流量

Let $\mathbf{\text{F}} = \left\langle - y,x \right\rangle$ be the vector field from Example 6.16 and let *C* represent the unit circle oriented counterclockwise. Calculate the circulation of F along *C*.

Solution 解

We use the standard parameterization of the unit circle: $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ t,\text{sin}\ t} \right\rangle,0 \leq t \leq 2\pi.$ Then, $\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) = \left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle$ and $\mathbf{r^{\prime}}(t) = \left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle.$ Therefore, the circulation of F along *C* is

$$\begin{array}{cl} {{\int_{C}\mathbf{\text{F}}} \cdot \mathbf{\text{T}}ds} & {= {\int_{0}^{2\pi}{\left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle \cdot \left\langle {\text{−}\text{sin}\ t,\text{cos}\ t} \right\rangle dt}}} \\ & {= {\int_{0}^{2\pi}\left( {\text{sin}^{2}t + \text{cos}^{2}t} \right)}\ dt} \\ & {= {\int_{0}^{2\pi}{dt}} = 2\pi.} \end{array}$$

Notice that the circulation is positive. The reason for this is that the orientation of *C* “flows” with the direction of F. At any point along the circle, the tangent vector and the vector from F form an angle of less than 90°, and therefore the corresponding dot product is positive.

In Example 6.25, what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by $\text{−}C.$ Then

$${\int_{\text{−}C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}}}ds = \text{−}{\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}}}ds = -2\pi.$$

Notice that the circulation is negative in this case. The reason for this is that the orientation of the curve flows against the direction of F.

Calculate the circulation of $\mathbf{\text{F}}(x,y) = \left\langle {- \frac{y}{x^{2} + y^{2}},\frac{x}{x^{2} + y^{2}}} \right\rangle$ along a unit circle oriented counterclockwise.

Calculating Work 计算功

Calculate the work done on a particle that traverses circle *C* of radius 2 centered at the origin, oriented counterclockwise, by field $\mathbf{\text{F}}(x,y) = \left\langle {-2,y} \right\rangle.$ Assume the particle starts its movement at $(1,0).$

Solution 解

The work done by F on the particle is the circulation of F along *C*: ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}}.$ We use the parameterization $\mathbf{\text{r}}(t) = \left\langle {2\ \text{cos}\ t,2\ \text{sin}\ t} \right\rangle,0 \leq t \leq 2\pi$ for *C*. Then, $\mathbf{r^{\prime}}(t) = \left\langle {-2\ \text{sin}\ t,2\ \text{cos}\ t} \right\rangle$ and $\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) = \left\langle {-2,2\ \text{sin}\ t} \right\rangle.$ Therefore, the circulation of F along *C* is

$$\begin{array}{cl} {\int_{C}\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds} & {= {\int_{0}^{2\pi}{\left\langle {-2,2\ \text{sin}\ t} \right\rangle \cdot \left\langle {-2\ \text{sin}\ t,2\ \text{cos}\ t} \right\rangle dt}}} \\ & {= {\int_{0}^{2\pi}{\left( {4\ \text{sin}\ t + 4\ \text{sin}\ t\ \text{cos}\ t} \right)dt}}} \\ & {= \left\lbrack {-4\ \text{cos}\ t + 4\ \text{sin}^{2}t} \right\rbrack_{0}^{2\pi}} \\ & {= \left( {-4\ \text{cos}(2\pi) + 2\ \text{sin}^{2}(2\pi)} \right) - \left( {-4\ \text{cos}(0) + 4\ \text{sin}^{2}(0)} \right)} \\ & {= -4 + 4 = 0.} \end{array}$$

The force field does zero work on the particle.

Notice that the circulation of F along *C* is zero. Furthermore, notice that since F is the gradient of $f(x,y) = -2x + \frac{y^{2}}{2},$ F is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.

Calculate the work done by field $\mathbf{\text{F}}(x,y) = \left\langle {2x,3y} \right\rangle$ on a particle that traverses the unit circle. Assume the particle begins its movement at $(-1,0).$

Section 6.2 Exercises 6.2 节习题

39.

*True or False?* Line integral $\int_{C}^{}{f(x,y)ds}$ is equal to a definite integral if *C* is a smooth curve defined on $\left\lbrack {a,b} \right\rbrack$ and if function $f$ is continuous on some region that contains curve *C*.

40\.

*True or False?* Vector functions $\mathbf{\text{r}}_{1} = t\mathbf{\text{i}} + t^{2}\mathbf{\text{j}},$ $0 \leq t \leq 1,$ and $\mathbf{\text{r}}_{2} = (1 - t)\mathbf{\text{i}} + {(1 - t)}^{2}\mathbf{\text{j}},$ $0 \leq t \leq 1,$ define the same oriented curve.

41.

*True or False?* $\int_{\text{−}C}^{}{(Pdx + Qdy) = {\int_{C}^{}{(Pdx - Qdy)}}}$

42\.

*True or False?* A piecewise smooth curve *C* consists of a finite number of smooth curves that are joined together end to end.

43.

*True or False?* If *C* is given by $x(t) = t\text{,}\ y(t) = t\text{, 0} \leq \text{t} \leq 1,$ then ${\int_{C}^{}{xyds = {\int_{0}^{1}{t^{2}dt}}}}.$

For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path.

44\.

\[T\] $\int_{C}^{}{(x + y)ds}$

$C\text{:}\ x = t,y = (1 - t)\text{,}\ z = 0$ from (0, 1, 0) to (1, 0, 0)

45.

\[T\] $\int_{C}^{}{(x - y)ds}$

$C\text{:}\ \mathbf{\text{r}}(t) = 4t\mathbf{\text{i}} + 3t\mathbf{\text{j}}$ when $0 \leq t \leq 2$

46\.

\[T\] $\int_{C}^{}{(x^{2} + y^{2} + z^{2})ds}$

$C\text{:}\ \mathbf{\text{r}}(t) = \text{sin}\ t\mathbf{\text{i}} + \text{cos}\ t\mathbf{\text{j}} + 8t\mathbf{\text{k}}$ when $0 \leq t \leq \frac{\pi}{2}$

47.

\[T\] Evaluate ${\int_{C}^{}{xy^{4}ds}},$ where *C* is the right half of circle $x^{2} + y^{2} = 16$ and is traversed in the clockwise direction.

48\.

\[T\] Evaluate ${\int_{C}^{}{4x^{3}ds}},$ where *C* is the line segment from $(-2,-1)$ to (1, 2).

For the following exercises, find the work done.

49.

Find the work done by vector field $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + 3xy\mathbf{\text{j}} - (x + z)\mathbf{\text{k}}$ on a particle moving along a line segment that goes from $(1,4,2)$ to $(0,5,1).$

50\.

Find the work done by a person weighing 150 lb walking exactly one revolution up a circular, spiral staircase of radius 3 ft if the person rises 10 ft.

51.

Find the work done by force field $\mathbf{\text{F}}(x,y,z) = - \frac{1}{2}x\mathbf{\text{i}} - \frac{1}{2}y\mathbf{\text{j}} + \frac{1}{4}\mathbf{\text{k}}$ on a particle as it moves along the helix $\mathbf{\text{r}}(t) = \text{cos}\ t\mathbf{\text{i}} + \text{sin}\ t\mathbf{\text{j}} + t\mathbf{\text{k}}$ from point $(1,0,0)$ to point $\left( {-1,0,3\pi} \right).$

52\.

Find the work done by vector field $\mathbf{\text{F}}(x,y) = y\mathbf{\text{i}} + 2x\mathbf{\text{j}}$ in moving an object along path *C*, the straight line which joins points (1, 0) and (0, 1).

53.

Find the work done by force $\mathbf{\text{F}}(x,y) = 2y\mathbf{\text{i}} + 3x\mathbf{\text{j}} + (x + y)\mathbf{\text{k}}$ in moving an object along curve $\mathbf{\text{r}}(t) = \text{cos}(t)\mathbf{\text{i}} + \text{sin}(t)\mathbf{\text{j}} + \frac{1}{6}\mathbf{\text{k}},$ where $0 \leq t \leq 2\pi.$

54\.

Find the mass of a wire in the shape of a circle of radius 2 centered at (3, 4) with linear mass density $\rho(x,y) = y^{2}.$

For the following exercises, evaluate the line integrals.

55.

Evaluate ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = -1\mathbf{\text{j}},$ and *C* is the part of the graph of $y = \frac{1}{2}x^{3} - x$ from $(2,2)$ to $(-2,-2).$

56\.

Evaluate ${\int_{\gamma}^{}{\left( {x^{2} + y^{2} + z^{2}} \right)^{-1}ds}},$ where $\gamma$ is the helix $x = \text{cos}\ t,y = \text{sin}\ t,z = t(0 \leq t \leq T).$

57.

Evaluate $\int_{C}^{}{yz\mspace{2mu} dx + xz\mspace{2mu} dy + xy\mspace{2mu} dz}$ over the line segment from $(1,1,1)$ to $(3,2,0).$

58\.

Let *C* be the line segment from point (0, 1, 1) to point (2, 2, 3). Evaluate line integral ${\int_{C}^{}{yds}}.$

59.

\[T\] Use a computer algebra system to evaluate the line integral ${\int_{C}{y^{2}dx + xdy}},$ where *C* is the arc of the parabola $x = 4 - y^{2}$ from (−5, −3) to (0, 2).

60\.

\[T\] Use a computer algebra system to evaluate the line integral $\int_{C}^{}{\left( {x + 3y^{2}} \right)dy}$ over the path *C* given by $x = 2t\text{,}\ y = 10t\text{,}$ where $0 \leq t \leq 1.$

61.

\[T\] Use a CAS to evaluate line integral $\int_{C}^{}{xydx + ydy}$ over path *C* given by $x = 2t\text{,}\ y = 10t\text{,}$ where $0 \leq t \leq 1.$

62\.

Evaluate line integral ${\int_{C}^{}{\left( {2x - y} \right)dx + \left( {x + 3y} \right)dy}},$ where *C* lies along the *x*-axis from $x = 0\ \text{to}\ x = 5.$

63.

\[T\] Use a CAS to evaluate ${\int_{C}^{}{\frac{y}{2x^{2} - y^{2}}ds}},$ where *C* is $x = t\text{,}\ y = t\text{,}\ 1 \leq t \leq 5.$

64\.

\[T\] Use a CAS to evaluate $\int_{C}{xyds,}$ where *C* is $x = t^{2},y = 4t,0 \leq t \leq 1.$

In the following exercises, find the work done by force field F on an object moving along the indicated path.

65.

$\mathbf{\text{F}}(x,y) = \text{−}x\mathbf{\text{i}} - 2y\mathbf{\text{j}}$

$C\text{:}\ y = x^{3}\ \text{from (0, 0) to (2, 8)}$

66\.

$\mathbf{\text{F}}(x\text{,}\ y) = 2x\mathbf{i} + y\mathbf{\text{j}}$

*C*: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 1)

67.

$\mathbf{\text{F}}(x\text{,}\ y\text{,}\ z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} - 5z\mathbf{\text{k}}$

$\textit{C}\text{:}\ \mathbf{\text{r}}(t) = 2\ \text{cos}\ t\mathbf{\text{i}} + 2\ \text{sin}\ t\mathbf{\text{j}} + t\mathbf{\text{k}}\text{,}\ 0 \leq t \leq 2\pi$

68\.

Let F be vector field $\mathbf{\text{F}}(x,y) = \left( {y^{2} + 2xe^{y} + 1} \right)\mathbf{\text{i}} + \left( {2xy + x^{2}e^{y} + 2y} \right)\mathbf{\text{j}}.$ Compute the work of integral ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where *C* is the path $\mathbf{\text{r}}(t) = \text{sin}\ t\mathbf{\text{i}} + \text{cos}\ t\mathbf{\text{j}}\text{,}\ 0 \leq t \leq \frac{\pi}{2}.$

69.

Compute the work done by force $\mathbf{\text{F}}(x,y,z) = 2x\mathbf{\text{i}} + 3y\mathbf{\text{j}} - z\mathbf{\text{k}}$ along path $\mathbf{\text{r}}(t) = t\mathbf{\text{i}} + t^{2}\mathbf{\text{j}} + t^{3}\mathbf{\text{k}},$ where $0 \leq t \leq 1.$

70\.

Evaluate ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = \frac{1}{x + y}\mathbf{\text{i}} + \frac{1}{x + y}\mathbf{\text{j}}$ and *C* is the segment of the unit circle going counterclockwise from $(1,0)$ to (0, 1).

71.

Force $\mathbf{\text{F}}(x,y,z) = zy\mathbf{\text{i}} + x\mathbf{\text{j}} + z^{2}x\mathbf{\text{k}}$ acts on a particle that travels from the origin to point (1, 2, 3). Calculate the work done if the particle travels:

1. along the path $(0,0,0)\rightarrow(1,0,0)\rightarrow(1,2,0)\rightarrow(1,2,3)$ along straight-line segments joining each pair of endpoints;

2. along the straight line joining the initial and final points.

3. Is the work the same along the two paths?

72\.

Find the work done by vector field $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + 3xy\mathbf{\text{j}} - (x + z)\mathbf{\text{k}}$ on a particle moving along a line segment that goes from (1, 4, 2) to (0, 5, 1).

73.

How much work is required to move an object in vector field $\mathbf{\text{F}}(x,y) = y\mathbf{\text{i}} + 3x\mathbf{\text{j}}$ along the upper part of ellipse $\frac{x^{2}}{4} + y^{2} = 1$ from (2, 0) to $(-2,0)?$

74\.

A vector field is given by $\mathbf{\text{F}}(x,y) = (2x + 3y)\mathbf{\text{i}} + (3x + 2y)\mathbf{\text{j}}.$ Evaluate the line integral of the field around a circle of unit radius traversed in a clockwise fashion.

75.

Evaluate the line integral of scalar function $xy$ along parabolic path $y = x^{2}$ connecting the origin to point (1, 1).

76\.

Find $\int_{C}^{}y^{2}dx + \left( {xy - x^{2}} \right)dy$ along *C*: $y = 3x$ from (0, 0) to (1, 3).

77.

Find $\int_{C}^{}y^{2}dx + \left( {xy - x^{2}} \right)dy$ along *C*: $y^{2} = 9x$ from (0, 0) to (1, 3).

For the following exercises, use a CAS to evaluate the given line integrals.

78\.

\[T\] Evaluate $\mathbf{\text{F}}(x,y,z) = x^{2}z\mathbf{\text{i}} + 6y\mathbf{\text{j}} + yz^{2}\mathbf{\text{k}},$ where *C* is represented by $\mathbf{\text{r}}(t) = t\mathbf{\text{i}} + t^{2}\mathbf{\text{j}} + \text{ln}\ t\mathbf{\text{k}}\text{,}\ 1 \leq t \leq 3.$

79.

\[T\] Evaluate line integral $\int_{\gamma}^{}{xe^{y}ds}$ where, $\gamma$ is the arc of curve $x = e^{y}$ from $(1,0)$ to $(e,1).$

80\.

\[T\] Evaluate the integral ${\int_{\gamma}^{}{xy^{2}ds}},$ where $\gamma$ is a triangle with vertices (0, 1, 2), (1, 0, 3), and $(0,-1,0).$

81.

\[T\] Evaluate line integral ${\int_{\gamma}^{}{\left( {y^{2} - xy} \right)ds}},$ where $\gamma$ is curve $y = \text{ln}\ x$ from (1, 0) toward $(e\text{,}\ 1).$

82\.

\[T\] Evaluate line integral ${\int_{\gamma}^{}{xy^{4}ds}},$ where $\gamma$ is the right half of circle $x^{2} + y^{2} = 16.$

83.

\[T\] Evaluate ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y,z) = x^{2}y\mathbf{\text{i}} + (x - z)\mathbf{\text{j}} + xyz\mathbf{\text{k}}$ and

*C*: $\mathbf{\text{r}}(t) = t\mathbf{\text{i}} + t^{2}\mathbf{\text{j}} + 2\mathbf{\text{k}}\text{,}\ 0 \leq t \leq 1.$

84\.

Evaluate ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = 2x\ \text{sin}(y)\mathbf{\text{i}} + \left( {x^{2}\text{cos}(y) - 3y^{2}} \right)\mathbf{\text{j}}$ and

*C* is any path from $(-1,0)$ to (5, 1).

85.

Find the line integral of $\mathbf{\text{F}}(x,y,z) = 12x^{2}\mathbf{\text{i}} - 5xy\mathbf{\text{j}} + xz\mathbf{\text{k}}$ over path *C* defined by $y = x^{2},$ $z = x^{3}$ from point (0, 0, 0) to point (2, 4, 8).

86\.

Find the line integral of ${\int_{C}^{}{\left( {1 + x^{2}y} \right)ds}},$ where *C* is ellipse $\mathbf{\text{r}}(t) = 2\ \text{cos}\ t\mathbf{\text{i}} + 3\ \text{sin}\ t\mathbf{\text{j}}$ from $0 \leq t \leq \pi.$

For the following exercises, find the flux.

87.

Compute the flux of $\mathbf{\text{F}} = x^{2}\mathbf{\text{i}} + y\mathbf{\text{j}}$ across a line segment from (0, 0) to (1, 2).

88\.

Let $\mathbf{\text{F}} = 5\mathbf{\text{i}}$ and let *C* be curve $y = 0,0 \leq x \leq 4.$ Find the flux across *C*.

89.

Let $\mathbf{\text{F}} = 5\mathbf{\text{j}}$ and let *C* be curve $y = 0,0 \leq x \leq 4.$ Find the flux across *C*.

90\.

Let $\mathbf{\text{F}} = \text{−}y\mathbf{\text{i}} + x\mathbf{\text{j}}$ and let *C*: $\mathbf{\text{r}}(t) = \text{cos}\ t\mathbf{\text{i}} + \text{sin}\ t\mathbf{\text{j}}$ $(0 \leq t \leq 2\pi).$ Calculate the flux across *C*.

91.

Let $\mathbf{\text{F}} = \left( {x^{2} + y^{3}} \right)\mathbf{\text{i}} + (2xy)\mathbf{\text{j}}.$ Calculate flux F orientated counterclockwise across curve *C*: $x^{2} + y^{2} = 9.$

92\.

Find the line integral of ${\int_{C}^{}{z^{2}dx + ydy + 2ydz}},$ where *C* consists of two parts: $C_{1}$ and $C_{2}.$ $C_{1}$ is the intersection of cylinder $x^{2} + y^{2} = 16$ and plane $z = 3$ from (0, 4, 3) to $(-4,0,3).$ $C_{2}$ is a line segment from $(-4,0,3)$ to (0, 1, 5).

93.

A spring is made of a thin wire twisted into the shape of a circular helix $x = 2\ \text{cos}\ t\text{,}\ y = 2\ \text{sin}\ t\text{,}\ z = t.$ Find the mass of two turns of the spring if the wire has constant mass density.

94\.

A thin wire is bent into the shape of a semicircle of radius *a*. If the linear mass density at point *P* is directly proportional to its distance from the line through the endpoints, find the mass of the wire.

95.

An object moves in force field $\mathbf{\text{F}}(x,y,z) = y^{2}\mathbf{\text{i}} + 2(x + 1)y\mathbf{\text{j}}$ counterclockwise from point (2, 0) along elliptical path $x^{2} + 4y^{2} = 4$ to $(-2,0),$ and back to point (2, 0) along the *x*-axis. How much work is done by the force field on the object?

96\.

Find the work done when an object moves in force field $\mathbf{\text{F}}(x,y,z) = 2x\mathbf{\text{i}} - (x + z)\mathbf{\text{j}} + (y - x)\mathbf{\text{k}}$ along the path given by $\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + \left( {t^{2} - t} \right)\mathbf{\text{j}} + 3\mathbf{\text{k}},$ $0 \leq t \leq 1.$

97.

If an inverse force field F is given by $\mathbf{\text{F}}(x,y,z) = \frac{\mathbf{k}}{\left\| \mathbf{\text{r}} \right\|^{3}}\mathbf{\text{r}},$ where *k* is a constant, find the work done by F as its point of application moves along the *x*-axis from $A(1,0,0)\ \text{to}\ B(2,0,0).$

98\.

David and Sandra plan to evaluate line integral $\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ along a path in the *xy*-plane from (0, 0) to (1, 1). The force field is $\mathbf{\text{F}}(x,y) = (x + 2y)\mathbf{\text{i}} + (\text{−}x + y^{2})\mathbf{\text{j}}.$ David chooses the path that runs along the *x*-axis from (0, 0) to (1, 0) and then runs along the vertical line $x = 1$ from (1, 0) to the final point (1, 1). Sandra chooses the direct path along the diagonal line $y = x$ from (0, 0) to (1, 1). Whose line integral is larger and by how much?

6.3 Conservative Vector Fields 6.3 保守向量场

• 6.3.1 Describe simple and closed curves; define connected and simply connected regions.
• 6.3.2 Explain how to find a potential function for a conservative vector field.
• 6.3.3 Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.
• 6.3.4 Explain how to test a vector field to determine whether it is conservative.

• 6.3.1 描述简单曲线与闭曲线；定义连通区域与单连通区域。
• 6.3.2 说明如何求保守向量场的势函数。
• 6.3.3 用线积分基本定理计算向量场中的线积分。
• 6.3.4 说明如何检验一个向量场是否保守。

In this section, we continue the study of conservative vector fields. We examine the Fundamental Theorem for Line Integrals, which is a useful generalization of the Fundamental Theorem of Calculus to line integrals of conservative vector fields. We also show how to test whether a given vector field is conservative, and determine how to build a potential function for a vector field known to be conservative.

Curves and Regions 曲线与区域

Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here.

We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve (Figure 6.25).

Curve *C* is a closed curve if there is a parameterization $\mathbf{\text{r}}(t),a \leq t \leq b$ of *C* such that the parameterization traverses the curve exactly once and $\mathbf{\text{r}}(a) = \mathbf{\text{r}}(b).$ Curve *C* is a simple curve if *C* does not cross itself. That is, *C* is simple if there exists a parameterization $\mathbf{\text{r}}(t),a \leq t \leq b$ of *C* such that r is one-to-one over $\left( {a,b} \right).$ It is possible for $\mathbf{\text{r}}(a) = \mathbf{\text{r}}(b),$ meaning that the simple curve is also closed.

Determining Whether a Curve Is Simple and Closed 判断曲线是否为简单闭曲线

Is the curve with parameterization $\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ t,\frac{\text{sin}(2t)}{2}} \right\rangle,0 \leq t \leq 2\pi$ a simple closed curve?

Solution 解

Note that $\mathbf{\text{r}}(0) = \left\langle {1,0} \right\rangle = \mathbf{\text{r}}\left( {2\pi} \right);$ therefore, the curve is closed. The curve is not simple, however. To see this, note that $\mathbf{\text{r}}\left( \frac{\pi}{2} \right) = \left\langle {0,0} \right\rangle = \mathbf{\text{r}}\left( \frac{3\pi}{2} \right),$ and therefore the curve crosses itself at the origin (Figure 6.26).

Is the curve given by parameterization $\mathbf{\text{r}}(t) = \left\langle {2\ \text{cos}\ t,3\ \text{sin}\mspace{2mu} t} \right\rangle,0 \leq t \leq 6\pi,$ a simple closed curve?

Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region’s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.

A region *D* is a connected region if, for any two points $P_{1}$ and $P_{2},$ there is a path from $P_{1}$ to $P_{2}$ with a trace contained entirely inside *D*. A region *D* is a simply connected region if *D* is connected for any simple closed curve *C* that lies inside *D*, and curve *C* can be shrunk continuously to a point while staying entirely inside *D*. In two dimensions, a region is simply connected if it is connected and has no holes.

All simply connected regions are connected, but not all connected regions are simply connected (Figure 6.27).

Is the region in the below image connected? Is the region simply connected?

Fundamental Theorem for Line Integrals 线积分基本定理

Now that we understand some basic curves and regions, let’s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function $f$ has an antiderivative *F*, then the integral of $f$ from *a* to *b* depends only on the values of *F* at *a* and at *b*—that is,

$$\int_{a}^{b}f(x)dx = \textit{F}(b) - \textit{F}(a).$$

If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.

Evaluating a Line Integral and the Antiderivatives of the Endpoints 计算线积分与端点处的原函数

Let $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {2x,4y} \right\rangle.$ Calculate $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}$ where *C* is the line segment from (0,0) to (2,2)(Figure 6.28).

Solution 解

We use Equation 6.9 to calculate $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}.}$ Curve *C* can be parameterized by $\mathbf{\text{r}}(t) = \left\langle {2t,2t} \right\rangle,0 \leq t \leq 1.$ Then, $\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) = \left\langle {4t,8t} \right\rangle$ and $\mathbf{\text{r}}^{\prime}(t) = \left\langle {2,2} \right\rangle,$ which implies that

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{0}^{1}\left\langle {4t,8t} \right\rangle} \cdot \left\langle {2,2} \right\rangle dt} \\ & {= {\int_{0}^{1}{(8t + 16t)}}dt = {\int_{0}^{1}{24t\mspace{2mu} dt}}} \\ & {= \left\lbrack {12t^{2}} \right\rbrack_{0}^{1} = 12.} \end{array}$$

Notice that $F = \text{∇}f,$ where $f\left( {x,y} \right) = x^{2} + 2y^{2}.$ If we think of the gradient as a derivative, then $f$ is an “antiderivative” of F. In the case of single-variable integrals, the integral of derivative $g^{\prime}(x)$ is $g(b) - g(a),$ where *a* is the start point of the interval of integration and *b* is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral F to be $f\left( P_{1} \right) - f\left( P_{0} \right),$ where $P_{1}$ is the endpoint of the curve of integration and $P_{0}$ is the start point. Notice that this is the case for this example:

$$\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} = {\int_{C}{\text{∇}f.d\mathbf{\text{r}} = 12}}}$$

and

$$f\left( {2,2} \right) - f\left( {0,0} \right) = 4 + 8 - 0 = 12.$$

In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.

The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the Fundamental Theorem for Line Integrals.

The Fundamental Theorem for Line Integrals 线积分基本定理

Let *C* be a piecewise smooth curve with parameterization $\mathbf{\text{r}}(t),a \leq t \leq b.$ Let $f$ be a function of two or three variables with first-order partial derivatives that exist and are continuous on *C*. Then,

$$\int_{C}{\text{∇}f.d\mathbf{\text{r}} = f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(a)} \right).}$$ (6.12)

Proof 证明

By Equation 6.9,

$$\int_{C}{\text{∇}f.d\mathbf{\text{r}} = {\int_{a}^{b}{\text{∇}f\left( {\mathbf{\text{r}}(t)} \right).\mathbf{\text{r}}^{\prime}(t)dt.}}}$$

By the chain rule,

$$\frac{d}{dt}\left( {f\left( {\mathbf{\text{r}}(t)} \right) = \text{∇}f\left( {\mathbf{\text{r}}(t)} \right).\mathbf{\text{r}}^{\prime}(t).} \right.$$

Therefore, by the Fundamental Theorem of Calculus,

$$\begin{array}{cl} {{\int_{C}\nabla}f.d\mathbf{\text{r}}} & {= {\int_{a}^{b}{\text{∇}f\left( {\mathbf{\text{r}}(t)} \right).\mathbf{\text{r}}^{\prime}(t)dt}}} \\ & {= {\int_{a}^{b}{\frac{d}{dt}\left( {f\left( {\mathbf{\text{r}}(t)} \right)} \right.dt}}} \\ & {= \left\lbrack {f\left( {\mathbf{\text{r}}(t)} \right)} \right\rbrack_{t = a}^{t = b}} \\ & {= f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(a)} \right).} \end{array}$$

□

We know that if F is a conservative vector field, there are potential functions $f$ such that $\text{∇}f = \mathbf{\text{F}}.$ Therefore ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}} = f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(a)} \right).$ In other words, just as with the Fundamental Theorem of Calculus, computing the line integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where F is conservative, is a two-step process: (1) find a potential function (“antiderivative”) $f$ for F and (2) compute the value of $f$ at the endpoints of *C* and calculate their difference $f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(a)} \right).$ Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. *A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.*

Applying the Fundamental Theorem 应用基本定理

Calculate integral $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}$ where $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {2x\ \text{ln}\mspace{2mu} y,\frac{x^{2}}{y} + z^{2},2yz} \right\rangle$ and *C* is a curve with parameterization $\mathbf{\text{r}}(t) = \left\langle {t^{2},t,t} \right\rangle,1 \leq t \leq e$

1. without using the Fundamental Theorem of Line Integrals and

2. using the Fundamental Theorem of Line Integrals.

Solution 解

1. First, let’s calculate the integral without the Fundamental Theorem for Line Integrals and instead use Equation 6.9:

$$\begin{matrix} {\int_{C}\textbf{F} \cdot d\textbf{r}} & {= \int_{1}^{e}\textbf{F}\left( \textbf{r}(t) \right).\textbf{r}^{'}(t)dt} \\ & {= \int_{1}^{e}\left\langle 2t^{2}\text{ln}\mspace{2mu} t,\frac{t^{4}}{t} + t^{2},2t^{2} \right\rangle.\left\langle 2t,1,1 \right\rangle dt} \\ & {= \int_{1}^{e}\left( 4t^{3}\text{ln}\mspace{2mu} t + t^{3} + 3t^{2} \right)dt} \\ & {= \int_{1}^{e}4t^{3}\text{ln}\mspace{2mu} tdt + \int_{1}^{e}\left( t^{3} + 3t^{2} \right)dt} \\ & {= \int_{1}^{e}4t^{3}\text{ln}\mspace{2mu} tdt + \left\lbrack \frac{t^{4}}{4} + t^{3} \right\rbrack_{1}^{e}} \\ & {= 2\int_{1}^{e}t^{3}\text{ln}\mspace{2mu} tdt + \frac{e^{4}}{4} + e^{3} - \frac{5}{4}.} \end{matrix}$$

Integral $\int_{1}^{e}{t^{3}\text{ln}\mspace{2mu} tdt}$ requires integration by parts. Let $u = \text{ln}\mspace{2mu} t$ and $dv = t^{3}.$ Then $u = \text{ln}\mspace{2mu} t,dv = t^{3}$

and

$$du = \frac{1}{t}dt,v = \frac{t^{4}}{4}.$$

Therefore,

$$\begin{array}{cl} {\int_{1}^{e}{t^{3}\text{ln}\mspace{2mu} tdt}} & {= \left\lbrack {\frac{t^{4}}{4}\text{ln}\mspace{2mu} t} \right\rbrack_{1}^{e} - \frac{1}{4}{\int_{1}^{e}{t^{3}dt}}} \\ & {= \frac{e^{4}}{4} - \frac{1}{4}\left( {\frac{e^{4}}{4} - \frac{1}{4}} \right).} \end{array}$$

Thus,

$$\begin{array}{cl} {{\int_{C}\mathbf{\text{F}}} \cdot dr} & {= 4{\int_{1}^{e}{t^{3}\text{ln}\mspace{2mu} tdt + \frac{e^{4}}{4} + e^{3} - \frac{5}{4}}}} \\ & {= 4\left( {\frac{e^{4}}{4} - \frac{1}{4}\left( {\frac{e^{4}}{4} - \frac{1}{4}} \right)} \right) + \frac{e^{4}}{4} + e^{3} - \frac{5}{4}} \\ & {= e^{4} - \frac{e^{4}}{4} + \frac{1}{4} + \frac{e^{4}}{4} + e^{3} - \frac{5}{4}} \\ & {= e^{4} + e^{3} - 1.} \end{array}$$

2. Given that $f\left( {x,y,z} \right) = x^{2}\text{ln}\mspace{2mu} y + yz^{2}$ is a potential function for F, let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that

$$\begin{array}{cl} {{\int_{C}\mathbf{\text{F}}} \cdot d\mathbf{\text{r}}} & {= {\int_{C}{\text{∇}f.d\mathbf{\text{r}}}}} \\ & {= f\left( {\mathbf{\text{r}}(e)} \right) - f\left( {\mathbf{\text{r}}(1)} \right)} \\ & {= f\left( {e^{2},e,e} \right) - f\left( {1,1,1} \right)} \\ & {= e^{4} + e^{3} - 1.} \end{array}$$

This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.

Example 6.29 illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.

Given that $f\left( {x,y} \right) = \left( {x - 1} \right)^{2}y + \left( {y + 1} \right)^{2}x$ is a potential function for $\mathbf{\text{F}} = \left\langle {2xy - 2y + \left( {y + 1} \right)^{2},\left( {x - 1} \right)^{2} + 2yx + 2x} \right\rangle,$ calculate integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where *C* is the lower half of the unit circle oriented counterclockwise.

The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if F is conservative and *C* is a closed curve, then the circulation of F along *C* is zero—that is, ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = 0.$ To see why this is true, let $f$ be a potential function for F. Since *C* is a closed curve, the terminal point r(b) of *C* is the same as the initial point r(a) of *C*—that is, $\mathbf{\text{r}}(a) = \mathbf{\text{r}}(b).$ Therefore, by the Fundamental Theorem for Line Integrals,

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}}} \\ & {= f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(a)} \right)} \\ & {= f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(b)} \right)} \\ & {= 0.} \end{array}$$

Recall that the reason a conservative vector field F is called “conservative” is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field F in integral $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ as a gravitational field, then the equation ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = 0$ follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.

The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path—meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.

Let F be a vector field with domain *D*. The vector field F is independent of path (or path independent) if ${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}$ for any paths $C_{1}$ and $C_{2}$ in *D* with the same initial and terminal points.

The second consequence is stated formally in the following theorem.

Path Independence of Conservative Fields 保守场的路径无关性

If F is a conservative vector field, then F is independent of path.

Proof 证明

Let *D* denote the domain of F and let $C_{1}$ and $C_{2}$ be two paths in *D* with the same initial and terminal points (Figure 6.29). Call the initial point $P_{1}$ and the terminal point $P_{2}.$ Since F is conservative, there is a potential function $f$ for F. By the Fundamental Theorem for Line Integrals,

$${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = f(P_{2}) - f(P_{1}) = {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$$

Therefore, ${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}$ and F is independent of path.

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To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.

We have shown that if F is conservative, then F is independent of path. It turns out that if the domain of F is open and connected, then the converse is also true. That is, if F is independent of path and the domain of F is open and connected, then F is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.

The Path Independence Test for Conservative Fields 保守场的路径无关性检验

If F is a continuous vector field that is independent of path and the domain *D* of F is open and connected, then F is conservative.

Proof 证明

We prove the theorem for vector fields in $\mathbb{R}^{2}.$ The proof for vector fields in $\mathbb{R}^{3}$ is similar. To show that $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is conservative, we must find a potential function $f$ for F. To that end, let *X* be a fixed point in *D*. For any point $\left( {x,y} \right)$ in *D*, let *C* be a path from *X* to $\left( {x,y} \right).$ Define $f$$\left( {x,y} \right)$ by $f(x,y) = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}.}}$ (Note that this definition of $f$ makes sense only because F is independent of path. If F was not independent of path, then it might be possible to find another path $C^{\prime}$ from *X* to $\left( {x,y} \right)$ such that ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} \neq {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ and in such a case $f$$\left( {x,y} \right)$ would not be a function.) We want to show that $f$ has the property $\text{∇}f = \mathbf{\text{F}}.$

Since domain *D* is open, it is possible to find a disk centered at $\left( {x,y} \right)$ such that the disk is contained entirely inside *D*. Let $\left( {a,y} \right)$ with $a < x$ be a point in that disk. Let *C* be a path from *X* to $\left( {x,y} \right)$ that consists of two pieces: $C_{1}$ and $C_{2}.$ The first piece, $C_{1},$ is any path from *X* to $\left( {a,y} \right)$ that stays inside *D*; $C_{2}$ is the horizontal line segment from $\left( {a,y} \right)$ to $\left( {x,y} \right)$ (Figure 6.30). Then

$$f(x,y) = {\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}.}}$$

The first integral does not depend on *x*, so

$$f_{x} = \frac{\partial}{\partial x}{\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}.}}$$

If we parameterize $C_{2}$ by $\mathbf{\text{r}}(t) = \left\langle {t,y} \right\rangle,a \leq t \leq x,$ then

$$\begin{matrix} f_{x} & {= \frac{\partial}{\partial x}\int_{C_{2}}\textbf{F} \cdot d\textbf{r}} \\ & {= \frac{\partial}{\partial x}\int_{a}^{x}\textbf{F}\left( \textbf{r}(t) \right).\textbf{r}^{'}(t)dt} \\ & {= \frac{\partial}{\partial x}\int_{a}^{x}\textbf{F}\left( \textbf{r}(t) \right).\frac{d}{dt}\left( \left\langle t,y \right\rangle \right)dt} \\ & {= \frac{\partial}{\partial x}\int_{a}^{x}\textbf{F}\left( \textbf{r}(t) \right).\left\langle 1,0 \right\rangle dt} \\ & {= \frac{\partial}{\partial x}\int_{a}^{x}P(t,y)dt.} \end{matrix}$$

By the Fundamental Theorem of Calculus (part 1),

$$f_{x} = \frac{\partial}{\partial x}{\int_{a}^{x}P}\left( {t,y} \right)dt = P\left( {x,y} \right).$$

A similar argument using a vertical line segment rather than a horizontal line segment shows that $f_{y} = Q\left( {x,y} \right).$

Therefore $\text{∇}f = \mathbf{\text{F}}$ and F is conservative.

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We have spent a lot of time discussing and proving Path Independence of Conservative Fields and The Path Independence Test for Conservative Fields, but we can summarize them simply: a vector field F on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.

Showing That a Vector Field Is Not Conservative 证明一个向量场不是保守场

Use path independence to show that vector field $\mathbf{\text{F}}(x,y) = \left\langle {x^{2}y,y + 5} \right\rangle$ is not conservative.

Solution 解

We can indicate that F is not conservative by showing that F is not path independent. We do so by giving two different paths, $C_{1}$ and $C_{2},$ that both start at $(0,0)$ and end at $(1,1),$ and yet ${\int_{C_{1}}\mathbf{\text{F}}} \cdot dr \neq {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}.}}$

Let $C_{1}$ be the curve with parameterization $\textbf{r}_{1}(t) = \left\langle {t,t} \right\rangle,0 \leq t \leq 1$ and let $C_{2}$ be the curve with parameterization $\textbf{r}_{2}(t) = \left\langle {t,t^{2}} \right\rangle,0 \leq t \leq 1$ (Figure 6.31). Then

$$\begin{array}{cl} {\int_{C_{1}}{\mathbf{\text{F}} \cdot d\textbf{r}}} & {= {\int_{0}^{1}{\mathbf{\text{F}}\left( {\textbf{r}_{1}(t)} \right) \cdot \textbf{r}_{1}{}^{\prime}(t)dt}}} \\ & {= {\int_{0}^{1}{\left\langle {t^{3},t + 5} \right\rangle \cdot \left\langle {1,1} \right\rangle dt}} = {\int_{0}^{1}{\left( {t^{3} + t + 5} \right)dt}}} \\ & {= \left\lbrack {\frac{t^{4}}{4} + \frac{t^{2}}{2} + 5t} \right\rbrack_{0}^{1} = \frac{23}{4}} \end{array}$$

and

$$\begin{array}{cl} {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\textbf{r}}} & {= {\int_{0}^{1}{\mathbf{\text{F}}\left( {\textbf{r}_{2}(t)} \right) \cdot \textbf{r}_{2}\prime(t)dt}}} \\ & {= {\int_{0}^{1}{\left\langle {t^{4},t^{2} + 5} \right\rangle \cdot \left\langle {1,2t} \right\rangle dt}} = {\int_{0}^{1}{\left( {t^{4} + 2t^{3} + 10t} \right)dt}}} \\ & {= \left\lbrack {\frac{t^{5}}{5} + \frac{t^{4}}{2} + 5t^{2}} \right\rbrack_{0}^{1} = \frac{57}{10}.} \end{array}$$

Since ${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} \neq {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}}$ the value of a line integral of F depends on the path between two given points. Therefore, F is not independent of path, and F is not conservative.

Show that $\mathbf{\text{F}}(x,y) = \left\langle {xy,x^{2}y^{2}} \right\rangle$ is not path independent by considering the line segment from $(0,0)$ to $(2,2)$ and the piece of the graph of $y = \frac{x^{2}}{2}$ that goes from $(0,0)$ to $(2,2).$

Conservative Vector Fields and Potential Functions 保守向量场与势函数

As we have learned, the Fundamental Theorem for Line Integrals says that if F is conservative, then calculating $\int_{C}{\mathbf{\text{F}} \cdot dr}$ has two steps: first, find a potential function $f$ for F and, second, calculate $f(P_{1}) - f(P_{0}),$ where $P_{1}$ is the endpoint of *C* and $P_{0}$ is the starting point. To use this theorem for a conservative field F, we must be able to find a potential function $f$ for F. Therefore, we must answer the following question: Given a conservative vector field F, how do we find a function $f$ such that $\text{∇}f = \mathbf{\text{F}}?$ Before giving a general method for finding a potential function, let’s motivate the method with an example.

Finding a Potential Function 求势函数

Find a potential function for $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {2xy^{3},3x^{2}y^{2} + \text{cos}(y)} \right\rangle,$ thereby showing that F is conservative.

Solution 解

Suppose that $f(x,y)$ is a potential function for F. Then, $\text{∇}f = \mathbf{\text{F}},$ and therefore

$$f_{x} = 2xy^{3}\ \text{and}\ f_{y} = 3x^{2}y^{2} + \text{cos}\mspace{2mu} y.$$

Integrating the equation $f_{x} = 2xy^{3}$ with respect to *x* yields the equation

$$f(x,y) = x^{2}y^{3} + h(y).$$

Notice that since we are integrating a two-variable function with respect to *x*, we must add a constant of integration that is a constant with respect to *x*, but may still be a function of *y*. The equation $f(x,y) = x^{2}y^{3} + h(y)$ can be confirmed by taking the partial derivative with respect to *x*:

$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left( {x^{2}y^{3}} \right) + \frac{\partial}{\partial x}\left( {h(y)} \right) = 2xy^{3} + 0 = 2xy^{3}.$$

Since $f$ is a potential function for F,

$$f_{y} = 3x^{2}y^{2} + \text{cos}(y),$$

and therefore

$$3x^{2}y^{2} + h^{\prime}(y) = 3x^{2}y^{2} + \text{cos}(y).$$

This implies that $h\prime(y) = \text{cos}\mspace{2mu} y,$ so $h(y) = \text{sin}\mspace{2mu} y + C.$ Therefore, *any* function of the form $f\left( {x,y} \right) = x^{2}y^{3} + \text{sin}(y) + C$ is a potential function. Taking, in particular, $C = 0$ gives the potential function $f\left( {x,y} \right) = x^{2}y^{3} + \text{sin}(y).$

To verify that $f$ is a potential function, note that $\text{∇}f = \left\langle {2xy^{3},3x^{2}y^{2} + \text{cos}\mspace{2mu} y} \right\rangle = \mathbf{\text{F}}.$

Find a potential function for $\mathbf{\text{F}}(x,y) = \left\langle {e^{x}y^{3} + y,3e^{x}y^{2} + x} \right\rangle.$

The logic of the previous example extends to finding the potential function for any conservative vector field in $\mathbb{R}^{2}.$ Thus, we have the following problem-solving strategy for finding potential functions:

Problem-Solving Strategy: Finding a Potential Function for a Conservative Vector Field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {P\left( {x,y} \right),Q\left( {x,y} \right)} \right\rangle$ 问题解决策略：求保守向量场 $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {P\left( {x,y} \right),Q\left( {x,y} \right)} \right\rangle$ 的势函数

1. Integrate *P* with respect to *x*. This results in a function of the form $g\left( {x,y} \right) + h(y),$ where $h(y)$ is unknown.

2. Take the partial derivative of $g\left( {x,y} \right) + h(y)$ with respect to *y*, which results in the function $g_{y}\left( {x,y} \right) + h^{\prime}(y).$

3. Use the equation $g_{y}\left( {x,y} \right) + h^{\prime}(y) = Q\left( {x,y} \right)$ to find $h^{\prime}(y).$

4. Integrate $h^{\prime}(y)$ to find $h(y).$

5. Any function of the form $f\left( {x,y} \right) = g\left( {x,y} \right) + h(y) + C,$ where *C* is a constant, is a potential function for F.

We can adapt this strategy to find potential functions for vector fields in $\mathbb{R}^{3},$ as shown in the next example.

Finding a Potential Function in $\mathbb{R}^{3}$ 在 $\mathbb{R}^{3}$ 中求势函数

Find a potential function for $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {2xy,x^{2} + 2yz^{3},3y^{2}z^{2} + 2z} \right\rangle,$ thereby showing that $\mathbf{\text{F}}$ is conservative.

Solution 解

Suppose that $f$ is a potential function. Then, $\text{∇}f = \mathbf{\text{F}}$ and therefore $f_{x} = 2xy.$ Integrating this equation with respect to *x* yields the equation $f\left( {x,y,z} \right) = x^{2}y + g\left( {y,z} \right)$ for some function *g*. Notice that, in this case, the constant of integration with respect to *x* is a function of *y* and *z*.

Since $f$ is a potential function,

$$x^{2} + 2yz^{3} = f_{y} = x^{2} + g_{y}.$$

Therefore,

$$g_{y} = 2yz^{3}.$$

Integrating this function with respect to *y* yields

$$g\left( {y,z} \right) = y^{2}z^{3} + h(z)$$

for some function $h(z)$ of *z* alone. (Notice that, because we know that *g* is a function of only *y* and *z*, we do not need to write $g\left( {y,z} \right) = y^{2}z^{3} + h\left( {x,z} \right).)$ Therefore,

$$f\left( {x,y,z} \right) = x^{2}y + g\left( {y,z} \right) = x^{2}y + y^{2}z^{3} + h(z).$$

To find $f$, we now must only find *h*. Since $f$ is a potential function,

$$3y^{2}z^{2} + 2z = g_{z} = 3y^{2}z^{2} + h^{\prime}(z).$$

This implies that $h^{\prime}(z) = 2z,$ so $h(z) = z^{2} + C.$ Letting $C = 0$ gives the potential function

$$f\left( {x,y,z} \right) = x^{2}y + y^{2}z^{3} + z^{2}.$$

To verify that $f$ is a potential function, note that $\text{∇}f = \left\langle {2xy,x^{2} + 2yz^{3},3y^{2}z^{2} + 2z} \right\rangle = \mathbf{\text{F}}.$

Find a potential function for $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {12x^{2},\text{cos}\mspace{2mu} y\ \text{cos}\ z,1 - \text{sin}\mspace{2mu} y\ \text{sin}\mspace{2mu} z} \right\rangle.$

We can apply the process of finding a potential function to a gravitational force. Recall that, if an object has unit mass and is located at the origin, then the gravitational force in $\mathbb{R}^{2}$ that the object exerts on another object of unit mass at the point $\left( {x,y} \right)$ is given by vector field

$$\mathbf{\text{F}}\left( {x,y} \right) = \text{−}G\left\langle {\frac{x}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}},\frac{y}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}} \right\rangle,$$

where *G* is the universal gravitational constant. In the next example, we build a potential function for F, thus confirming what we already know: that gravity is conservative.

Finding a Potential Function 求势函数

Find a potential function $f$ for $\mathbf{\text{F}}\left( {x,y} \right) = \text{−}G\left\langle {\frac{x}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}},\frac{y}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}} \right\rangle.$

Solution 解

Suppose that $f$ is a potential function. Then, $\text{∇}f = \mathbf{\text{F}}$ and therefore

$$f_{x} = \frac{\text{−}Gx}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}.$$

To integrate this function with respect to *x,* we can use *u*-substitution. If $u = x^{2} + y^{2},$ then $\frac{du}{2} = xdx,$ so

$$\begin{array}{cl} {{\int\frac{\text{−}Gx}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}}dx} & {= {\int\frac{\text{−}G}{2u^{3\text{/}2}}}du} \\ & {= \frac{G}{\sqrt{u}} + h(y)} \\ & {= \frac{G}{\sqrt{x^{2} + y^{2}}} + h(y)} \end{array}$$

for some function $h(y).$ Therefore,

$$f(x,y) = \frac{G}{\sqrt{x^{2} + y^{2}}} + h(y).$$

Since $f$ is a potential function for F,

$$f_{y} = \frac{\text{−}Gy}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}.$$

Since $f(x,y) = \frac{G}{\sqrt{x^{2} + y^{2}}} + h(y),$ $f_{y}$ also equals $\frac{\text{−}Gy}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}} + h^{\prime}(y).$

Therefore,

$$\frac{\text{−}Gy}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}} + h\prime(y) = \frac{\text{−}Gy}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}},$$

which implies that $h^{\prime}(y) = 0.$ Thus, we can take $h(y)$ to be any constant; in particular, we can let $h(y) = 0.$ The function

$$f\left( {x,y} \right) = \frac{G}{\sqrt{x^{2} + y^{2}}}$$

is a potential function for the gravitational field F. To confirm that $f$ is a potential function, note that

$$\begin{array}{cl} {\text{∇}f} & {= \left\langle {- \frac{1}{2}\ \frac{G}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}\left( {2x} \right), - \frac{1}{2}\ \frac{G}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}\left( {2y} \right)} \right\rangle} \\ & {= \left\langle {\frac{\text{−}Gx}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}},\frac{\text{−}Gy}{\left( {x^{2} + y^{2}} \right)^{3\text{/}2}}} \right\rangle} \\ & {= \mathbf{\text{F}}.} \end{array}$$

Find a potential function $f$ for the three-dimensional gravitational force $\mathbf{\text{F}}(x,y,z) = \left\langle {\frac{\text{−}Gx}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}},\frac{\text{−}Gy}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}},\frac{\text{−}Gz}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}}} \right\rangle.$

Testing a Vector Field 检验向量场

Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if F is conservative, then F has the cross-partial property (see The Cross-Partial Property of Conservative Vector Fields). That is, if $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ is conservative, then $P_{y} = Q_{x},P_{z} = R_{x},$ and $Q_{z} = R_{y}.$ So, if F has the cross-partial property, then is F conservative? If the domain of F is open and simply connected, then the answer is yes.

The Cross-Partial Test for Conservative Fields 保守场的交叉偏导检验

If $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ is a vector field on an open, simply connected region *D* and $P_{y} = Q_{x},P_{z} = R_{x},$ and $Q_{z} = R_{y}$ throughout *D*, then F is conservative.

Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.

Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:

Cross-Partial Property of Conservative Fields 保守场的交叉偏导性质

Let $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ be a vector field on an open, simply connected region *D.* Then $P_{y} = Q_{x},P_{z} = R_{x},$ and $Q_{z} = R_{y}$ throughout *D* if and only if F is conservative.

The version of this theorem in $\mathbb{R}^{2}$ is also true. If $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a vector field on an open, simply connected domain in $\mathbb{R}^{2},$ then F is conservative if and only if $P_{y} = Q_{x}.$

Determining Whether a Vector Field Is Conservative 判断一个向量场是否保守

Determine whether vector field $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {xy^{2}z,x^{2}yz,z^{2}} \right\rangle$ is conservative.

Solution 解

Note that the domain of F is all of $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ is simply connected. Therefore, we can use Cross-Partial Property of Conservative Fields to determine whether F is conservative. Let

$$P\left( {x,y,z} \right) = xy^{2}z,Q\left( {x,y,z} \right) = x^{2}yz,\ \text{and}\ R\left( {x,y,z} \right) = z^{2}.$$

Since $Q_{z} = x^{2}y$ and $R_{y} = 0,$ the vector field is not conservative.

Determining Whether a Vector Field Is Conservative 判断一个向量场是否保守

Determine vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {x\ \text{ln}(y),\frac{x^{2}}{2y}} \right\rangle$ is conservative.

Solution 解

Note that the domain of F is the part of $\mathbb{R}^{2}$ in which $y > 0.$ Thus, the domain of F is part of a plane above the *x*-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use Cross-Partial Property of Conservative Fields to determine whether F is conservative. Let

$$P\left( {x,y} \right) = x\ \text{ln}(y)\ \text{and}\ Q\left( {x,y} \right) = \frac{x^{2}}{2y}.$$

Then $P_{y} = \frac{x}{y} = Q_{x}$ and thus F is conservative.

Determine whether $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {\text{sin}\mspace{2mu} x\ \text{cos}\mspace{2mu} y,\text{cos}\mspace{2mu} x\ \text{sin}\mspace{2mu} y} \right\rangle$ is conservative.

When using Cross-Partial Property of Conservative Fields, it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of Cross-Partial Property of Conservative Fields, the theorem can be applied only if the domain of the vector field is simply connected.

To see what can go wrong when misapplying the theorem, consider the vector field:

$$\mathbf{\text{F}}\left( {x,y} \right) = \frac{y}{x^{2} + y^{2}}\mathbf{\text{i}} + \frac{\text{−}x}{x^{2} + y^{2}}\mathbf{\text{j}}.$$

This vector field satisfies the cross-partial property, since

$$\frac{\partial}{\partial y}\left( \frac{y}{x^{2} + y^{2}} \right) = \frac{\left( {x^{2} + y^{2}} \right) - y\left( {2y} \right)}{\left( {x^{2} + y^{2}} \right)^{2}} = \frac{x^{2} - y^{2}}{\left( {x^{2} + y^{2}} \right)^{2}}$$

and

$$\frac{\partial}{\partial x}\left( \frac{\text{−}x}{x^{2} + y^{2}} \right) = \frac{\text{−}\left( {x^{2} + y^{2}} \right) + x\left( {2x} \right)}{\left( {x^{2} + y^{2}} \right)^{2}} = \frac{x^{2} - y^{2}}{\left( {x^{2} + y^{2}} \right)^{2}}.$$

Since F satisfies the cross-partial property, we might be tempted to conclude that F is conservative. However, F is not conservative. To see this, let

$$\mathbf{\text{r}}(t) = \left\langle {\text{cos}\mspace{2mu} t,\text{sin}\mspace{2mu} t} \right\rangle,0 \leq t \leq \pi$$

be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this $C_{1})$ and let

$$\textbf{s}(t) = \left\langle {\text{cos}\mspace{2mu} t,\text{−}\text{sin}\mspace{2mu} t} \right\rangle,0 \leq t \leq \pi$$

be a parameterization of the lower half of a unit circle oriented clockwise (denote this $C_{2}).$ Notice that $C_{1}$ and $C_{2}$ have the same starting point and endpoint. Since $\text{sin}^{2}t + \text{cos}^{2}t = 1,$

$$\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right).\mathbf{\text{r}}^{\prime}(t) = \left\langle {\text{sin}(t),\text{−}\text{cos}(t)} \right\rangle.\left\langle {\text{−}\text{sin}(t),\text{cos}(t)} \right\rangle = -1$$

and

$$\begin{array}{cl}{\mathbf{\text{F}}\left( {s(t)} \right) \cdot \mathbf{\textbf{s}}\prime(t)} & {= \left\langle {\text{−}\text{sin}\mspace{2mu} t,\text{−}\text{cos}\mspace{2mu} t} \right\rangle \cdot \left\langle {\text{−}\text{sin}\mspace{2mu} t,\text{−}\text{cos}\mspace{2mu} t} \right\rangle} \\ & {= \text{sin}^{2}t + \text{cos}^{2}t} \\ & {= 1.}\end{array}$$

Therefore,

$${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{0}^{\pi}{-1dt}} = \text{−}\pi\ \text{and}\ {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\textbf{r} =}}{\int_{0}^{\pi}{1dt}} = \pi.$$

Thus, $C_{1}$ and $C_{2}$ have the same starting point and endpoint, but ${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} \neq {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$ Therefore, F is not independent of path and F is not conservative.

To summarize: F satisfies the cross-partial property and yet F is not conservative. What went wrong? Does this contradict Cross-Partial Property of Conservative Fields? The issue is that the domain of F is all of $\mathbb{R}^{2}$ except for the origin. In other words, the domain of F has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, Cross-Partial Property of Conservative Fields does not apply to F.

We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ then our first question should be: Is F conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can’t help us and we have to use other methods, such as using Equation 6.9.

Using the Fundamental Theorem for Line Integrals 使用线积分基本定理

Calculate line integral ${\int_{C}{\mathbf{\text{F}} \cdot d\textbf{r}}},$ where $\mathbf{\text{F}}(x,y,z) = \left\langle {2xe^{y}z + e^{x}z,x^{2}e^{y}z,x^{2}e^{y} + e^{x}} \right\rangle$ and *C* is any smooth curve that goes from the origin to $(1,1,1).$

Solution 解

Before trying to compute the integral, we need to determine whether F is conservative and whether the domain of F is simply connected. The domain of F is all of $\mathbb{R}^{3},$ which is connected and has no holes. Therefore, the domain of F is simply connected. Let

$$P(x,y,z) = 2xe^{y}z + e^{x}z,Q(x,y,z) = x^{2}e^{y}z,\ \text{and}\ R(x,y,z) = x^{2}e^{y} + e^{x}$$

so that $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle.$ Since the domain of F is simply connected, we can check the cross partials to determine whether F is conservative. Note that

$$\begin{array}{rll}P_{y} & = & {2xe^{y}z = Q_{x}} \\P_{z} & = & {2xe^{y} + e^{x} = R_{x}} \\Q_{z} & = & {x^{2}e^{y} = R_{y}.}\end{array}$$

Therefore, F is conservative.

To evaluate $\int_{C}{\mathbf{\text{F}} \cdot d\textbf{r}}$ using the Fundamental Theorem for Line Integrals, we need to find a potential function $f$ for F. Let $f$ be a potential function for F. Then, $\text{∇}f = \mathbf{\text{F}},$ and therefore $f_{x} = 2xe^{y}z + e^{x}z.$ Integrating this equation with respect to *x* gives $f\left( {x,y,z} \right) = x^{2}e^{y}z + e^{x}z + h\left( {y,z} \right)$ for some function *h*. Differentiating this equation with respect to *y* gives $x^{2}e^{y}z + h_{y} = Q = x^{2}e^{y}z,$ which implies that $h_{y} = 0.$ Therefore, *h* is a function of *z* only, and $f(x,y,z) = x^{2}e^{y}z + e^{x}z + h(z).$ To find *h*, note that $f_{z} = x^{2}e^{y} + e^{x} + h\prime(z) = R = x^{2}e^{y} + e^{x}.$ Therefore, $h\prime(z) = 0$ and we can take $h(z) = 0.$ A potential function for F is $f(x,y,z) = x^{2}e^{y}z + e^{x}z.$

Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,

$$\begin{array}{cl}{\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}}} \\ & {= f(1,1,1) - f(0,0,0)} \\ & {= 2e.}\end{array}$$

Analysis 分析

Notice that if we hadn’t recognized that F is conservative, we would have had to parameterize *C* and use Equation 6.9. Since curve *C* is unknown, using the Fundamental Theorem for Line Integrals is much simpler.

Calculate integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = \left\langle {\text{sin}\mspace{2mu} x\ \text{sin}\mspace{2mu} y,5 - \text{cos}\mspace{2mu} x\ \text{cos}\mspace{2mu} y} \right\rangle$ and *C* is a semicircle with starting point $(0,\pi)$ and endpoint $(0,\text{−}\pi).$

Work Done on a Particle 粒子上所做的功

Let $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {2xy^{2},2x^{2}y} \right\rangle$ be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the *x*-axis or the *y*-axis. Furthermore, the particle’s motion can be modeled with a smooth parameterization. Show that F does positive work on the particle.

Solution 解

We show that F does positive work on the particle by showing that F is conservative and then by using the Fundamental Theorem for Line Integrals.

To show that F is conservative, suppose $f\left( {x,y} \right)$ were a potential function for F. Then, $\text{∇}f = \mathbf{\text{F}} = \left\langle {2xy^{2},2x^{2}y} \right\rangle$ and therefore $f_{x} = 2xy^{2}$ and $f_{y} = 2x^{2}y.$ Equation $f_{x} = 2xy^{2}$ implies that $f\left( {x,y} \right) = x^{2}y^{2} + h(y).$ Deriving both sides with respect to *y* yields $f_{y} = 2x^{2}y + h^{\prime}(y).$ Therefore, $h^{\prime}(y) = 0$ and we can take $h(y) = 0.$

If $f(x,y) = x^{2}y^{2},$ then note that $\text{∇}f = \left\langle {2xy^{2},2x^{2}y} \right\rangle = \mathbf{\text{F}},$ and therefore $f$ is a potential function for F.

Let $(a,b)$ be the point at which the particle stops is motion, and let *C* denote the curve that models the particle’s motion. The work done by F on the particle is ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$ By the Fundamental Theorem for Line Integrals,

$$\begin{array}{cl}{\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}}} \\ & {= f(a,b) - f(0,0)} \\ & {= a^{2}b^{2}.}\end{array}$$

Since $a \neq 0$ and $b \neq 0,$ by assumption, $a^{2}b^{2} > 0.$ Therefore, ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} > 0,$ and F does positive work on the particle.

Analysis 分析

Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use Equation 6.9. Since the path of motion *C* can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.

Let $\mathbf{\text{F}}(x,y) = \left\langle {4x^{3}y^{4},4x^{4}y^{3}} \right\rangle,$ and suppose that a particle moves from point $(4,4)$ to $(1,1)$ along any smooth curve. Is the work done by F on the particle positive, negative, or zero?

Section 6.3 Exercises 6.3 节习题

99.

*True* or *False?* If vector field F is conservative on the open and connected region *D*, then line integrals of F are path independent on *D*, regardless of the shape of *D*.

100\.

*True* or *False?* Function $\mathbf{\text{r}}(t) = \mathbf{\text{a}} + t\left( {\mathbf{\text{b}} - \mathbf{\text{a}}} \right),$ where $0 \leq t \leq 1,$ parameterizes the straight-line segment from $\mathbf{\text{a}}\ \text{to}\ \mathbf{\text{b}}.$

101.

*True* or *False?* Vector field $\mathbf{\text{F}}\left( {x,y,z} \right) = \left( {y\ \text{sin}\mspace{2mu} z} \right)\mathbf{\text{i}} + \left( {x\ \text{sin}\mspace{2mu} z} \right)\mathbf{\text{j}} + \left( {xy\ \text{cos}\mspace{2mu} z} \right)\mathbf{\text{k}}$ is conservative.

102\.

*True* or *False?* Vector field $\mathbf{\text{F}}\left( {x,y,z} \right) = y\mathbf{\text{i}} + \left( {x + z} \right)\mathbf{\text{j}} - y\mathbf{\text{k}}$ is conservative.

103.

Verify the Fundamental Theorem of Line Integrals for $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ by computing the integral using a parameterization and, separately, by finding a potential function for the case when $\mathbf{\text{F}}\left( {x,y} \right) = \left( {2x + 2y} \right)\mathbf{\text{i}} + \left( {2x + 2y} \right)\mathbf{\text{j}}$ and *C* is a portion of circle $x^{2} + y^{2} = 25$ oriented counterclockwise from (5, 0) to (3, 4).

104\.

\[T\] Find ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = \left( {ye^{xy} + \text{cos}\mspace{2mu} x} \right)\mathbf{\text{i}} + \left( {xe^{xy} + \frac{1}{y^{2} + 1}} \right)\mathbf{\text{j}}$ and *C* is a portion of curve $y = \text{sin}\mspace{2mu} x$ from $x = 0$ to $x = \frac{\pi}{2}.$

105.

\[T\] Evaluate line integral ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = \left( {e^{x}\text{sin}\mspace{2mu} y - y} \right)\mathbf{\text{i}} + \left( {e^{x}\text{cos}\mspace{2mu} y - x - 2} \right)\mathbf{\text{j}},$ and *C* is the path given by $\mathbf{r}(t) = \left\lbrack {t^{3}\text{sin}\ \frac{\pi t}{2}} \right\rbrack\mathbf{\text{i}} - \left\lbrack {\frac{\pi}{2}\text{cos}\left( {\frac{\pi t}{2} + \frac{\pi}{2}} \right)} \right\rbrack\mathbf{\text{j}}$ for $0 \leq t \leq 1.$

For the following exercises, determine whether the vector field is conservative and, if it is, find the potential function.

106\.

$\mathbf{\text{F}}(x,y) = 2xy^{3}\mathbf{\text{i}} + 3y^{2}x^{2}\mathbf{\text{j}}$

107.

$\mathbf{\text{F}}(x,y) = \left( {\text{−}y + e^{x}\text{sin}\mspace{2mu} y} \right)\mathbf{\text{i}} + \left\lbrack {\left( {x + 2} \right)e^{x}\text{cos}\mspace{2mu} y} \right\rbrack\mathbf{\text{j}}$

108\.

$\mathbf{\text{F}}(x,y) = \left( {e^{2x}\text{sin}\mspace{2mu} y} \right)\mathbf{\text{i}} + \left\lbrack {e^{2x}\text{cos}\mspace{2mu} y} \right\rbrack\mathbf{\text{j}}$

109.

$\mathbf{\text{F}}(x,y) = (6x + 5y)\mathbf{\text{i}} + (5x + 4y)\mathbf{\text{j}}$

110\.

$\mathbf{\text{F}}(x,y) = \left\lbrack {2x\ \text{cos}(y) - y\ \text{cos}(x)} \right\rbrack\mathbf{\text{i}} + \left\lbrack {\text{−}x^{2}\text{sin}(y) - \text{sin}(x)} \right\rbrack\mathbf{\text{j}}$

111.

$\mathbf{\text{F}}(x,y) = \left\lbrack {ye^{x} + \text{sin}(y)} \right\rbrack\mathbf{\text{i}} + \left\lbrack {e^{x} + x\ \text{cos}(y)} \right\rbrack\mathbf{\text{j}}$

For the following exercises, evaluate the line integrals using the Fundamental Theorem of Line Integrals.

112\.

${\int_{C}{(y\mathbf{\text{i}} + x\mathbf{\text{j}}) \cdot d\mathbf{\text{r}}}},$ where *C* is any path from (0, 0) to (2, 4)

113.

${\int_{C}{(2ydx + 2xdy)}},$ where *C* is the line segment from (0, 0) to (4, 4)

114\.

\[T\] ${\int_{C}{\left\lbrack {\text{arctan}\ \frac{y}{x} - \frac{xy}{x^{2} + y^{2}}} \right\rbrack dx + \left\lbrack {\frac{x^{2}}{x^{2} + y^{2}} + e^{\text{−}y}(1 - y)} \right\rbrack dy}},$ where *C* is any smooth curve from (1, 1) to $\left( {-1,2} \right)$

115.

Find the conservative vector field for the potential function

$$f(x,y) = 5x^{2} + 3xy + {10y}^{2}.$$

For the following exercises, determine whether the vector field is conservative and, if so, find a potential function.

116\.

$\mathbf{\text{F}}(x,y) = \left( {12xy} \right)\mathbf{\text{i}} + 6\left( {x^{2} + y^{2}} \right)\mathbf{\text{j}}$

117.

$\mathbf{\text{F}}(x,y) = \left( {e^{x}\text{cos}\mspace{2mu} y} \right)\mathbf{\text{i}} + 6\left( {e^{x}\text{sin}\mspace{2mu} y} \right)\mathbf{\text{j}}$

118\.

$\mathbf{\text{F}}(x,y) = \left( {2xye^{x^{2}y}} \right)\mathbf{\text{i}} + \left( {x^{2}e^{x^{2}y}} \right)\mathbf{\text{j}}$

119.

$\mathbf{\text{F}}(x,y\text{,}\ z) = \left( {ye^{z}} \right)\mathbf{\text{i}} + \left( {xe^{z}} \right)\mathbf{\text{j}} + \left( {xye^{z}} \right)\mathbf{\text{k}}$

120\.

$\mathbf{\text{F}}(x,y\text{,}\ z) = \left( {\text{sin}\mspace{2mu} y} \right)\mathbf{\text{i}} - \left( {x\ \text{cos}\mspace{2mu} y} \right)\mathbf{\text{j}} + \mathbf{\text{k}}$

121.

$\mathbf{\text{F}}(x,y\text{,}\ z) = \left( {–\frac{1}{y}} \right)\mathbf{\text{i}} + \left( \frac{x}{y^{2}} \right)\mathbf{\text{j}} + \left( {2z - 1} \right)\mathbf{\text{k}}$

122\.

$\mathbf{\text{F}}(x,y,z) = 3z^{2}\mathbf{\text{i}} - \text{cos}\mspace{2mu} y\mathbf{\text{j}} + 2xz\mathbf{\text{k}}$

123.

$\mathbf{\text{F}}(x,y\text{,}\ z) = \left( {2xy} \right)\mathbf{\text{i}} + \left( {x^{2} + 2yz} \right)\mathbf{\text{j}} + y^{2}\mathbf{\text{k}}$

124\.

$\mathbf{\text{F}}\left( {x,y} \right) = \left( 4{e^{x}\text{cos}\mspace{2mu} y} \right)\mathbf{\text{i}}–\left( 4{e^{x}\text{sin}\mspace{2mu} y} \right)\mathbf{\text{j}}$

125.

$\mathbf{\text{F}}\left( {x,y} \right) = \left( {ye^{x^{2}}} \right)\mathbf{\text{i}} + \left( {x^{2}e^{y^{2}}} \right)\mathbf{\text{j}}$

For the following exercises, evaluate the integral using the Fundamental Theorem of Line Integrals.

126\.

Evaluate ${\int_{C}{\nabla f \cdot d\mathbf{\text{r}}}},$ where $f(x,y\text{,}\ z) = \text{cos}(\pi x) + \text{sin}(\pi y) - xyz$ and *C* is any path that starts at $\left( {1,\frac{1}{2},2} \right)$ and ends at $\left( {2,1,-1} \right).$

127.

\[T\] Evaluate ${\int_{C}{\nabla f \cdot d\mathbf{\text{r}}}},$ where $f(x,y) = xy + e^{x}$ and *C* is a straight line from $\left( {0,0} \right)$ to $\left( {2,1} \right).$

128\.

\[T\] Evaluate ${\int_{C}{\nabla f \cdot d\mathbf{\text{r}}}},$ where $f(x,y) = x^{2}y - x$ and *C* is any path in a plane from (1, 2) to (3, 2).

129.

Evaluate ${\int_{C}{\nabla f \cdot d\mathbf{\text{r}}}},$ where $f(x,y\text{,}\ z) = xyz^{2} - yz$ and *C* has initial point (1, 2, 3) and terminal point (3, 5, 1).

For the following exercises, let $\mathbf{\text{F}}(x,y) = 2xy^{2}\mathbf{\text{i}} + \left( {2yx^{2} + 2y} \right)\mathbf{\text{j}}$ and $\mathbf{G}(x,y) = (y + x)\mathbf{\text{i}} + (y - x)\mathbf{\text{j}},$ and let *C*₁ be the curve consisting of the circle of radius 2, centered at the origin and oriented counterclockwise, and *C*₂ be the curve consisting of a line segment from (0, 0) to (1, 1) followed by a line segment from (1, 1) to (3, 1).

130\.

Calculate the line integral of F over *C*₁.

131.

Calculate the line integral of G over *C*₁.

132\.

Calculate the line integral of F over *C*₂.

133.

Calculate the line integral of G over *C*₂.

134\.

\[T\] Let $\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + z\ \text{sin}(yz)\mathbf{\text{j}} + y\ \text{sin}(yz)\mathbf{\text{k}}.$ Calculate ${\int_{C}{\mathbf{\text{F}} \cdot dr}},$ where *C* is a path from $A = (0,0,1)$ to $B = (3,1,2).$

135.

\[T\] Find line integral $\int_{C}{\mathbf{\text{F}} \cdot dr}$ of vector field $\mathbf{\text{F}}(x,y,z) = 3x^{2}z\mathbf{\text{i}} + z^{2}\mathbf{\text{j}} + \left( {x^{3} + 2yz} \right)\mathbf{\text{k}}$ along curve *C* parameterized by $\mathbf{r}(t) = \left( \frac{\text{ln}\mspace{2mu} t}{\text{ln}\ 2} \right)\mathbf{\text{i}} + t^{3\text{/}2}\mathbf{\text{j}} + t\ \text{cos}{\left( {\pi t} \right)\mathbf{k}}\text{,}\ 1 \leq t \leq 4.$

For the following exercises, show that the following vector fields are conservative by using a computer. Calculate $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ for the given curve.

136\.

$\mathbf{\text{F}} = \left( {xy^{2} + 3x^{2}y} \right)\mathbf{\text{i}} + \left( {x + y} \right)x^{2}\mathbf{\text{j}};$ *C* is the curve consisting of line segments from $(1,1)$ to $(0,2)$ to $(3,0).$

137.

$\mathbf{\text{F}} = \frac{2x}{y^{2} + 1}\mathbf{\text{i}} - \frac{2y\left( {x^{2} + 1} \right)}{\left( {y^{2} + 1} \right)^{2}}\mathbf{\text{j}};$ *C* is parameterized by $x = t^{3} - 1,y = t^{6} - t,0 \leq t \leq 1.$

138\.

\[T\] $\mathbf{\text{F}} = \left\lbrack {\text{cos}\left( {xy^{2}} \right) - xy^{2}\text{sin}\left( {xy^{2}} \right)} \right\rbrack\mathbf{\text{i}} - 2x^{2}y\ \text{sin}\left( {xy^{2}} \right)\mathbf{\text{j}};$ *C* is curve ${\mathbf{r}(t) = {< {e^{t},e^{t + 1}} >}},-1 \leq t \leq 0.$

139.

The mass of Earth is approximately $6\ \times \ 10^{27}\text{g}$ and that of the Sun is 330,000 times as much. The gravitational constant is $6.7\ \times \ 10^{-8}{\text{cm}^{3}\text{/}{\left( \text{s}^{2} \cdot \text{g} \right).}}$ The distance of Earth from the Sun is about $1.5\ \times \ 10^{12}\text{cm}.$ Compute, approximately, the work necessary to increase the distance of Earth from the Sun by $1\ \text{cm}.$

140\.

\[T\] Let $\mathbf{\text{F}} = \left( {e^{x}\text{sin}\mspace{2mu} y} \right)\mathbf{\text{i}} + \left( {e^{x}\text{cos}\mspace{2mu} y} \right)\mathbf{\text{j}} + z^{2}\mathbf{\text{k}}.$ Evaluate the integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{r}}},$ where $C$ is the curve ${\mathbf{r}(t) = {< {\sqrt{t},t^{3},e^{\sqrt{t}}} >}},0 \leq t \leq 1.$

141.

\[T\] Let $\mathbf{\text{r}}:\left\lbrack {1,2} \right\rbrack\rightarrow\mathbb{R}^{2}$ be given by $x = e^{t - 1},y = \text{sin}\left( \frac{\pi}{t} \right).$ Use a computer to compute the integral $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} = {\int_{C}{2x\ \text{cos}\mspace{2mu} ydx - x^{2}\text{sin}\mspace{2mu} ydy,}}}$ where $\mathbf{\text{F}} = \left( {2x\ \text{cos}\mspace{2mu} y} \right)\mathbf{\text{i}} - \left( {x^{2}\text{sin}\mspace{2mu} y} \right)\mathbf{\text{j}}.$

142\.

\[T\] Use a computer algebra system to find the mass of a wire that lies along curve $\mathbf{\text{r}}(t) = \left( {t^{2} - 1} \right)\mathbf{\text{j}} + 2t\mathbf{\text{k}},0 \leq t \leq 1,$ if the density is $\frac{3}{2}t.$

143.

Find the circulation and flux of field $\mathbf{\text{F}} = \text{−}y\mathbf{\text{i}} + x\mathbf{\text{j}}$ around and across the closed semicircular path that consists of semicircular arch $\mathbf{\text{r}}_{1}(t) = \left( {a\mspace{2mu}\text{cos}\mspace{2mu} t} \right)\mathbf{\text{i}} + \left( {a\mspace{2mu}\text{sin}\mspace{2mu} t} \right)\mathbf{\text{j}},0 \leq t \leq \pi,$ followed by line segment $\mathbf{\text{r}}_{2}(t) = t\mathbf{\text{i}},\text{−}a \leq t \leq a.$

144\.

Compute ${\int_{C}{\text{cos}\mspace{2mu} x\ \text{cos}\mspace{2mu} ydx - \text{sin}\mspace{2mu} x\ \text{sin}\mspace{2mu} ydy}},$ where $\textbf{c}(t) = {< {t,t^{2}} >},0 \leq t \leq 1.$

145.

Complete the proof of The Path Independence Test for Conservative Fields by showing that $f_{y} = Q\left( {x,y} \right).$

6.4 Green's Theorem 6.4 格林定理

• 6.4.1 Apply the circulation form of Green's theorem.
• 6.4.2 Apply the flux form of Green's theorem.
• 6.4.3 Calculate circulation and flux on more general regions.

• 6.4.1 应用格林定理的环流量形式。
• 6.4.2 应用格林定理的通量形式。
• 6.4.3 在更一般的区域上计算环流量与通量。

In this section, we examine Green's theorem, which is an extension of the Fundamental Theorem of Calculus to two dimensions. Green's theorem has two forms: a circulation form and a flux form, both of which require region *D* in the double integral to be simply connected. However, we will extend Green's theorem to regions that are not simply connected.

Put simply, Green's theorem relates a line integral around a simply closed plane curve *C* and a double integral over the region enclosed by *C*. The theorem is useful because it allows us to translate difficult line integrals into more simple double integrals, or difficult double integrals into more simple line integrals.

Extending the Fundamental Theorem of Calculus 推广微积分基本定理

Recall that the Fundamental Theorem of Calculus says that

$${\int_{a}^{b}{F^{\prime}(x)dx = F(b) - F(a)}}.$$

As a geometric statement, this equation says that the integral over the region below the graph of $F^{\prime}(x)$ and above the line segment $\lbrack a,b\rbrack$ depends only on the value of *F* at the endpoints *a* and *b* of that segment. Since the numbers *a* and *b* are the boundary of the line segment $\lbrack a,b\rbrack,$ the theorem says we can calculate integral $\int_{a}^{b}{F\text{'}(x)dx}$ based on information about the boundary of line segment $\lbrack a,b\rbrack$ (Figure 6.32). The same idea is true of the Fundamental Theorem for Line Integrals:

$${\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}} = f(\mathbf{\text{r}}(b)) - f(\mathbf{\text{r}}(a)).$$

When we have a potential function (an "antiderivative"), we can calculate the line integral based solely on information about the boundary of curve *C*.

Green's theorem takes this idea and extends it to calculating double integrals. Green's theorem says that we can calculate a double integral over region *D* based solely on information about the boundary of *D*. Green's theorem also says we can calculate a line integral over a simple closed curve *C* based solely on information about the region that *C* encloses. In particular, Green's theorem connects a double integral over region *D* to a line integral around the boundary of *D*.

Circulation Form of Green's Theorem 格林定理的环流量形式

The first form of Green's theorem that we examine is the circulation form. This form of the theorem relates the vector line integral over a simple, closed plane curve *C* to a double integral over the region enclosed by *C*. Therefore, the circulation of a vector field along a simple closed curve can be transformed into a double integral and vice versa.

Green's Theorem, Circulation Form 格林定理（环流量形式）

Let *D* be an open, simply connected region with a boundary curve *C* that is a piecewise smooth, simple closed curve oriented counterclockwise (Figure 6.33). Let $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ be a vector field with component functions that have continuous partial derivatives on *D*. Then,

$${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} =}}{\int_{C}{Pdx + Qdy = {\iint_{D}{(Q_{x} - P_{y})dA}}}}.$$ (6.13)

Notice that Green's theorem can be used only for a two-dimensional vector field F. If F is a three-dimensional field, then Green's theorem does not apply. Since

$${\int_{C}{Pdx + Qdy = {\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}}}},$$

this version of Green's theorem is sometimes referred to as the *tangential form* of Green's theorem.

The proof of Green's theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that *D* is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is conservative. In this case,

$$\int_{C}{Pdx + Qdy = 0}$$

because the circulation is zero in conservative vector fields. By Cross-Partial Property of Conservative Fields, F satisfies the cross-partial condition, so $P_{y} = Q_{x}.$ Therefore,

$${\iint_{D}{(Q_{x} - P_{y})dA}} = {\iint_{D}{0dA = 0 = {\int_{C}{Pdx + Qdy}}}},$$

which confirms Green's theorem in the case of conservative vector fields.

Proof 证明

Let's now prove that the circulation form of Green's theorem is true when the region *D* is a rectangle. Let *D* be the rectangle $\left\lbrack {a,b} \right\rbrack\ \times \ \left\lbrack {c,d} \right\rbrack$ oriented counterclockwise. Then, the boundary *C* of *D* consists of four piecewise smooth pieces $C_{1},$ $C_{2},$ $C_{3},$ and $C_{4}$ (Figure 6.34). We parameterize each side of *D* as follows:

$$\begin{array}{rll} {C_{1}\text{:}\ \mathbf{\text{r}}_{1}(t)} & = & {\left\langle {t,c} \right\rangle,a \leq t \leq b} \\ {C_{2}\text{:}\ \mathbf{\text{r}}_{2}(t)} & = & {\left\langle {b,t} \right\rangle,c \leq t \leq d} \\ {\text{−}C_{3}\text{:}\ \mathbf{\text{r}}_{3}(t)} & = & {\left\langle {t,d} \right\rangle,a \leq t \leq b} \\ {\text{−}C_{4}\text{:}\ \mathbf{\text{r}}_{4}(t)} & = & {\left\langle {a,t} \right\rangle,c \leq t \leq d.} \end{array}$$

Then,

$$\begin{matrix} {\int_{C}\textbf{F} \cdot d\textbf{r}} & {= \int_{C_{1}}\textbf{F} \cdot d\textbf{r} + \int_{C_{2}}\textbf{F} \cdot d\textbf{r} + \int_{C_{3}}\textbf{F} \cdot d\textbf{r} + \int_{C_{4}}\textbf{F} \cdot d\textbf{r}} \\ & {= \int_{C_{1}}\textbf{F} \cdot d\textbf{r} + \int_{C_{2}}\textbf{F} \cdot d\textbf{r} - \int_{\text{−}C_{3}}\textbf{F} \cdot d\textbf{r} - \int_{\text{−}C_{4}}\textbf{F} \cdot d\textbf{r}} \\ & {= \int_{a}^{b}\textbf{F}\left( \textbf{r}_{1}(t) \right) \cdot \textbf{r}_{1'}(t)dt + \int_{c}^{d}\textbf{F}\left( \textbf{r}_{2}(t) \right) \cdot \textbf{r}_{2'}(t)dt} \\ & {\mspace{9mu}\text{−}\int_{a}^{b}\textbf{F}\left( \textbf{r}_{3}(t) \right) \cdot \textbf{r}_{3'}(t)dt - \int_{c}^{d}\textbf{F}\left( \textbf{r}_{4}(t) \right) \cdot \textbf{r}_{4'}(t)dt} \\ & {= \int_{a}^{b}P(t,c)dt + \int_{c}^{d}Q(b,t)dt - \int_{a}^{b}P(t,d)dt - \int_{c}^{d}Q(a,t)dt} \\ & {= \int_{a}^{b}\left( P(t,c) - P(t,d) \right)dt + \int_{c}^{d}\left( Q(b,t) - Q(a,t) \right)dt} \\ & {= \text{−}\int_{a}^{b}\left( P(t,d) - P(t,c) \right)dt + \int_{c}^{d}\left( Q(b,t) - Q(a,t) \right)dt.} \end{matrix}$$

By the Fundamental Theorem of Calculus,

$$P\left( {t,d} \right) - P\left( {t,c} \right) = {\int_{c}^{d}{\frac{\partial}{\partial y}P\left( {t,y} \right)dy}}\ \text{and}\ Q\left( {b,t} \right) - Q\left( {a,t} \right) = {\int_{a}^{b}{\frac{\partial}{\partial x}Q\left( {x,t} \right)dx.}}$$

Therefore,

$$\begin{array}{l} \\ \\ \\ \\ {\mspace{9mu}\text{−}{\int_{a}^{b}{\left( {P\left( {t,d} \right) - P\left( {t,c} \right)} \right)dt +}}{\int_{c}^{d}{\left( {Q\left( {b,t} \right) - Q\left( {a,t} \right)} \right)dt}}} \\ {= \text{−}{\int_{a}^{b}{\int_{c}^{d}{\frac{\partial}{\partial y}P\left( {t,y} \right)dydt +}}}{\int_{c}^{d}{\int_{a}^{b}{\frac{\partial}{\partial x}Q\left( {x,t} \right)dxdt.}}}} \end{array}$$

But,

$$\begin{array}{cl} {- {\int_{a}^{b}{\int_{c}^{d}\frac{\partial}{\partial y}}}P\left( {t,y} \right)dydt + {\int_{c}^{d}{\int_{a}^{b}\frac{\partial}{\partial x}}}Q\left( {x,t} \right)dxdt} & {= \text{−}{\int_{a}^{b}{\int_{c}^{d}{\frac{\partial}{\partial y}P\left( {x,y} \right)dydx + {\int_{c}^{d}{\int_{a}^{b}{\frac{\partial}{\partial x}Q\left( {x,y} \right)dxdy}}}}}}} \\ & {= {\int_{a}^{b}{\int_{c}^{d}{\left( {Q_{x} - P_{y}} \right)dydx}}}} \\ & {= {\int{\int_{D}{\left( {Q_{x} - P_{y}} \right)dA.}}}} \end{array}$$

Therefore, $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} = {\int{\int_{D}{\left( {Q_{x} - P_{y}} \right)dA}}}}$ and we have proved Green's theorem in the case of a rectangle.

To prove Green's theorem over a general region *D*, we can decompose *D* into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.

□

Applying Green's Theorem over a Rectangle 在矩形上应用格林定理

Calculate the line integral

$${\int_{C}{x^{2}ydx + (y - 3)dy}},$$

where *C* is a rectangle with vertices $(1,1),$ $(4,1),$ $(4,5),$ and $(1,5)$ oriented counterclockwise.

Solution 解

Let $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {P\left( {x,y} \right),Q\left( {x,y} \right)} \right\rangle = \left\langle {x^{2}y,y - 3} \right\rangle.$ Then, $Q_{x} = 0$ and $P_{y} = x^{2}.$ Therefore, $Q_{x} - P_{y} = \text{−}x^{2}.$

Let *D* be the rectangular region enclosed by *C* (Figure 6.35). By Green's theorem,

$$\begin{array}{cl} {\int_{C}{x^{2}ydx + (y - 3)dy}} & {= {\iint_{D}{\left( {Q_{x} - P_{y}} \right)dA}}} \\ & {= {\int{\int_{D}{\text{−}x^{2}dA}}} = {\int_{1}^{5}{\int_{1}^{4}{\text{−}x^{2}dxdy}}}} \\ & {= {\int_{1}^{5}{-21dy}} = -84.} \end{array}$$

Analysis 分析

If we were to evaluate this line integral without using Green's theorem, we would need to parameterize each side of the rectangle, break the line integral into four separate line integrals, and use the methods from Line Integrals to evaluate each integral. Furthermore, since the vector field here is not conservative, we cannot apply the Fundamental Theorem for Line Integrals. Green's theorem makes the calculation much simpler.

Applying Green's Theorem to Calculate Work 应用格林定理计算功

Calculate the work done on a particle by force field

$$\mathbf{\text{F}}(x,y) = \left\langle {y + \text{sin}\ x,e^{y} - x} \right\rangle$$

as the particle traverses circle $x^{2} + y^{2} = 4$ exactly once in the counterclockwise direction, starting and ending at point $(2,0).$

Solution 解

Let *C* denote the circle and let *D* be the disk enclosed by *C*. The work done on the particle is

$$W = {\int_{C}{\left( {y + \text{sin}\ x} \right)dx + (e^{y} - x)dy}}.$$

As with Example 6.38, this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green's theorem (Figure 6.36).

Let $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {P\left( {x,y} \right),Q\left( {x,y} \right)} \right\rangle = \left\langle {y + \text{sin}\ x,e^{y} - x} \right\rangle.$ Then, $Q_{x} = -1$ and $P_{y} = 1.$ Therefore, $Q_{x} - P_{y} = -2.$

By Green's theorem,

$$\begin{array}{cl} W & {= {\int_{C}{(y + \text{sin}(x))dx + (e^{y} - x)dy}}} \\ & {= {\iint_{D}{\left( {Q_{x} - P_{y}} \right)dA}} = {\iint_{D}{-2dA}}} \\ & {= -2\left( {\text{area}(D)} \right) = -2\pi\left( 2^{2} \right) = -8\pi.} \end{array}$$

Use Green's theorem to calculate line integral

$${\int_{C}{\text{sin}(x^{2})dx + (3x - y)dy}},$$

where *C* is a right triangle with vertices $(-1,2),$ $(4,2),$ and $(4,5)$ oriented counterclockwise.

In the preceding two examples, the double integral in Green's theorem was easier to calculate than the line integral, so we used the theorem to calculate the line integral. In the next example, the double integral is more difficult to calculate than the line integral, so we use Green's theorem to translate a double integral into a line integral.

Applying Green's Theorem over an Ellipse 在椭圆上应用格林定理

Calculate the area enclosed by ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ (Figure 6.37).

Solution 解

Let *C* denote the ellipse and let *D* be the region enclosed by *C*. Recall that ellipse *C* can be parameterized by

$$x = a\ \text{cos}\ t,y = b\ \text{sin}\ t,0 \leq t \leq 2\pi.$$

Calculating the area of *D* is equivalent to computing double integral ${\iint_{D}{dA}}.$ To calculate this integral without Green's theorem, we would need to divide *D* into two regions: the region above the *x*-axis and the region below. The area of the ellipse is

$$\int_{\text{−}a}^{a}{\int_{0}^{\sqrt{b^{2} - {({{bx}\text{/}a})}^{2}}}{dydx + {\int_{\text{−}a}^{a}{\int_{\text{−}\sqrt{b^{2} - {({{bx}\text{/}a})}^{2}}}^{0}\mspace{2mu}{dydx.}}}}}$$

These two integrals are not straightforward to calculate (although when we know the value of the first integral, we know the value of the second by symmetry). Instead of trying to calculate them, we use Green's theorem to transform $\iint_{D}{dA}$ into a line integral around the boundary *C*.

Consider vector field

$$\mathbf{\text{F}}(x,y) = \left\langle {P,Q} \right\rangle = \left\langle {- \frac{y}{2},\frac{x}{2}} \right\rangle.$$

Then, $Q_{x} = \frac{1}{2}$ and $P_{y} = - \frac{1}{2},$ and therefore $Q_{x} - P_{y} = 1.$ Notice that F was chosen to have the property that $Q_{x} - P_{y} = 1.$ Since this is the case, Green's theorem transforms the line integral of F over *C* into the double integral of 1 over *D*.

By Green's theorem,

$$\begin{array}{cl} & \\ & \\ & \\ {\iint_{D}{dA}} & {= {\iint_{D}{\left( {Q_{x} - P_{y}} \right)dA}}} \\ & {= {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = \frac{1}{2}{\int_{C}{\text{−}ydx + xdy}}} \\ & {= \frac{1}{2}{\int_{0}^{2\pi}{\text{−}b\ \text{sin}\ t(\text{−}a\ \text{sin}\ t) + a\left( {\text{cos}\ t} \right)b\ \text{cos}\ tdt}}} \\ & {= \frac{1}{2}{\int_{0}^{2\pi}{ab\ \text{cos}^{2}t + ab\ \text{sin}^{2}tdt}} = \frac{1}{2}{\int_{0}^{2\pi}{abdt}} = \pi ab.} \end{array}$$

Therefore, the area of the ellipse is $\pi ab.$

In Example 6.40, we used vector field $\mathbf{\text{F}}(x,y) = \left\langle {P,Q} \right\rangle = \left\langle {- \frac{y}{2},\frac{x}{2}} \right\rangle$ to find the area of any ellipse. The logic of the previous example can be extended to derive a formula for the area of any region *D*. Let *D* be any region with a boundary that is a simple closed curve *C* oriented counterclockwise. If $\mathbf{\text{F}}(x,y) = \left\langle {P,Q} \right\rangle = \left\langle {- \frac{y}{2},\frac{x}{2}} \right\rangle,$ then $Q_{x} - P_{y} = 1.$ Therefore, by the same logic as in Example 6.40,

$$\text{area of}\ D = {\iint_{D}{dA = \frac{1}{2}{\int_{C}{\text{−}ydx + xdy}}}}.$$ (6.14)

It's worth noting that if $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is any vector field with $Q_{x} - P_{y} = 1,$ then the logic of the previous paragraph works. So. Equation 6.14 is not the only equation that uses a vector field's mixed partials to get the area of a region.

Find the area of the region enclosed by the curve with parameterization $\mathbf{\text{r}}(t) = \left\langle {\text{sin}\ t\ \text{cos}\ t,\text{sin}\ t} \right\rangle,0 \leq t \leq \pi.$

Flux Form of Green's Theorem 格林定理的通量形式

The circulation form of Green's theorem relates a double integral over region *D* to line integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}},$ where *C* is the boundary of *D*. The flux form of Green's theorem relates a double integral over region *D* to the flux across boundary *C*. The flux of a fluid across a curve can be difficult to calculate using the flux line integral. This form of Green's theorem allows us to translate a difficult flux integral into a double integral that is often easier to calculate.

Green's Theorem, Flux Form 格林定理（通量形式）

Let *D* be an open, simply connected region with a boundary curve *C* that is a piecewise smooth, simple closed curve that is oriented counterclockwise (Figure 6.38). Let $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ be a vector field with component functions that have continuous partial derivatives on an open region containing *D*. Then,

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds = {\iint_{D}{P_{x} + Q_{y}dA}}}}.$$ (6.15)

Because this form of Green's theorem contains unit normal vector N, it is sometimes referred to as the *normal form* of Green's theorem.

Proof 证明

Recall that ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}} = {\int_{C}{\text{−}Qdx + Pdy}}.$ Let $M = \text{−}Q$ and $N = P.$ By the circulation form of Green's theorem,

$$\begin{array}{cl} {{\int_{C}{\text{−}Qdx}} + Pdy} & {= {\int_{C}{Mdx + Ndy}}} \\ & {= {\iint_{D}{N_{x} - M_{y}dA}}} \\ & {= {\iint_{D}{P_{x} - {(\text{−}Q)}_{y}dA}}} \\ & {= {\iint_{D}{P_{x} + Q_{y}dA}}.} \end{array}$$

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Applying Green's Theorem for Flux across a Circle 应用格林定理计算穿过圆的通量

Let *C* be a circle of radius *r* centered at the origin (Figure 6.39) and let $\mathbf{\text{F}}(x,y) = \left\langle {x,y} \right\rangle.$ Calculate the flux across *C*.

Solution 解

Let *D* be the disk enclosed by *C.* The flux across *C* is ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}.$ We could evaluate this integral using tools we have learned, but Green's theorem makes the calculation much more simple. Let $P\left( {x,y} \right) = x$ and $Q\left( {x,y} \right) = y$ so that $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle.$ Note that $P_{x} = 1 = Q_{y},$ and therefore $P_{x} + Q_{y} = 2.$ By Green's theorem,

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds = {\int{\int_{D}{2dA =}}}}}2{\int{\int_{D}{dA.}}}$$

Since $\int{\int_{D}{dA}}$ is the area of the circle, ${\int{\int_{D}{dA}}} = \pi r^{2}.$ Therefore, the flux across *C* is $2\pi r^{2}.$

Applying Green's Theorem for Flux across a Triangle 应用格林定理计算穿过三角形的通量

Let *S* be the triangle with vertices $(0,0),$ $(1,0),$ and $(0,3)$ oriented clockwise (Figure 6.40). Calculate the flux of $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {P\left( {x,y} \right),Q\left( {x,y} \right)} \right\rangle = \left\langle {x^{2} + e^{y},x + y} \right\rangle$ across *S*.

Solution 解

To calculate the flux without Green’s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green’s theorem to translate the flux line integral into a single double integral is much more simple.

Let *D* be the region enclosed by *S*. Note that $P_{x} = 2x$ and $Q_{y} = 1;$ therefore, $P_{x} + Q_{y} = 2x + 1.$ Green’s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds = \text{−}}}{\int_{\text{−}S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}$ and $\text{−}S$ is oriented counterclockwise. By Green’s theorem, the flux is

$$\begin{array}{cl} & \\ & \\ & \\ {\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}} & {= {\int_{\text{−}S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}} \\ & {= \text{−}{\iint_{D}{\left( {P_{x} + Q_{y}} \right)dA}}} \\ & {= \text{−}{\iint_{D}{\left( {2x + 1} \right)dA}}.} \end{array}$$

Notice that the top edge of the triangle is the line $y = -3x + 3.$ Therefore, in the iterated double integral, the *y*-values run from $y = 0$ to $y = -3x + 3,$ and we have

$$\begin{array}{cl} {\text{−}{\iint_{D}{\left( {2x + 1} \right)dA}}} & {= \text{−}{\int_{0}^{1}{\int_{0}^{-3x + 3}{\left( {2x + 1} \right)dydx}}}} \\ & {= \text{−}{\int_{0}^{1}{\left( {2x + 1} \right)\left( {-3x + 3} \right)dx}} = \text{−}{\int_{0}^{1}{\left( {-6x^{2} + 3x + 3} \right)dx}}} \\ & {= \text{−}\left\lbrack {-2x^{3} + \frac{3x^{2}}{2} + 3x} \right\rbrack_{0}^{1} = - \frac{5}{2}.} \end{array}$$

Calculate the flux of $\mathbf{\text{F}}(x,y) = \left\langle {x^{3},y^{3}} \right\rangle$ across a unit circle oriented counterclockwise.

Applying Green’s Theorem for Water Flow across a Rectangle 把格林定理用于矩形上的水流

Water flows from a spring located at the origin. The velocity of the water is modeled by vector field $\mathbf{\text{v}}\left( {x,y} \right) = \left\langle {5x + y,x + 3y} \right\rangle$ m/sec. Find the amount of water per second that flows across the rectangle with vertices $\left( {-1,-2} \right),\left( {1,-2} \right),\left( {1,3} \right),\ \text{and}\ \left( {-1,3} \right),$ oriented counterclockwise (Figure 6.41).

Solution 解

Let *C* represent the given rectangle and let *D* be the rectangular region enclosed by *C*. To find the amount of water flowing across *C*, we calculate flux $\int_{C}{\mathbf{\text{v}} \cdot \mathbf{\text{N}{ds}}.}$ Let $P\left( {x,y} \right) = 5x + y$ and $Q\left( {x,y} \right) = x + 3y$ so that $\mathbf{\text{v}} = \left( {P,Q} \right).$ Then, $P_{x} = 5$ and $Q_{y} = 3.$ By Green’s theorem,

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{v}} \cdot \mathbf{\text{N}{ds}}}} & {= {\iint_{D}{\left( {P_{x} + Q_{y}} \right)dA}}} \\ & {= {\iint_{D}{8dA}}} \\ & {= 8\left( {\text{area of}\ D} \right) = 80.} \end{array}$$

Therefore, the water flux is 80 m²/sec.

Recall that if vector field F is conservative, then F does no work around closed curves—that is, the circulation of F around a closed curve is zero. In fact, if the domain of F is simply connected, then F is conservative if and only if the circulation of F around any closed curve is zero. If we replace “circulation of F” with “flux of F,” then we get a definition of a source-free vector field. The following statements are all equivalent ways of defining a source-free field $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ on a simply connected domain (note the similarities with properties of conservative vector fields):

1. The flux $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}$ across any closed curve *C* is zero.

2. If $C_{1}$ and $C_{2}$ are curves in the domain of F with the same starting points and endpoints, then ${\int_{C_{1}}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}} = {\int_{C_{2}}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}.$ In other words, flux is independent of path.

3. There is a stream function $g(x,y)$ for F. A stream function for $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a function *g* such that $P = g_{y}$ and $Q = \text{−}g_{x}.$ Geometrically, $\mathbf{\text{F}}\left( {a,b} \right)$ is tangential to the level curve of *g* at $\left( {a,b} \right).$ Since the gradient of *g* is perpendicular to the level curve of *g* at $\left( {a,b} \right),$ stream function *g* has the property $\mathbf{\text{F}}\left( {a,b} \right) \cdot \text{∇}g\left( {a,b} \right) = 0$ for any point $\left( {a,b} \right)$ in the domain of *g*. (Stream functions play the same role for source-free fields that potential functions play for conservative fields.)

4. $P_{x} + Q_{y} = 0$

Finding a Stream Function 寻找流函数

Verify that rotation vector field $\mathbf{\text{F}}(x,y) = \left\langle {y,\text{−}x} \right\rangle$ is source free, and find a stream function for F.

Solution 解

Note that the domain of F is all of $\mathbb{R}^{2},$ which is simply connected. Therefore, to show that F is source free, we can show any of items 1 through 4 from the previous list to be true. In this example, we show that item 4 is true. Let $P\left( {x,y} \right) = y$ and $Q\left( {x,y} \right) = \text{−}x.$ Then $P_{x} + Q_{y} = 0 + 0 = 0.$ Thus, F is source free.

To find a stream function for F, proceed in the same manner as finding a potential function for a conservative field. Let *g* be a stream function for F. Then $g_{y} = y,$ which implies that

$$g\left( {x,y} \right) = \frac{y^{2}}{2} + h(x).$$

Since $\text{−}g_{x} = Q = \text{−}x,$ we have $h\text{'}(x) = x.$ Therefore,

$$h(x) = \frac{x^{2}}{2} + C.$$

Letting $C = 0$ gives stream function

$$g\left( {x,y} \right) = \frac{x^{2}}{2} + \frac{y^{2}}{2}.$$

To confirm that *g* is a stream function for F, note that $g_{y} = y = P$ and $\text{−}g_{x} = \text{−}x = Q.$

Notice that source-free rotation vector field $\mathbf{\text{F}}(x,y) = \left\langle {y,\text{−}x} \right\rangle$ is perpendicular to conservative radial vector field $\text{∇}g = \left\langle {x,y} \right\rangle$ (Figure 6.42).

Find a stream function for vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {x\ \text{sin}\ y,\text{cos}\ y} \right\rangle.$

Vector fields that are both conservative and source free are important vector fields. One important feature of conservative and source-free vector fields on a simply connected domain is that any potential function $f$ of such a field satisfies Laplace’s equation $f_{xx} + f_{yy} = 0.$ Laplace’s equation is foundational in the field of partial differential equations because it models such phenomena as gravitational and magnetic potentials in space, and the velocity potential of an ideal fluid. A function that satisfies Laplace’s equation is called a *harmonic* function. Therefore any potential function of a conservative and source-free vector field is harmonic.

To see that any potential function of a conservative and source-free vector field on a simply connected domain is harmonic, let $f$ be such a potential function of vector field $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle.$ Then, $f_{x} = P$ and $f_{x} = Q$ because $\text{∇}f = \mathbf{\text{F}}.$ Therefore, $f_{xx} = P_{x}$ and $f_{yy} = Q_{y}.$ Since F is source free, $f_{xx} + f_{yy} = P_{x} + Q_{y} = 0,$ and we have that $f$ is harmonic.

Satisfying Laplace’s Equation 满足拉普拉斯方程

For vector field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {e^{x}\text{sin}\ y,e^{x}\text{cos}\ y} \right\rangle,$ verify that the field is both conservative and source free, find a potential function for F, and verify that the potential function is harmonic.

Solution 解

Let $P\left( {x,y} \right) = e^{x}\text{sin}\ y$ and $Q\left( {x,y} \right) = e^{x}\text{cos}\ y.$ Notice that the domain of F is all of two-space, which is simply connected. Therefore, we can check the cross-partials of F to determine whether F is conservative. Note that $P_{y} = e^{x}\text{cos}\ y = Q_{x},$ so F is conservative. Since $P_{x} = e^{x}\text{sin}\ y$ and $Q_{y} = - e^{x}\text{sin}\ y,P_{x} + Q_{y} = 0$ and the field is source free.

To find a potential function for F, let $f$ be a potential function. Then, $\text{∇}f = \mathbf{\text{F}},$ so $f_{x} = e^{x}\text{sin}\ y.$ Integrating this equation with respect to *x* gives $f\left( {x,y} \right) = e^{x}\text{sin}\ y + h(y).$ Since $f_{y} = e^{x}\text{cos}\ y,$ differentiating $f$ with respect to *y* gives $e^{x}\text{cos}\ y = e^{x}\text{cos}\ y + h\text{'}(y).$ Therefore, we can take $h(y) = 0,$ and $f\left( {x,y} \right) = e^{x}\text{sin}\ y$ is a potential function for $f.$

To verify that $f$ is a harmonic function, note that $f_{xx} = \frac{\partial}{\partial x}\left( {e^{x}\text{sin}\ y} \right) = e^{x}\text{sin}\ y$ and

$f_{yy} = \frac{\partial}{\partial x}\left( {e^{x}\text{cos}\ y} \right) = \text{−}e^{x}\text{sin}\ y.$ Therefore, $f_{xx} + f_{yy} = 0,$ and $f$ satisfies Laplace’s equation.

Is the function $f\left( {x,y} \right) = e^{x + 5y}$ harmonic?

Green’s Theorem on General Regions 一般区域上的格林定理

Green’s theorem, as stated, applies only to regions that are simply connected—that is, Green’s theorem as stated so far cannot handle regions with holes. Here, we extend Green’s theorem so that it does work on regions with finitely many holes (Figure 6.43).

Before discussing extensions of Green’s theorem, we need to go over some terminology regarding the boundary of a region. Let *D* be a region and let *C* be a component of the boundary of *D*. We say that *C* is positively oriented if, as we walk along *C* in the direction of orientation, region *D* is always on our left. Therefore, the counterclockwise orientation of the boundary of a disk is a positive orientation, for example. Curve *C* is negatively oriented if, as we walk along *C* in the direction of orientation, region *D* is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation, for example.

Let *D* be a region with finitely many holes (so that *D* has finitely many boundary curves), and denote the boundary of *D* by $\partial D$ (Figure 6.44). To extend Green’s theorem so it can handle *D*, we divide region *D* into two regions, $D_{1}$ and $D_{2}$ (with respective boundaries $\partial D_{1}$ and $\partial D_{2}),$ in such a way that $D = D_{1} \cup D_{2}$ and neither $D_{1}$ nor $D_{2}$ has any holes (Figure 6.44).

Assume the boundary of *D* is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region $D_{1}$ and $D_{2}$ is positively oriented. If F is a vector field defined on *D*, then Green’s theorem says that

$$\begin{array}{cl} {\int_{\partial D}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{\partial D_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{\partial D_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}} \\ & {= {\iint_{D_{1}}{Q_{x} - P_{y}dA +}}{\iint_{D_{2}}{Q_{x} - P_{y}dA}}} \\ & {= {\iint_{D}{(Q_{x} - P_{y})dA}}.} \end{array}$$

Therefore, Green’s theorem still works on a region with holes.

To see how this works in practice, consider annulus *D* in Figure 6.45 and suppose that $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a vector field defined on this annulus. Region *D* has a hole, so it is not simply connected. Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure 6.45) so that, when we divide the region into $D_{1}$ and $D_{2},$ we are able to keep the region on our left as we walk along a path that traverses the boundary. Let $D_{1}$ be the upper half of the annulus and $D_{2}$ be the lower half. Neither of these regions has holes, so we have divided *D* into two simply connected regions.

We label each piece of these new boundaries as $P_{i}$ for some *i,* as in Figure 6.45. If we begin at *P* and travel along the oriented boundary, the first segment is $P_{1},$ then $P_{2},P_{3},$ and $P_{4}.$ Now we have traversed $D_{1}$ and returned to *P.* Next, we start at *P* again and traverse $D_{2}.$ Since the first piece of the boundary is the same as $P_{4}$ in $D_{1},$ but oriented in the opposite direction, the first piece of $D_{2}$ is $\text{−}P_{4}.$ Next, we have $P_{5},$ then $\text{−}P_{2},$ and finally $P_{6}.$

Figure 6.45 shows a path that traverses the boundary of *D*. Notice that this path traverses the boundary of region $D_{1},$ returns to the starting point, and then traverses the boundary of region $D_{2}.$ Furthermore, as we walk along the path, the region is always on our left. Notice that this traversal of the $P_{i}$ paths covers the entire boundary of region *D.* If we had only traversed one portion of the boundary of *D*, then we cannot apply Green’s theorem to *D*.

The boundary of the upper half of the annulus, therefore, is $P_{1} \cup P_{2} \cup P_{3} \cup P_{4}$ and the boundary of the lower half of the annulus is $\text{−}P_{4} \cup P_{5} \cup - P_{2} \cup P_{6}.$ Then, Green’s theorem implies

$$\begin{array}{cl} {\int_{\partial D}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{P_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{3}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{4}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{\text{−}P_{4}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{5}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} - {\int_{P_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{6}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}} \\ & {= {\int_{P_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{3}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{4}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} - {\int_{P_{4}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{5}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} - {\int_{P_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{6}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}} \\ & {= {\int_{P_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{3}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{5}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{P_{6}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}} \\ & {= {\int_{\partial D_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} + {\int_{\partial D_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}} \\ & {= {\iint_{D_{1}}{\left( {Q_{x} - P_{y}} \right)dA +}}{\iint_{D_{2}}{\left( {Q_{x} - P_{y}} \right)dA}}} \\ & {= {\iint_{D}{\left( {Q_{x} - P_{y}} \right)dA}}.} \end{array}$$

Therefore, we arrive at the equation found in Green’s theorem—namely,

$${\int_{\partial D}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{D}{\left( {Q_{x} - P_{y}} \right)dA}}.$$

The same logic implies that the flux form of Green’s theorem can also be extended to a region with finitely many holes:

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}} = {\iint_{D}{\left( {P_{x} + Q_{y}} \right)dA.}}$$

Using Green’s Theorem on a Region with Holes 在带孔区域上用格林定理

Calculate integral

$$\oint_{\partial D}\left( \text{sin}\ \text{x} - \frac{y^{3}}{3} \right)dx + \left( \frac{x^{3}}{3} + \text{sin}\ \text{y} \right)dy,$$

where *D* is the annulus given by the polar inequalities $1 \leq \mathbf{\text{r}} \leq 2,$ $0 \leq \theta \leq 2\pi.$

Solution 解

Although *D* is not simply connected, we can use the extended form of Green’s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:

$$\begin{array}{cl} {\int_{\partial D}{\left( {\text{sin}\ x - \frac{y^{3}}{3}} \right)dx + \left( {\frac{x^{3}}{3} + \text{sin}\ y} \right)dy}} & {= {\iint_{D}{\left( {Q_{x} - P_{y}} \right)dA}}} \\ & {= {\iint_{D}{\left( {x^{2} + y^{2}} \right)dA}}} \\ & {= {\int_{0}^{2\pi}{\int_{1}^{2}{r^{3}drd\theta}}} = {\int_{0}^{2\pi}{\frac{15}{4}d\theta}}} \\ & {= \frac{15\pi}{2}.} \end{array}$$

Using the Extended Form of Green’s Theorem 使用格林定理的推广形式

Let $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle = \left\langle {\frac{y}{x^{2} + y^{2}}, - \frac{x}{x^{2} + y^{2}}} \right\rangle$ and let *C* be any simple closed curve in a plane oriented counterclockwise. What are the possible values of $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}?}$

Solution 解

We use the extended form of Green’s theorem to show that $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ is either 0 or $-2\pi$—that is, no matter how crazy curve *C* is, the line integral of F along *C* can have only one of two possible values. We consider two cases: the case when *C* encompasses the origin and the case when *C* does not encompass the origin.

Case 1: *C* Does Not Encompass the Origin 情形 1：*C* 不包含原点

In this case, the region enclosed by *C* is simply connected because the only hole in the domain of F is at the origin. We showed in our discussion of cross-partials that F satisfies the cross-partial condition. If we restrict the domain of F just to *C* and the region it encloses, then F with this restricted domain is now defined on a simply connected domain. Since F satisfies the cross-partial property on its restricted domain, the field F is conservative on this simply connected region and hence the circulation $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ is zero.

Case 2: *C* Does Encompass the Origin 情形 2：曲线 C 包含原点

In this case, the region enclosed by *C* is not simply connected because this region contains a hole at the origin. Let $C_{1}$ be a circle of radius *a* centered at the origin so that $C_{1}$ is entirely inside the region enclosed by *C* (Figure 6.46). Give $C_{1}$ a clockwise orientation.

Let *D* be the region between $C_{1}$ and *C*, and *C* is orientated counterclockwise. By the extended version of Green’s theorem,

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} + {\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}}} & {= \left. \iint{}_{D}{Q_{x} - P_{y}dA} \right.} \\ & {= {\iint_{D}{- \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} +}}\frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}}dA} \\ & {= 0,} \end{array}$$

and therefore

$${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} =}} - {\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}.}}$$

Since $C_{1}$ is a specific curve, we can evaluate ${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$ Let

$$x = a\ \text{cos}\ t,y = {–a}\ \text{sin}\ t,0 \leq t \leq 2\pi$$

be a parameterization of $C_{1}.$ Then,

$$\begin{array}{cl} {\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{0}^{2\pi}{\mathbf{\text{F}}(\mathbf{\text{r}}(t)) \cdot \mathbf{\text{r}}\text{'}(t)dt}}} \\ & {= {\int_{0}^{2\pi}{\left\langle {- \frac{\text{sin}(t)}{a}, - \frac{\text{cos}(t)}{a}} \right\rangle \cdot \left\langle {\text{−}a\ \text{sin}(t),\text{−}a\ \text{cos}(t)} \right\rangle dt}}} \\ & {= {\int_{0}^{2\pi}{\text{sin}^{2}(t) + \text{cos}^{2}(t)dt}} = {\int_{0}^{2\pi}{dt}} = 2\pi.} \end{array}$$

Therefore, ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} =}} - 2\pi.$

Calculate integral $\int_{\partial D}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}$ where *D* is the annulus given by the polar inequalities $2 \leq r \leq 5,0 \leq \theta \leq 2\pi,$ and $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {x^{3},5x + e^{y}\text{sin}\ y} \right\rangle.$

Measuring Area from a Boundary: The Planimeter 由边界测量面积：面积仪

Imagine you are a doctor who has just received a magnetic resonance image of your patient’s brain. The brain has a tumor (Figure 6.47). How large is the tumor? To be precise, what is the area of the red region? The red cross-section of the tumor has an irregular shape, and therefore it is unlikely that you would be able to find a set of equations or inequalities for the region and then be able to calculate its area by conventional means. You could approximate the area by chopping the region into tiny squares (a Riemann sum approach), but this method always gives an answer with some error.

Instead of trying to measure the area of the region directly, we can use a device called a *rolling planimeter* to calculate the area of the region exactly, simply by measuring its boundary. In this project you investigate how a planimeter works, and you use Green’s theorem to show the device calculates area correctly.

A rolling planimeter is a device that measures the area of a planar region by tracing out the boundary of that region (Figure 6.48). To measure the area of a region, we simply run the tracer of the planimeter around the boundary of the region. The planimeter measures the number of turns through which the wheel rotates as we trace the boundary; the area of the shape is proportional to this number of wheel turns. We can derive the precise proportionality equation using Green’s theorem. As the tracer moves around the boundary of the region, the tracer arm rotates and the roller moves back and forth (but does not rotate).

Let *C* denote the boundary of region *D*, the area to be calculated. As the tracer traverses curve *C*, assume the roller moves along the *y*-axis (since the roller does not rotate, one can assume it moves along a straight line). Use the coordinates $\left( {x,y} \right)$ to represent points on boundary *C*, and coordinates $\left( {0,Y} \right)$ to represent the position of the pivot. As the planimeter traces *C*, the pivot moves along the *y*-axis while the tracer arm rotates on the pivot.

Watch a short animation of a planimeter in action.

Begin the analysis by considering the motion of the tracer as it moves from point $\left( {x,y} \right)$ counterclockwise to point $\left( {x + dx,y + dy} \right)$ that is close to $\left( {x,y} \right)$ (Figure 6.49). The pivot also moves, from point $\left( {0,Y} \right)$ to nearby point $\left( {0,Y + dY} \right).$ How much does the wheel turn as a result of this motion? To answer this question, break the motion into two parts. First, roll the pivot along the *y*-axis from $\left( {0,Y} \right)$ to $\left( {0,Y + dY} \right)$ without rotating the tracer arm. The tracer arm then ends up at point $\left( {x,y + dY} \right)$ while maintaining a constant angle $\phi$ with the *x*-axis. Second, rotate the tracer arm by an angle $d\theta$ without moving the roller. Now the tracer is at point $\left( {x + dx,y + dy} \right).$ Let $l$ be the distance from the pivot to the wheel and let *L* be the distance from the pivot to the tracer (the length of the tracer arm).

1. Explain why the total distance through which the wheel rolls the small motion just described is $\text{sin}\ \phi dY + ld\theta = \frac{x}{L}dY + ld\theta.$

2. Show that $\int_{C}{d\theta = 0.}$

3. Use step 2 to show that the total rolling distance of the wheel as the tracer traverses curve *C* is Total wheel roll $= \frac{1}{L}{\int_{C}{xdY.}}$ Now that you have an equation for the total rolling distance of the wheel, connect this equation to Green’s theorem to calculate area *D* enclosed by *C*.

4. Show that $x^{2} + \left( {y - Y} \right)^{2} = L^{2}.$

5. Assume the orientation of the planimeter is as shown in Figure 6.49. Explain why $Y \leq y,$ and use this inequality to show there is a unique value of *Y* for each point $\left( {x,y} \right)\text{:}$ $Y = y - \sqrt{L^{2} - x^{2}}.$

6. Use step 5 to show that $dY = dy + \frac{x}{\sqrt{L^{2} - x^{2}}}dx.$

7. Use Green’s theorem to show that $\int_{C}{\frac{x}{\sqrt{L^{2} - x^{2}}}dx = 0.}$

8. Use step 7 to show that the total wheel roll is Total wheel roll $= \frac{1}{L}{\int_{C}{xdy.}}$ It took a bit of work, but this equation says that the variable of integration *Y* in step 3 can be replaced with *y*.

9. Use Green’s theorem to show that the area of *D* is $\int_{C}{xdy.}$ The logic is similar to the logic used to show that the area of $D = \frac{1}{2}{\int_{C}{\text{−}ydx + xdy.}}$

10. Conclude that the area of *D* equals the length of the tracer arm multiplied by the total rolling distance of the wheel. You now know how a planimeter works and you have used Green’s theorem to justify that it works. To calculate the area of a planar region *D*, use a planimeter to trace the boundary of the region. The area of the region is the length of the tracer arm multiplied by the distance the wheel rolled.

Section 6.4 Exercises 6.4 节习题

For the following exercises, evaluate the line integrals by applying Green’s theorem.

146\.

${\int_{C}^{}{2xydx + (x + y)dy}},$ where *C* is the path from (0, 0) to (1, 1) along the graph of $y = x^{3}$ and from (1, 1) to (0, 0) along the graph of $y = x$ oriented in the counterclockwise direction

147.

${\int_{C}^{}{2xydx + (x + y)dy}},$ where *C* is the boundary of the region lying between the graphs of $y = 0$ and $y = 4 - x^{2}$ oriented in the counterclockwise direction

148\.

$\int_{C}^{}{2\ \text{arctan}\left( \frac{y}{x} \right)dx + \text{ln}\left( {x^{2} + y^{2}} \right)dy,}$ where *C* is defined by $x = 4 + 2\ \text{cos}\ \theta,y = 4\ \text{sin}\ \theta$ oriented in the counterclockwise direction

149.

${\int_{C}^{}{\text{sin}\ x\ \text{cos}\ ydx + (xy + \text{cos}\ x\ \text{sin}\ y)dy}}\text{,}$ where *C* is the boundary of the region lying between the graphs of $y = x$ and $y = \sqrt{x}$ oriented in the counterclockwise direction

150\.

${\int_{C}^{}{xydx + (x + y)dy}},$ where *C* is the boundary of the region lying between the graphs of $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} = 9$ oriented in the counterclockwise direction

151.

${\int_{C}{(\text{−}ydx + xdy)}},$ where *C* consists of line segment *C*₁ from $\left( {-1,0} \right)$ to (1, 0), followed by the semicircular arc *C*₂ from (1, 0) back to (–1, 0)

For the following exercises, use Green’s theorem.

152\.

Let *C* be the curve consisting of line segments from (0, 0) to (1, 1) to (0, 1) and back to (0, 0). Find the value of $\int_{C}{xydx + \sqrt{y^{2} + 1}dy.}$

153.

Evaluate line integral $\int_{C}{xe^{-2x}dx + \left( {x^{4} + 2x^{2}y^{2}} \right)dy,}$ where *C* is the boundary of the region between circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} = 4,$ and is a positively oriented curve.

154\.

Find the counterclockwise circulation of field $\mathbf{\text{F}}\left( {x,y} \right) = xy\mathbf{\text{i}} + y^{2}\mathbf{\text{j}}$ around and over the boundary of the region enclosed by curves $y = x^{2}$ and $y = x$ in the first quadrant and oriented in the counterclockwise direction.

155.

Evaluate ${\int_{C}{y^{3}dx - x^{3}y^{2}dy}},$ where *C* is the positively oriented circle of radius 2 centered at the origin.

156\.

Evaluate ${\int_{C}{y^{3}dx - x^{3}dy}},$ where *C* includes the two circles of radius 2 and radius 1 centered at the origin, both with positive orientation.

157.

Calculate ${\int_{C}{\text{−}x^{2}ydx + xy^{2}dy}},$ where *C* is a circle of radius 2 centered at the origin and oriented in the counterclockwise direction.

158\.

Calculate integral $\int_{C}{2\left\lbrack {y + x\ \text{sin}(y)} \right\rbrack dx + \left\lbrack {x^{2}\text{cos}(y) - 3y^{2}} \right\rbrack dy}$ along triangle *C* with vertices (0, 0), (1, 0) and (1, 1), oriented counterclockwise, using Green’s theorem.

159.

Evaluate integral ${\int_{C}{\left( {x^{2} + y^{2}} \right)dx + 2xydy}},$ where *C* is the curve that follows parabola $y = x^{2}\ \text{from}\ \left( {0,0} \right)\text{to}\left( {2,4} \right),$ then the line from (2, 4) to (2, 0), and finally the line from (2, 0) to (0, 0).

160\.

Evaluate line integral ${\int_{C}{(y - \text{sin}(y)\text{cos}(y))dx + 2x\ \text{sin}^{2}(y)dy}},$ where *C* is oriented in a counterclockwise path around the region bounded by $x = -1,x = 2,y = 4 - x^{2},$ and $y = x - 2.$

For the following exercises, use Green’s theorem to find the area.

161.

Find the area between ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ and circle $x^{2} + y^{2} = 25.$

162\.

Find the area of the region enclosed by parametric equation

$$p(\theta) = \left( {\text{cos}(\theta) - \text{cos}^{2}(\theta)} \right)\mathbf{\text{i}} + \left( {\text{sin}(\theta) - \text{cos}(\theta)\text{sin}(\theta)} \right)\mathbf{\text{j}}\ \text{for}\ 0 \leq \theta \leq 2\pi.$$ 163.

Find the area of the region bounded by hypocycloid $\mathbf{\text{r}}(t) = \text{cos}^{3}(t)\mathbf{\text{i}} + \text{sin}^{3}(t)\mathbf{\text{j}}.$ The curve is parameterized by $t \in \left\lbrack {0,2\pi} \right\rbrack.$

164\.

Find the area of a pentagon with vertices $(0,4),(4,1),(3,0),(-1,-1),$ and $(-2,2).$

165.

Use Green’s theorem to evaluate $\int_{C^{+}}\left( y^{2} + x^{3} \right)dx + x^{4}dy,$ where $C^{+}$ is the perimeter of square $\left\lbrack {0,1} \right\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack$ oriented counterclockwise.

166\.

Use Green’s theorem to prove the area of a disk with radius $a$ is $A = \pi a^{2}.$

167.

Use Green’s theorem to find the area of one loop of a four-leaf rose $r = 3\ \text{sin}\ 2\theta.$ (*Hint*: $xdy - ydx = r^{2}d\theta).$

168\.

Use Green’s theorem to find the area under one arch of the cycloid given by parametric plane $x = t - \text{sin}\ t,y = 1 - \text{cos}\ t,t \geq 0.$

169.

Use Green’s theorem to find the area of the region enclosed by curve

$$\mathbf{\text{r}}(t) = t^{2}\mathbf{\text{i}} + \left( {\frac{t^{3}}{3} - t} \right)\mathbf{\text{j}}\text{,}\ - \sqrt{3} \leq t \leq \sqrt{3}.$$ 170.

\[T\] Verify Green’s theorem by using a computer algebra system to evaluate the integral $\int_{C}{xe^{y}dx + e^{x}dy,}$ where *C* is the circle given by $x^{2} + y^{2} = 4$ and is oriented in the counterclockwise direction.

171.

Evaluate $\int_{C}{\left( {x^{2}y - 2xy + y^{2}} \right)ds,}$ where *C* is the boundary of the unit square $0 \leq x \leq 1\text{,}\ 0 \leq y \leq 1\text{,}$ traversed counterclockwise. Use $ds~ = ~dx~ + ~dy$ for this exercise.

172\.

Evaluate ${\int_{C}^{}\frac{\text{−}(y + 2)dx + (x - 1)dy}{{(x - 1)}^{2} + {(y + 2)}^{2}}}\text{,}$ where *C* is any simple closed curve with an interior that does not contain point $(1,-2)$ traversed counterclockwise.

173.

Evaluate ${\int_{C}^{}\frac{xdx + ydy}{x^{2} + y^{2}}},$ where *C* is any piecewise, smooth simple closed curve enclosing the origin, traversed counterclockwise.

For the following exercises, use Green’s theorem to calculate the work done by force F on a particle that is moving counterclockwise around closed path *C*.

174\.

$\mathbf{\text{F}}(x,y) = xy\mathbf{\text{i}} + (x + y)\mathbf{\text{j}},$ $C:x^{2} + y^{2} = 4$

175.

$\mathbf{\text{F}}(x,y) = \left( {x^{3\text{/}2} - 3y} \right)\mathbf{\text{i}} + \left( {6x + 5\sqrt{y}} \right)\mathbf{\text{j}},$ *C* : boundary of a triangle with vertices (0, 0), (5, 0), and (0, 5)

176\.

Evaluate $\int_{C}{\left( {2x^{3} - y^{3}} \right)dx + \left( {x^{3} + y^{3}} \right)dy,}$ where *C* is a unit circle oriented in the counterclockwise direction.

177.

A particle starts at point $(-2,0),$ moves along the *x*-axis to (2, 0), and then travels along semicircle $y = \sqrt{4 - x^{2}}$ to the starting point. Use Green’s theorem to find the work done on this particle by force field $\mathbf{\text{F}}(x,y) = x\mathbf{\text{i}} + \left( {x^{3} + 3xy^{2}} \right)\mathbf{\text{j}}.$

178\.

David and Sandra are skating on a frictionless pond in the wind. David skates on the inside, going along a circle of radius 2 in a counterclockwise direction. Sandra skates once around a circle of radius 3, also in the counterclockwise direction. Suppose the force of the wind at point $\left( {x,y} \right)$ $\left( {x,y} \right)$ $\left( {x,y} \right)$ is $\mathbf{\text{F}}(x,y) = \left( {x^{2}y + 10y} \right)\mathbf{\text{i}} + \left( {x^{3} + 2xy^{2}} \right)\mathbf{\text{j}}.$ Use Green’s theorem to determine who does more work.

179.

Use Green’s theorem to find the work done by force field $\mathbf{\text{F}}(x,y) = (3y - 4x)\mathbf{\text{i}} + (4x - y)\mathbf{\text{j}}$ when an object moves once counterclockwise around ellipse $4x^{2} + y^{2} = 4.$

180\.

Use Green’s theorem to evaluate line integral ${\int_{C}{e^{2x}\text{sin}\ 2ydx + e^{2x}\text{cos}\ 2ydy}},$ where *C* is ellipse $9{(x - 1)}^{2} + 4{(y - 3)}^{2} = 36$ oriented counterclockwise.

181.

Evaluate line integral $\int_{C}{y^{2}dx + x^{2}dy,}$ where *C* is the boundary of a triangle with vertices $\left( {0,0} \right),\left( {1,1} \right),\ \text{and}\ \left( {1,0} \right),$ with the counterclockwise orientation.

182\.

Use Green’s theorem to evaluate line integral $\int_{C}{\mathbf{\text{h}} \cdot d\mathbf{\text{r}}}$ if $\mathbf{\text{h}}\left( {x,y} \right) = e^{y}\mathbf{\text{i}} - \text{sin}\ \pi x\mathbf{\text{j}},$ where *C* is a triangle with vertices (1, 0), (0, 1), and (–1, 0) traversed counterclockwise.

183.

Use Green’s theorem to evaluate line integral $\int_{C}^{}{\sqrt{1 + x^{3}}dx + 2xydy}$ where *C* is a triangle with vertices (0, 0), (1, 0), and (1, 3) oriented clockwise.

184\.

Use Green’s theorem to evaluate line integral $\int_{C}^{}{x^{2}ydx - xy^{2}dy}$ where *C* is a circle $x^{2} + y^{2} = 4$ oriented counterclockwise.

185.

Use Green’s theorem to evaluate line integral ${\int_{C}^{}{\left( {3y - e^{\text{sin}\ x}} \right)dx}} + \left( {7x + \sqrt{y^{4} + 1}} \right)dy$ where *C* is circle $x^{2} + y^{2} = 9$ oriented in the counterclockwise direction.

186\.

Use Green’s theorem to evaluate line integral ${\int_{C}^{}{(3x - 5y)dx}} + (x - 6y)dy\text{,}$ where *C* is ellipse $\frac{x^{2}}{4} + y^{2} = 1$ and is oriented in the counterclockwise direction.

187.

Let *C* be a triangular closed curve from (0, 0) to (1, 0) to (1, 1) and finally back to (0, 0). Let $\mathbf{\text{F}}(x,y) = 4y\mathbf{\text{i}} + 6x^{2}\mathbf{\text{j}}.$ Use Green’s theorem to evaluate ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$

188\.

Use Green’s theorem to evaluate line integral ${\int_{C}{ydx - xdy}}\text{,}$ where *C* is circle $x^{2} + y^{2} = a^{2}$ oriented in the clockwise direction.

189.

Use Green’s theorem to evaluate line integral ${\int_{C}{(y + x)dx + (x + \text{sin}\ y)dy}}\text{,}$ where *C* is any smooth simple closed curve joining the origin to itself oriented in the counterclockwise direction.

190\.

Use Green’s theorem to evaluate line integral ${\int_{C}^{}{\left( {y - \text{ln}\left( {x^{2} + y^{2}} \right)} \right)dx + \left( {2\ \text{arctan}\ \frac{y}{x}} \right)dy}},$ where C is the positively oriented circle $\left( {x - 2} \right)^{2} + \left( {y - 3} \right)^{2} = 1.$

191.

Use Green’s theorem to evaluate $\int_{C}{xydx + x^{3}y^{3}dy,}$ where *C* is a triangle with vertices (0, 0), (1, 0), and (1, 2) with positive orientation.

192\.

Use Green’s theorem to evaluate line integral ${\int_{C}^{}{\text{sin}\ ydx + x\ \text{cos}\ ydy}},$ where *C* is ellipse $x^{2} + xy + y^{2} = 1$ oriented in the counterclockwise direction.

193.

Let $\mathbf{\text{F}}(x,y) = \left( {\text{cos}\left( x^{5} \right)} \right) - \frac{1}{3}y^{3}\mathbf{\text{i}} + \frac{1}{3}x^{3}\mathbf{\text{j}}.$ Find the counterclockwise circulation ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where *C* is a curve consisting of the line segment joining $(-2,0)\ \text{and}\ (-1,0)\text{,}$ half circle $y = \sqrt{1 - x^{2}},$ the line segment joining (1, 0) and (2, 0), and half circle $y = \sqrt{4 - x^{2}}.$

194\.

Use Green’s theorem to evaluate line integral ${\int_{C}^{}{\text{sin}\left( x^{3} \right)dx + 2ye^{x^{2}}dy}},$ where *C* is a triangular closed curve that connects the points (0, 0), (2, 2), and (0, 2) counterclockwise.

195.

Let *C* be the boundary of square $0 \leq x \leq \pi\text{,}\ 0 \leq y \leq \pi\text{,}$ traversed counterclockwise. Use Green’s theorem to find ${\int_{C}^{}{\text{sin}(x + y)dx + \text{cos}(x + y)dy}}.$

196\.

Use Green’s theorem to evaluate line integral ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = \left( {y^{2} - x^{2}} \right)\mathbf{\text{i}} + \left( {x^{2} + y^{2}} \right)\mathbf{\text{j}},$ and *C* is a triangle bounded by $y = 0\text{,}\ x = 3,\ \text{and}\ y = x\text{,}$ oriented counterclockwise.

197.

Use Green’s Theorem to evaluate integral ${\int_{C}^{}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}}(x,y) = \left( {xy^{2}} \right)\mathbf{\text{i}} + x\mathbf{\text{j}},$ and *C* is a unit circle oriented in the counterclockwise direction.

198\.

Use Green’s theorem in a plane to evaluate line integral ${\int_{C}{\left( {xy + y^{2}} \right)dx + x^{2}dy}},$ where *C* is a closed curve of a region bounded by $y = x\ \text{and}\ y = x^{2}$ oriented in the counterclockwise direction.

199.

Calculate the outward flux of $\mathbf{\text{F}} = \text{−}x\mathbf{\text{i}} + 2y\mathbf{\text{j}}$ over a square with corners $\left( {\pm 1,\pm 1} \right),$ where the unit normal is outward pointing and oriented in the counterclockwise direction.

200\.

\[T\] Let *C* be circle $x^{2} + y^{2} = 4$ oriented in the counterclockwise direction. Evaluate $\int_{C}\left\lbrack {\left( {3y - e^{\text{tan} - 1_{x}}} \right)dx + \left( {7x + \sqrt{y^{4} + 1}} \right)dy} \right\rbrack$ using a computer algebra system.

201.

Find the flux of field $\mathbf{\text{F}} = \text{−}x\mathbf{\text{i}} + y\mathbf{\text{j}}$ across $x^{2} + y^{2} = 16$ oriented in the counterclockwise direction.

202\.

Let $\mathbf{\text{F}} = \left( {y^{2} - x^{2}} \right)\mathbf{\text{i}} + \left( {x^{2} + y^{2}} \right)\mathbf{\text{j}},$ and let *C* be a triangle bounded by $y = 0,x = 3,$ and $y = x$ oriented in the counterclockwise direction. Find the outward flux of F through *C*.

203.

\[T\] Let *C* be unit circle $x^{2} + y^{2} = 1$ traversed once counterclockwise. Evaluate $\int_{C}{\left\lbrack {\text{−}y^{3} + \text{sin}\left( {xy} \right) + xy\ \text{cos}\left( {xy} \right)} \right\rbrack dx + \left\lbrack {x^{3} + x^{2}\text{cos}\left( {xy} \right)} \right\rbrack dy}$ by using a computer algebra system.

204\.

\[T\] Find the outward flux of vector field $\mathbf{\text{F}} = xy^{2}\mathbf{\text{i}} + x^{2}y\mathbf{\text{j}}$ across the boundary of annulus $R = \left\{ {\left( {x,y} \right):1 \leq x^{2} + y^{2} \leq 4} \right\} = \left\{ {\left( {r,\theta} \right):1 \leq r \leq 2,0 \leq \theta \leq 2\pi} \right\}$ using a computer algebra system.

205.

Consider region *R* bounded by parabolas $y = x^{2}\ \text{and}\ x = y^{2}.$ Let *C* be the boundary of *R* oriented counterclockwise. Use Green’s theorem to evaluate $\int_{C}{\left( {y + e^{\sqrt{x}}} \right)dx + \left( {2x + \text{cos}\left( y^{2} \right)} \right)dy.}$

6.5 Divergence and Curl 6.5 散度与旋度

• 6.5.1 Determine divergence from the formula for a given vector field.
• 6.5.2 Determine curl from the formula for a given vector field.
• 6.5.3 Use the properties of curl and divergence to determine whether a vector field is conservative.

• 6.5.1 根据给定向量场的公式求散度。
• 6.5.2 根据给定向量场的公式求旋度。
• 6.5.3 利用旋度与散度的性质判断一个向量场是否保守。

In this section, we examine two important operations on a vector field: divergence and curl. They are important to the field of calculus for several reasons, including the use of curl and divergence to develop some higher-dimensional versions of the Fundamental Theorem of Calculus. In addition, curl and divergence appear in mathematical descriptions of fluid mechanics, electromagnetism, and elasticity theory, which are important concepts in physics and engineering. We can also apply curl and divergence to other concepts we already explored. For example, under certain conditions, a vector field is conservative if and only if its curl is zero.

In addition to defining curl and divergence, we look at some physical interpretations of them, and show their relationship to conservative and source-free vector fields.

Divergence 散度

Divergence is an operation on a vector field that tells us how the field behaves toward or away from a point. Locally, the divergence of a vector field F in $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$ at a particular point *P* is a measure of the “outflowing-ness” of the vector field at *P*. If F represents the velocity of a fluid, then the divergence of F at *P* measures the net rate of change with respect to time of the amount of fluid flowing away from *P* (the tendency of the fluid to flow “out of” *P*). In particular, if the amount of fluid flowing into *P* is the same as the amount flowing out, then the divergence at *P* is zero.

If $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ is a vector field in $\mathbb{R}^{3}$ and $P_{x},Q_{y},$ and $R_{z}$ all exist, then the divergence of F is defined by

$$\text{div}\ \mathbf{\text{F}} = P_{x}\text{+}Q_{y} + R_{z} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}.$$ (6.16)

Note the divergence of a vector field is not a vector field, but a scalar function. In terms of the gradient operator $\nabla = \left\langle {\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}} \right\rangle,$ divergence can be written symbolically as the dot product

$$\text{div}\ \mathbf{\text{F}} = \nabla \cdot \mathbf{\text{F}}.$$

Note this is merely helpful notation, because the dot product of a vector of operators and a vector of functions is not meaningfully defined given our current definition of dot product.

If $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a vector field in $\mathbb{R}^{2},$ and $P_{x}$ and $Q_{y}$ both exist, then the divergence of F is defined similarly as

$$\text{div}\ \mathbf{\text{F}} = P_{x} + Q_{y} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = \nabla \cdot \mathbf{\text{F}}.$$

To illustrate this point, consider the two vector fields in Figure 6.50. At any particular point, the amount flowing in is the same as the amount flowing out, so at every point the “outflowing-ness” of the field is zero. Therefore, we expect the divergence of both fields to be zero, and this is indeed the case, as

$$\text{div}\left( \left\langle {1,2} \right\rangle \right) = \frac{\partial}{\partial x}(1) + \frac{\partial}{\partial y}(2) = 0\ \text{and}\ \text{div}\left( \left\langle {\text{−}y,x} \right\rangle \right) = \frac{\partial}{\partial x}\left( {\text{−}y} \right) + \frac{\partial}{\partial y}(x) = 0.$$

By contrast, consider radial vector field $\mathbf{\text{R}}\left( {x,y} \right) = \left\langle {\text{−}x,\text{−}y} \right\rangle$ in Figure 6.51. At any given point, more fluid is flowing in than is flowing out, and therefore the “outgoingness” of the field is negative. We expect the divergence of this field to be negative, and this is indeed the case, as $\text{div}\left( \mathbf{\text{R}} \right) = \frac{\partial}{\partial x}\left( {\text{−}x} \right) + \frac{\partial}{\partial y}\left( {\text{−}y} \right) = -2.$

To get a global sense of what divergence is telling us, suppose that a vector field in $\mathbb{R}^{2}$ represents the velocity of a fluid. Imagine taking an elastic circle (a circle with a shape that can be changed by the vector field) and dropping it into a fluid. If the circle maintains its exact area as it flows through the fluid, then the divergence is zero. This would occur for both vector fields in Figure 6.50. On the other hand, if the circle’s shape is distorted so that its area shrinks or expands, then the divergence is not zero. Imagine dropping such an elastic circle into the radial vector field in Figure 6.51 so that the center of the circle lands at point (3, 3). The circle would flow toward the origin, and as it did so the front of the circle would travel more slowly than the back, causing the circle to “scrunch” and lose area. This is how you can see a negative divergence.

Calculating Divergence at a Point 在某点计算散度

If $\mathbf{\text{F}}\left( {x,y,z} \right) = e^{x}\mathbf{\text{i}} + yz\mathbf{\text{j}} - yz^{2}\mathbf{\text{k}},$ then find the divergence of F at $\left( {0,2,-1} \right).$

Solution 解

The divergence of F is

$$\frac{\partial}{\partial x}\left( e^{x} \right) + \frac{\partial}{\partial y}\left( {yz} \right) - \frac{\partial}{\partial z}\left( {yz^{2}} \right) = e^{x} + z - 2yz.$$

Therefore, the divergence at $\left( {0,2,-1} \right)$ is $e^{0} - 1 + 4 = 4.$ If F represents the velocity of a fluid, then more fluid is flowing out than flowing in at point $\left( {0,2,-1} \right).$

Find $\text{div}\ \mathbf{\text{F}}$ for $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {xy,5 - z^{2}y,x^{2} + y^{2}} \right\rangle.$

One application for divergence occurs in physics, when working with magnetic fields. A magnetic field is a vector field that models the influence of electric currents and magnetic materials. Physicists use divergence in Gauss’s law for magnetism, which states that if B is a magnetic field, then $\nabla \cdot \mathbf{\text{B}} = 0;$ in other words, the divergence of a magnetic field is zero.

Determining Whether a Field Is Magnetic 判断一个场是否为磁场

Is it possible for $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {x^{2}y,y - xy^{2}} \right\rangle$ to be a magnetic field?

Solution 解

If F were magnetic, then its divergence would be zero. The divergence of F is

$$\frac{\partial}{\partial x}\left( {x^{2}y} \right) + \frac{\partial}{\partial y}\left( {y - xy^{2}} \right) = 2xy + 1 - 2xy = 1$$

and therefore F cannot model a magnetic field (Figure 6.52).

Another application for divergence is detecting whether a field is source free. Recall that a source-free field is a vector field that has a stream function; equivalently, a source-free field is a field with a flux that is zero along any closed curve. The next two theorems say that, under certain conditions, source-free vector fields are precisely the vector fields with zero divergence.

Divergence of a Source-Free Vector Field 无源向量场的散度

If $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a source-free continuous vector field with differentiable component functions, then $\text{div}\ \mathbf{\text{F}} = 0.$

Proof 证明

Since F is source free, there is a function $g\left( {x,y} \right)$ with $g_{y} = P$ and $\text{−}g_{x} = Q.$ Therefore, $\mathbf{\text{F}} = \left\langle {g_{y},\text{−}g_{x}} \right\rangle$ and $\text{div}\ \mathbf{\text{F}} = g_{yx} - g_{xy} = 0$ by Clairaut’s theorem.

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The converse of Divergence of a Source-Free Vector Field is true on simply connected regions, but the proof is too technical to include here. Thus, we have the following theorem, which can test whether a vector field in $\mathbb{R}^{2}$ is source free.

Divergence Test for Source-Free Vector Fields 无源向量场的散度判别法

Let $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ be a continuous vector field with differentiable component functions with a domain that is simply connected. Then, $\text{div}\ \mathbf{\text{F}} = 0$ if and only if F is source free.

Determining Whether a Field Is Source Free 判断一个场是否无源

Is field $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {x^{2}y,5 - xy^{2}} \right\rangle$ source free?

Solution 解

Note the domain of F is $\mathbb{R}^{2},$ which is simply connected. Furthermore, F is continuous with differentiable component functions. Therefore, we can use Divergence Test for Source-Free Vector Fields to analyze F. The divergence of F is

$$\frac{\partial}{\partial x}\left( {x^{2}y} \right) + \frac{\partial}{\partial y}\left( {5 - xy^{2}} \right) = 2xy - 2xy = 0.$$

Therefore, F is source free by Divergence Test for Source-Free Vector Fields.

Let $\mathbf{\text{F}}\left( {x,y} \right) = \left\langle {\text{−}ay,bx} \right\rangle$ be a rotational field where *a* and *b* are positive constants. Is F source free?

Recall that the flux form of Green’s theorem says that

$${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds = {\iint_{D}{P_{x} + Q_{y}dA}}}},$$

where *C* is a simple closed curve and *D* is the region enclosed by *C*. Since $P_{x} + Q_{y} = \text{div}\ \mathbf{\text{F}},$ Green’s theorem is sometimes written as

$$\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds = {\iint_{D}{\text{div}\ \mathbf{\text{F}}dA.}}}$$

Therefore, Green’s theorem can be written in terms of divergence. If we think of divergence as a derivative of sorts, then Green’s theorem says the “derivative” of F on a region can be translated into a line integral of F along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function $f$ on a line segment $\lbrack a,b\rbrack$ can be translated into a statement about $f$ on the boundary of $\lbrack a,b\rbrack.$ Using divergence, we can see that Green’s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.

We can use all of what we have learned in the application of divergence. Let v be a vector field modeling the velocity of a fluid. Since the divergence of v at point *P* measures the “outflowing-ness” of the fluid at *P*, $\text{div}\ \mathbf{\text{v}}(P) > 0$ implies that more fluid is flowing out of *P* than flowing in. Similarly, $\text{div}\ \mathbf{\text{v}}(P) < 0$ implies the more fluid is flowing in to *P* than is flowing out, and $\text{div}\ \mathbf{\text{v}}(P) = 0$ implies the same amount of fluid is flowing in as flowing out.

Determining Flow of a Fluid 判断流体的流动

Suppose $\mathbf{\text{v}}(x,y) = \left\langle {\text{−}xy,y} \right\rangle,y > 0$ models the flow of a fluid. Is more fluid flowing into point $(1,4)$ than flowing out?

Solution 解

To determine whether more fluid is flowing into $(1,4)$ than is flowing out, we calculate the divergence of v at ${(1,4)}\text{:}$

$$\text{div}\left( \mathbf{\text{v}} \right) = \frac{\partial}{\partial x}\left( {\text{−}xy} \right) + \frac{\partial}{\partial y}(y) = \text{−}y + 1.$$

To find the divergence at $(1,4),$ substitute the point into the divergence: $-4 + 1 = -3.$ Since the divergence of v at $(1,4)$ is negative, more fluid is flowing in than flowing out (Figure 6.53).

For vector field $\mathbf{\text{v}}(x,y) = \left\langle {\text{−}xy,y} \right\rangle,y > 0,$ find all points *P* such that the amount of fluid flowing in to *P* equals the amount of fluid flowing out of *P*.

Curl 旋度

The second operation on a vector field that we examine is the curl, which measures the extent of rotation of the field about a point. Suppose that F represents the velocity field of a fluid. Then, the curl of F at point *P* is a vector that measures the tendency of particles near *P* to rotate about the axis that points in the direction of this vector. The magnitude of the curl vector at *P* measures how quickly the particles rotate around this axis. In other words, the curl at a point is a measure of the vector field’s “spin” at that point. Visually, imagine placing a paddlewheel into a fluid at *P*, with the axis of the paddlewheel aligned with the curl vector (Figure 6.54). The curl measures the tendency of the paddlewheel to rotate.

Consider the vector fields in Figure 6.50. In part (a), the vector field is constant and there is no spin at any point. Therefore, we expect the curl of the field to be zero, and this is indeed the case. Part (b) shows a rotational field, so the field has spin. In particular, if you place a paddlewheel into a field at any point so that the axis of the wheel is perpendicular to a plane, the wheel rotates counterclockwise. Therefore, we expect the curl of the field to be nonzero, and this is indeed the case (the curl is $2\mathbf{\text{k}}).$

To see what curl is measuring globally, imagine dropping a leaf into the fluid. As the leaf moves along with the fluid flow, the curl measures the tendency of the leaf to rotate. If the curl is zero, then the leaf doesn’t rotate as it moves through the fluid.

If $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ is a vector field in $\mathbb{R}^{3},$ and $P_{x},~P_{y},~P_{z},~Q_{y},~Q_{x},~Q_{z},~R_{z},~R_{x},~{and}~R_{y}$ all exist, then the curl of F is defined by

$$\begin{array}{cl} {\text{curl}\ \mathbf{\text{F}}} & {= \left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}}} \\ & {= \left( {\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}} \right)\mathbf{\text{i}} + \left( {\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}} \right)\mathbf{\text{j}} + \left( {\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}} \right)\mathbf{\text{k}}.} \end{array}$$ (6.17)

Note that the curl of a vector field is a vector field, in contrast to divergence.

The definition of curl can be difficult to remember. To help with remembering, we use the notation $\nabla\ \times \ \mathbf{\text{F}}$ to stand for a “determinant” that gives the curl formula:

$$\left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} \right|.$$

The determinant of this matrix is

$$\left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} - \left( {R_{x} - P_{z}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}} = \left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}} = \text{curl}\ \mathbf{\text{F}}.$$

Thus, this matrix is a way to help remember the formula for curl. Keep in mind, though, that the word *determinant* is used very loosely. A determinant is not really defined on a matrix with entries that are three vectors, three operators, and three functions.

If $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a vector field in $\mathbb{R}^{2},$ then the curl of F, by definition, is

$$\text{curl}\ \mathbf{\text{F}} = \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}} = \left( {\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}} \right)\mathbf{\text{k}}.$$

Finding the Curl of a Three-Dimensional Vector Field 求三维向量场的旋度

Find the curl of $\mathbf{\text{F}}\left( {P,Q,R} \right) = \left\langle {x^{2}z,e^{y} + xz,xyz} \right\rangle.$

Solution 解

The curl is

$$\begin{array}{cl} {\text{curl}\ \mathbf{\text{F}}} & {= \nabla\ \times \ \mathbf{\text{F}}} \\ & {= \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ {\partial\text{/}{\partial x}} & {\partial\text{/}{\partial y}} & {\partial\text{/}{\partial z}} \\ P & Q & R \end{matrix} \right|} \\ & {= \left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}}} \\ & {= \left( {xz - x} \right)\mathbf{\text{i}} + \left( {x^{2} - yz} \right)\mathbf{\text{j}} + z\mathbf{\text{k}}.} \end{array}$$

Find the curl of $\mathbf{\text{F}} = \left\langle {\text{sin}\mspace{2mu} x\ \text{cos}\mspace{2mu} z,\text{sin}\mspace{2mu} y\ \text{sin}\mspace{2mu} z,\text{cos}\mspace{2mu} x\ \text{cos}\mspace{2mu} y} \right\rangle$ at point $\left( {0,\frac{\pi}{2},\frac{\pi}{2}} \right).$

Finding the Curl of a Two-Dimensional Vector Field 求二维向量场的旋度

Find the curl of $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle = \left\langle {y,0} \right\rangle.$

Solution 解

Notice that this vector field consists of vectors that are all parallel. In fact, each vector in the field is parallel to the *x*-axis. This fact might lead us to the conclusion that the field has no spin and that the curl is zero. To test this theory, note that

$$\text{curl}\ \mathbf{\text{F}} = \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}} = \text{−}\mathbf{\text{k}} \neq 0.$$

Therefore, this vector field does have spin. To see why, imagine placing a paddlewheel at any point in the first quadrant (Figure 6.55). The larger magnitudes of the vectors at the top of the wheel cause the wheel to rotate. The wheel rotates in the clockwise (negative) direction, causing the coefficient of the curl to be negative.

Note that if $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a vector field in a plane, then $\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{k}} = \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}} \cdot \mathbf{\text{k}} = Q_{x} - P_{y}.$ Therefore, the circulation form of Green’s theorem is sometimes written as

$${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} =}}{\iint_{D}{\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{k}}dA}},$$

where *C* is a simple closed curve and *D* is the region enclosed by *C*. Therefore, the circulation form of Green’s theorem can be written in terms of the curl. If we think of curl as a derivative of sorts, then Green’s theorem says that the “derivative” of F on a region can be translated into a line integral of F along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function $f$ on line segment $\lbrack a,b\rbrack$ can be translated into a statement about $f$ on the boundary of $\lbrack a,b\rbrack.$ Using curl, we can see the circulation form of Green’s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.

We can now use what we have learned about curl to show that gravitational fields have no “spin.” Suppose there is an object at the origin with mass $m_{1}$ at the origin and an object with mass $m_{2}.$ Recall that the gravitational force that object 1 exerts on object 2 is given by field

$$\mathbf{\text{F}}(x,y,z) = \text{−}Gm_{1}m_{2}\left\langle {\frac{x}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}},\frac{y}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}},\frac{z}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}}} \right\rangle.$$

Determining the Spin of a Gravitational Field 确定引力场的自旋

Show that a gravitational field has no spin.

Solution 解

To show that F has no spin, we calculate its curl. Let $P(x,y,z) = \frac{x}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}},$ $Q(x,y,z) = \frac{y}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}},$ and $R(x,y,z) = \frac{z}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}}.$ Then,

$$\begin{array}{cl} {\text{curl}\ \mathbf{\text{F}}} & {= \text{−}Gm_{1}m_{2}\left\lbrack {\left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}}} \right\rbrack} \\ & {= \text{−}Gm_{1}m_{2}\left\lbrack \begin{array}{l} {\left( {\frac{-3yz}{\left( {x^{2} + y^{2} + z^{2}} \right)^{5\text{/}2}} - \left( \frac{-3yz}{\left( {x^{2} + y^{2} + z^{2}} \right)^{5\text{/}2}} \right)} \right)\mathbf{\text{i}}} \\ {+ \left( {\frac{-3xz}{\left( {x^{2} + y^{2} + z^{2}} \right)^{5\text{/}2}} - \left( \frac{-3xz}{\left( {x^{2} + y^{2} + z^{2}} \right)^{5\text{/}2}} \right)} \right)\mathbf{\text{j}}} \\ {+ \left( {\frac{-3xy}{\left( {x^{2} + y^{2} + z^{2}} \right)^{5\text{/}2}} - \left( \frac{-3xy}{\left( {x^{2} + y^{2} + z^{2}} \right)^{5\text{/}2}} \right)} \right)\mathbf{\text{k}}} \end{array} \right\rbrack} \\ & {= 0.} \end{array}$$

Since the curl of the gravitational field is zero, the field has no spin.

Field $\mathbf{\text{v}}(x,y) = \left\langle {- \frac{y}{x^{2} + y^{2}},\frac{x}{x^{2} + y^{2}}} \right\rangle$ models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.

Using Divergence and Curl 利用散度与旋度

Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.

If F is a vector field in $\mathbb{R}^{3},$ then the curl of F is also a vector field in $\mathbb{R}^{3}.$ Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field. To give this result a physical interpretation, recall that divergence of a velocity field v at point *P* measures the tendency of the corresponding fluid to flow out of *P*. Since $\text{div}\ \text{curl}\ \left( \mathbf{\text{v}} \right) = 0,$ the net rate of flow in vector field curl(v) at any point is zero. Taking the curl of vector field F eliminates whatever divergence was present in F.

Divergence of the Curl 旋度的散度

Let $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ be a vector field in $\mathbb{R}^{3}$ such that the component functions all have continuous second-order partial derivatives. Then, $\text{div}\ \text{curl}\ \left( \mathbf{\text{F}} \right) = \nabla \cdot \left( {\nabla\ \times \ \mathbf{\text{F}}} \right) = 0.$

Proof 证明

By the definitions of divergence and curl, and by Clairaut’s theorem,

$$\begin{array}{cl} {\text{div curl}\ \mathbf{\text{F}}} & {= \text{div}\left\lbrack {\left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}}} \right\rbrack} \\ & {= R_{yx} - Q_{xz} + P_{yz} - R_{yx} + Q_{zx} - P_{zy} \\ & {= 0.} \end{array}}$$}

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Showing That a Vector Field Is Not the Curl of Another 证明一个向量场不是另一向量场的旋度

Show that $\mathbf{\text{F}}\left( {x,y,z} \right) = e^{x}\mathbf{\text{i}} + yz\mathbf{\text{j}} + xz^{2}\mathbf{\text{k}}$ is not the curl of another vector field. That is, show that there is no other vector G with $\text{curl}\ \mathbf{\text{G}} = \mathbf{\text{F}}.$

Solution 解

Notice that the domain of F is all of $\mathbb{R}^{3}$ and the second-order partials of F are all continuous. Therefore, we can apply the previous theorem to F.

The divergence of F is $e^{x} + z + 2xz.$ If F were the curl of vector field G, then $\text{div}\ \mathbf{\text{F}} = \text{div curl}\ \mathbf{\text{G}} = 0.$ But, the divergence of F is not zero, and therefore F is not the curl of any other vector field.

Is it possible for $\mathbf{\text{G}}(x,y,z) = \left\langle {\text{sin}\mspace{2mu} x,\text{cos}\mspace{2mu} y,\text{sin}\left( {xyz} \right)} \right\rangle$ to be the curl of a vector field?

With the next two theorems, we show that if F is a conservative vector field then its curl is zero, and if the domain of F is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.

Curl of a Conservative Vector Field 保守向量场的旋度

If $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ is conservative, then $\text{curl}\ \textbf{F} = 0.$

Proof 证明

Since conservative vector fields satisfy the cross-partials property, all the cross-partials of F are equal. Therefore,

$$\begin{array}{cl} {\text{curl}\ \mathbf{\text{F}}} & {= \left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}}} \\ & {= 0.} \end{array}$$

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The same theorem is true for vector fields in a plane.

Since a conservative vector field is the gradient of a scalar function, the previous theorem says that $\text{curl}\ \left( {\text{∇}f} \right) = 0$ for any scalar function $f.$ In terms of our curl notation, $\nabla\ \times \ \nabla(f) = 0.$ This equation makes sense because the cross product of a vector with itself is always the zero vector. Sometimes equation $\nabla\ \times \ \nabla(f) = 0$ is simplified as $\nabla\ \times \ \nabla = 0.$

Curl Test for a Conservative Field 保守场的旋度判别法

Let $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ be a vector field in space on a simply connected domain. If $\text{curl}\ \mathbf{\text{F}} = 0,$ then F is conservative.

Proof 证明

Since $\text{curl}\ \mathbf{\text{F}} = 0,$ we have that $R_{y} = Q_{z},P_{z} = R_{x},$ and $Q_{x} = P_{y}.$ Therefore, F satisfies the cross-partials property on a simply connected domain, and Cross-Partial Property of Conservative Fields implies that F is conservative.

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The same theorem is also true in a plane. Therefore, if F is a vector field in a plane or in space and the domain is simply connected, then F is conservative if and only if $\text{curl}\ \mathbf{\text{F}} = 0.$

Testing Whether a Vector Field Is Conservative 检验向量场是否保守

Use the curl to determine whether $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {yz,xz,xy} \right\rangle$ is conservative.

Solution 解

Note that the domain of F is all of $\mathbb{R}^{3},$ which is simply connected (Figure 6.56). Therefore, we can test whether F is conservative by calculating its curl.

The curl of F is

$$\left( {\frac{\partial}{\partial y}xy - \frac{\partial}{\partial z}xz} \right)\mathbf{\text{i}} + \left( {\frac{\partial}{\partial y}yz - \frac{\partial}{\partial z}xy} \right)\mathbf{\text{j}} + \left( {\frac{\partial}{\partial y}xz - \frac{\partial}{\partial z}yz} \right)\mathbf{\text{k}} = \left( {x - x} \right)\mathbf{\text{i}} + \left( {y - y} \right)\mathbf{\text{j}} + \left( {z - z} \right)\mathbf{\text{k}} = 0.$$

Thus, F is conservative.

We have seen that the curl of a gradient is zero. What is the divergence of a gradient? If $f$ is a function of two variables, then $\text{div}(\text{∇}f) = \nabla \cdot (\text{∇}f) = f_{xx} + f_{yy}.$ We abbreviate this “double dot product” as $\nabla^{2}.$ This operator is called the *Laplace operator,* and in this notation Laplace’s equation becomes $\nabla^{2}f = 0.$ Therefore, a harmonic function is a function that becomes zero after taking the divergence of a gradient.

Similarly, if $f$ is a function of three variables then

$$\text{div}(\text{∇}f) = \nabla \cdot (\text{∇}f) = f_{xx} + f_{yy} + f_{zz}.$$

Using this notation we get Laplace’s equation for harmonic functions of three variables:

$$\nabla^{2}f = 0.$$

Harmonic functions arise in many applications. For example, the potential function of an electrostatic field in a region of space that has no static charge is harmonic.

Analyzing a Function 分析一个函数

Is it possible for $f(x,y) = x^{2} + x - y$ to be the potential function of an electrostatic field that is located in a region of $\mathbb{R}^{2}$ free of static charge?

Solution 解

If $f$ were such a potential function, then $f$ would be harmonic. Note that $f_{xx} = 2$ and $f_{yy} = 0,$ and so $f_{xx} + f_{yy} \neq 0.$ Therefore, $f$ is not harmonic and $f$ cannot represent an electrostatic potential.

Is it possible for function $f(x,y) = x^{2} - y^{2} + x$ to be the potential function of an electrostatic field located in a region of $\mathbb{R}^{2}$ free of static charge?

Section 6.5 Exercises 6.5 节习题

For the following exercises, determine whether the statement is *true or false*.

206\.

If the coordinate functions of $\mathbf{\text{F}}:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ have continuous second partial derivatives, then $\text{curl}\ (\text{div}(\mathbf{\text{F}}))$ equals zero.

207.

$\nabla \cdot \left( {x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}} \right) = 1.$

208\.

All vector fields of the form $\mathbf{\text{F}}\left( {x,y,z} \right) = f(x)\mathbf{\text{i}} + g(y)\mathbf{\text{j}} + h(z)\mathbf{\text{k}}$ are conservative.

209.

If $\text{curl}\ \mathbf{\text{F}} = 0,$ then F is conservative.

210\.

If F is a constant vector field then $\text{div}\ \mathbf{\text{F}} = 0.$

211.

If F is a constant vector field then $\text{curl}\ \mathbf{\text{F}} = 0.$

For the following exercises, find the curl of F.

212\.

$\mathbf{\text{F}}\left( {x,y,z} \right) = xy^{2}z^{4}\mathbf{\text{i}} + \left( {2x^{2}y + z} \right)\mathbf{\text{j}} + y^{3}z^{2}\mathbf{\text{k}}$

213.

$\mathbf{\text{F}}\left( {x,y,z} \right) = x^{2}z\mathbf{\text{i}} + y^{2}x\mathbf{\text{j}} + \left( {y + 2z} \right)\mathbf{\text{k}}$

214\.

$\mathbf{\text{F}}\left( {x,y,z} \right) = 3xyz^{2}\mathbf{\text{i}} + y^{2}\text{sin}\mspace{2mu} z\mathbf{\text{j}} + xe^{2z}\mathbf{\text{k}}$

215.

$\mathbf{\text{F}}(x,y,z) = x^{2}yz\mathbf{\text{i}} + xy^{2}z\mathbf{\text{j}} + xyz^{2}\mathbf{\text{k}}$

216\.

$\mathbf{\text{F}}(x,y,z) = (x\ \text{cos}\mspace{2mu} y)\mathbf{\text{i}} + xy^{2}\mathbf{\text{j}}$

217.

$\mathbf{\text{F}}(x,y,z) = (x - y)\mathbf{\text{i}} + (y - z)\mathbf{\text{j}} + (z - x)\mathbf{\text{k}}$

218\.

$\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + x^{2}y^{2}z^{2}\mathbf{\text{j}} + y^{2}z^{3}\mathbf{\text{k}}$

219.

$\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} + yz\mathbf{\text{j}} + xz\mathbf{\text{k}}$

220\.

$\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + y^{2}\mathbf{\text{j}} + z^{2}\mathbf{\text{k}}$

221.

$\mathbf{\text{F}}(x,y,z) = ax\mathbf{\text{i}} + by\mathbf{\text{j}} + c\mathbf{\text{k}}$ for constants *a*, *b*, *c*

For the following exercises, find the divergence of F.

222\.

$\mathbf{\text{F}}(x,y,z) = x^{2}z\mathbf{\text{i}} + y^{2}x\mathbf{\text{j}} + \left( {y + 2z} \right)\mathbf{\text{k}}$

223.

$\mathbf{\text{F}}(x,y,z) = 3xyz^{2}\mathbf{\text{i}} + y^{2}\text{sin}\mspace{2mu} z\mathbf{\text{j}} + xe^{2z}\mathbf{\text{k}}$

224\.

$\mathbf{\text{F}}(x,y) = (\text{sin}\mspace{2mu} x)\mathbf{\text{i}} + (\text{cos}\mspace{2mu} y)\mathbf{\text{j}}$

225.

$\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + y^{2}\mathbf{\text{j}} + z^{2}\mathbf{\text{k}}$

226\.

$\mathbf{\text{F}}(x,y,z) = (x - y)\mathbf{\text{i}} + (y - z)\mathbf{\text{j}} + \left( {z - x} \right)\mathbf{\text{k}}$

227.

$\mathbf{\text{F}}(x,y) = \frac{x}{\sqrt{x^{2} + y^{2}}}\mathbf{\text{i}} + \frac{y}{\sqrt{x^{2} + y^{2}}}\mathbf{\text{j}}$

228\.

$\mathbf{\text{F}}(x,y) = x\mathbf{\text{i}} - y\mathbf{\text{j}}$

229.

$\mathbf{\text{F}}(x,y,z) = ax\mathbf{\text{i}} + by\mathbf{\text{j}} + c\mathbf{\text{k}}$ for constants *a*, *b*, *c*

230\.

$\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + x^{2}y^{2}z^{2}\mathbf{\text{j}} + y^{2}z^{3}\mathbf{\text{k}}$

231.

$\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} + yz\mathbf{\text{j}} + xz\mathbf{\text{k}}$

For the following exercises, determine whether each of the given scalar functions is harmonic.

232\.

$u(x,y,z) = e^{\text{−}x}(\text{cos}\mspace{2mu} y - \text{sin}\mspace{2mu} y)$

233.

$w(x,y,z) = \left( {x^{2} + y^{2} + z^{2}} \right)^{\text{−}{1\text{/}2}}$

234\.

If $\mathbf{\text{F}}(x,y,z) = 2\mathbf{\text{i}} + 2x\mathbf{\text{j}} + 3y\mathbf{\text{k}}$ and $\mathbf{\text{G}}(x,y,z) = x\mathbf{\text{i}} - y\mathbf{\text{j}} + z\mathbf{\text{k}},$ find $\text{curl}\ (\mathbf{\text{F}}\ \times \ \mathbf{\text{G}}).$

235.

If $\mathbf{\text{F}}(x,y,z) = 2\mathbf{\text{i}} + 2x\mathbf{\text{j}} + 3y\mathbf{\text{k}}$ and $\mathbf{\text{G}}(x,y,z) = x\mathbf{\text{i}} - y\mathbf{\text{j}} + z\mathbf{\text{k}},$ find $\text{div}\ (\mathbf{\text{F}}\ \times \ \mathbf{\text{G}}).$

236\.

Find $\text{div}\ \mathbf{\text{F}},$ given that $\mathbf{\text{F}} = \text{∇}f,$ where $f(x,y,z) = xy^{3}z^{2}.$

237.

Find the divergence of F for vector field $\mathbf{\text{F}}(x,y,z) = \left( {y^{2} + z^{2}} \right)\left( {x + y} \right)\mathbf{\text{i}} + \left( {z^{2} + x^{2}} \right)\left( {y + z} \right)\mathbf{\text{j}} + \left( {x^{2} + y^{2}} \right)\left( {z + x} \right)\mathbf{\text{k}}.$

238\.

Find the divergence of F for vector field $\mathbf{\text{F}}(x,y,z) = f_{1}(y,z)\mathbf{\text{i}} + f_{2}(x,z)\mathbf{\text{j}} + f_{3}(x,y)\mathbf{\text{k}}.$

For the following exercises, use $r = \left. ||\mathbf{\text{r}} \right.||$ and $\textbf{r} = {< {x,y,z} >}.$

239.

Find $\text{curl}\ \mathbf{\text{r}}.$

240\.

Find $\text{curl}\ \frac{\mathbf{\text{r}}}{r}.$

241.

Find $\text{curl}\ \frac{\mathbf{\text{r}}}{r^{3}}.$

242\.

Let $\mathbf{\text{F}}(x,y) = \frac{\text{−}y\mathbf{\text{i}} + x\mathbf{\text{j}}}{x^{2} + y^{2}},$ where F is defined on $\left\{ (x,y) \in \mathbb{R}^{2} \middle| (x,y) \neq (0,0) \right\}.$ Find $\text{curl}\ \mathbf{\text{F}}.$

For the following exercises, use a computer algebra system to find the curl of the given vector fields.

243.

\[T\] $\mathbf{\text{F}}(x,y,z) = \text{arctan}\left( \frac{x}{y} \right)\mathbf{\text{i}} + \text{ln}\sqrt{x^{2} + y^{2}}\mathbf{\text{j}} + \mathbf{\text{k}}$

244\.

\[T\] $\mathbf{\text{F}}(x,y,z) = \text{sin}(x - y)\mathbf{\text{i}} + \text{sin}(y - z)\mathbf{\text{j}} + \text{sin}(z - x)\mathbf{\text{k}}$

For the following exercises, find the divergence of F at the given point.

245.

$\mathbf{\text{F}}(x,y,z) = \mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}$ at $(2,-1,3)$

246\.

$\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + y\mathbf{\text{j}} + x\mathbf{\text{k}}$ at $(1,2,3)$

247.

$\mathbf{\text{F}}(x,y,z) = e^{\text{−}xy}\mathbf{\text{i}} + e^{xz}\mathbf{\text{j}} + e^{yz}\mathbf{\text{k}}$ at $(3,2,0)$

248\.

$\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}$ at (1, 2, 1)

249.

$\mathbf{\text{F}}(x,y,z) = e^{x}\text{sin}\mspace{2mu} y\mathbf{\text{i}} - e^{x}\text{cos}\mspace{2mu} y\mathbf{\text{j}}$ at (0, 0, 3)

For the following exercises, find the curl of F at the given point.

250\.

$\mathbf{\text{F}}(x,y,z) = \mathbf{\text{i}} + \mathbf{\text{j}} + \mathbf{\text{k}}$ at $(2,-1,3)$

251.

$\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + y\mathbf{\text{j}} + x\mathbf{\text{k}}$ at $(1,2,3)$

252\.

$\mathbf{\text{F}}(x,y,z) = e^{\text{−}xy}\mathbf{\text{i}} + e^{xz}\mathbf{\text{j}} + e^{yz}\mathbf{\text{k}}$ at (3, 2, 0)

253.

$\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}$ at (1, 2, 1)

254\.

$\mathbf{\text{F}}(x,y,z) = e^{x}\text{sin}\mspace{2mu} y\mathbf{\text{i}} - e^{x}\text{cos}\mspace{2mu} y\mathbf{\text{j}}$ at (0, 0, 3)

255.

Let $\mathbf{\text{F}}(x,y,z) = \left( {3x^{2}y + az} \right)\mathbf{\text{i}} + x^{3}\mathbf{\text{j}} + \left( {3x + 3z^{2}} \right)\mathbf{\text{k}}.$ For what value of *a* is F conservative?

256\.

Given vector field $\textbf{F}(x,y) = \frac{1}{x^{2} + y^{2}}{< {\text{−}y,x} >}$ on domain $D = {\mathbb{R}}^{2} - \left\{ (0,0) \right\} = \left\{ (x,y) \in \mathbb{R}^{2} \middle| (x,y) \neq (0,0) \right\},$ is F conservative?

257.

Given vector field $\textbf{F}(x,y) = \frac{1}{x^{2} + y^{2}}{< {x,y} >}$ on domain $D = {\mathbb{R}}^{2} - \left\{ (0,0) \right\} = \left\{ (x,y) \in \mathbb{R}^{2} \middle| (x,y) \neq (0,0) \right\},$ is F conservative?

258\.

Find the work done by force field $\mathbf{\text{F}}(x,y) = e^{\text{−}y}\mathbf{\text{i}} - xe^{\text{−}y}\mathbf{\text{j}}$ in moving an object from *P*(0, 1) to *Q*(2, 0). Is the force field conservative?

259.

Compute the divergence of $\mathbf{\text{F}} = \left( {\text{sinh}\ x} \right)\mathbf{\text{i}} + \left( {\text{cosh}\ y} \right)\mathbf{\text{j}} - xyz\mathbf{\text{k}}.$

260\.

Compute the curl of $\mathbf{\text{F}} = \left( {\text{sinh}\ x} \right)\mathbf{\text{i}} + \left( {\text{cosh}\ y} \right)\mathbf{\text{j}} - xyz\mathbf{\text{k}}.$

For the following exercises, consider a rigid body that is rotating about the *x*-axis counterclockwise with constant angular velocity $\omega = \left\langle {a,b,c} \right\rangle.$ If *P* is a point in the body located at $\mathbf{\text{r}} = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}\text{,}$ the velocity at *P* is given by vector field $\mathbf{\text{F}} = \omega\ \times \ \mathbf{\text{r}}.$

261.

Express F in terms of i, j, and k vectors.

262\.

Find $\text{the div of}\ \mathbf{\text{F}}.$

263.

Find $\text{the curl of}\ \mathbf{\text{F}}$

In the following exercises, suppose that $\nabla \cdot \mathbf{\text{F}} = 0$ and $\nabla \cdot \mathbf{\text{G}} = 0.$

264\.

Does $\mathbf{\text{F}} + \mathbf{\text{G}}$ necessarily have zero divergence?

265.

Does $\mathbf{\text{F}}\ \times \ \mathbf{\text{G}}$ necessarily have zero divergence?

In the following exercises, suppose a solid object in $\mathbb{R}^{3}$ has a temperature distribution given by $T\left( {x,y,z} \right) = 100e^{- x^{2} + y^{2} + z^{2}}$. The heat flow vector field in the object is $\mathbf{\text{F}} = \text{−}k\text{∇}T,$ where $k > 0$ is a property of the material. The heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is $\nabla \cdot \mathbf{\text{F}} = \text{−}k\nabla \cdot \text{∇}T = \text{−}k\nabla^{2}T.$

266\.

Compute the heat flow vector field.

267.

Compute the divergence.

268\.

\[T\] Consider rotational velocity field $\mathbf{\text{v}} = \left\langle {0,10z,-10y} \right\rangle.$ If a paddlewheel is placed in plane $x + y + z = 1$ with its axis normal to this plane, using a computer algebra system, calculate how fast the paddlewheel spins in revolutions per unit of time.

6.6 Surface Integrals 6.6 曲面积分

• 6.6.1 Find the parametric representations of a cylinder, a cone, and a sphere.
• 6.6.2 Describe the surface integral of a scalar-valued function over a parametric surface.
• 6.6.3 Use a surface integral to calculate the area of a given surface.
• 6.6.4 Explain the meaning of an oriented surface, giving an example.
• 6.6.5 Describe the surface integral of a vector field.
• 6.6.6 Use surface integrals to solve applied problems.

• 6.6.1 求柱面、锥面与球面的参数表示。
• 6.6.2 描述标量值函数在参数曲面上的曲面积分。
• 6.6.3 用曲面积分计算给定曲面的面积。
• 6.6.4 解释有向曲面的含义，并举例说明。
• 6.6.5 描述向量场的曲面积分。
• 6.6.6 用曲面积分解决实际问题。

We have seen that a line integral is an integral over a path in a plane or in space. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. We can extend the concept of a line integral to a surface integral to allow us to perform this integration.

Surface integrals are important for the same reasons that line integrals are important. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus. In particular, surface integrals allow us to generalize Green’s theorem to higher dimensions, and they appear in some important theorems we discuss in later sections.

Parametric Surfaces 参数曲面

A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. In this sense, surface integrals expand on our study of line integrals. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field.

However, before we can integrate over a surface, we need to consider the surface itself. Recall that to calculate a scalar or vector line integral over curve *C*, we first need to parameterize *C*. In a similar way, to calculate a surface integral over surface *S*, we need to parameterize *S*. That is, we need a working concept of a parameterized surface (or a parametric surface), in the same way that we already have a concept of a parameterized curve.

A parameterized surface is given by a description of the form

$$\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle.$$

Notice that this parameterization involves two parameters, *u* and *v*, because a surface is two-dimensional, and therefore two variables are needed to trace out the surface. The parameters *u* and *v* vary over a region called the parameter domain, or parameter space—the set of points in the *uv*-plane that can be substituted into r. Each choice of *u* and *v* in the parameter domain gives a point on the surface, just as each choice of a parameter *t* gives a point on a parameterized curve. The entire surface is created by making all possible choices of *u* and *v* over the parameter domain.

Given a parameterization of surface $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle,$ the parameter domain of the parameterization is the set of points in the *uv*-plane that can be substituted into r.

Parameterizing a Cylinder 参数化柱面

Describe surface *S* parameterized by

$$\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {\text{cos}\ u,\text{sin}\ u,v} \right\rangle,\text{−}\infty < u < \infty,\text{−}\infty < v < \infty.$$

Solution 解

To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of $\mathbb{R}^{2},$ we can choose any value for *u* and *v* and plot the corresponding point. If $u = v = 0,$ then $\mathbf{\text{r}}\left( {0,0} \right) = \left\langle {1,0,0} \right\rangle,$ so point (1, 0, 0) is on *S*. Similarly, points $\mathbf{\text{r}}\left( {\pi,2} \right) = \left( {-1,0,2} \right)$ and $\mathbf{\text{r}}\left( {\frac{\pi}{2},4} \right) = \left( {0,1,4} \right)$ are on *S*.

Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize *S*, we visualize two families of curves that lie on *S.* In the first family of curves we hold *u* constant; in the second family of curves we hold *v* constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.

First, suppose that *u* is a constant *K*. Then the curve traced out by the parameterization is $\left\langle {\text{cos}\ K,\text{sin}\ K,v} \right\rangle,$ which gives a vertical line that goes through point $\left( {\text{cos}\ K,\text{sin}\ K,v} \right)$ in the *xy*-plane.

Now suppose that *v* is a constant *K.* Then the curve traced out by the parameterization is $\left\langle {\text{cos}\ u,\text{sin}\ u,K} \right\rangle,$ which gives a circle in plane $z = K$ with radius 1 and center (0, 0, *K*).

If *u* is held constant, then we get vertical lines; if *v* is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder $x^{2} + y^{2} = 1$ (Figure 6.57).

Notice that if $x = \text{cos}\ u$ and $y = \text{sin}\ u,$ then $x^{2} + y^{2} = 1,$ so points from *S* do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle $\left\langle {\text{cos}\ u,\text{sin}\ u,k} \right\rangle$ for some *k*, and therefore each point on the cylinder is contained in the parameterized surface (Figure 6.58).

Analysis 分析

Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to $0 \leq u \leq \pi,0 < v < 6,$ then the surface would be a half-cylinder of height 6.

Describe the surface with parameterization $\mathbf{\text{r}}(u,v) = \left\langle {2\ \text{cos}\ u,2\ \text{sin}\ u,v} \right\rangle,0 \leq u < 2\pi,\text{−}\infty < v < \infty.$

It follows from Example 6.58 that we can parameterize all cylinders of the form $x^{2} + y^{2} = R^{2}.$ If *S* is a cylinder given by equation $x^{2} + y^{2} = R^{2},$ then a parameterization of *S* is

$$\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {R\ \text{cos}\ u,R\ \text{sin}\ u,v} \right\rangle,0 \leq u < 2\pi,\text{−}\infty < v < \infty.$$

We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.

Describing a Surface 描述一个曲面

Describe surface *S* parameterized by

$$\mathbf{\text{r}}(u,v) = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,u^{2}} \right\rangle,0 \leq u < \infty,0 \leq v < 2\pi.$$

Solution 解

Notice that if *u* is held constant, then the resulting curve is a circle of radius *u* in plane $z = u^{2}.$ Therefore, as *u* increases, the radius of the resulting circle increases. If *v* is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that

$$\begin{array}{cl} {x^{2} + y^{2}} & {= \left( {u\ \text{cos}\ v} \right)^{2} + \left( {u\ \text{sin}\ v} \right)^{2}} \\ & {= u^{2}\text{cos}^{2}v + u^{2}\text{sin}^{2}v} \\ & {= u^{2}} \\ & {= z.} \end{array}$$

Therefore, the surface is elliptic paraboloid $x^{2} + y^{2} = z$ (Figure 6.59).

Describe the surface parameterized by $\mathbf{\text{r}}(u,v) = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,u} \right\rangle,\text{−}\infty < u < \infty,0 \leq v < 2\pi.$

Finding a Parameterization 求参数化

Give a parameterization of the cone $x^{2} + y^{2} = z^{2}$ lying on or above the plane $z = -2.$

Solution 解

The horizontal cross-section of the cone at height $z = u$ is circle $x^{2} + y^{2} = u^{2}.$ Therefore, a point on the cone at height *u* has coordinates $\left( {u\ \text{cos}\ v,u\ \text{sin}\ v,u} \right)$ for angle *v*. Hence, a parameterization of the cone is $\mathbf{\text{r}}(u,v) = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,u} \right\rangle.$ Since we are not interested in the entire cone, only the portion on or above plane $z = -2,$ the parameter domain is given by $-2 \leq u < \infty,0 \leq v < 2\pi$ (Figure 6.60).

Give a parameterization for the portion of cone $x^{2} + y^{2} = z^{2}$ lying in the first octant.

We have discussed parameterizations of various surfaces, but two cases deserve separate attention: spheres and graphs of two-variable functions. To parameterize a sphere, it is easiest to use spherical coordinates. The sphere of radius $\rho$ centered at the origin is given by the parameterization

$$\mathbf{\text{r}}\left( {\phi,\theta} \right) = \left\langle {\rho\ \text{cos}\ \theta\ \text{sin}\ \phi,\rho\ \text{sin}\ \theta\ \text{sin}\ \phi,\rho\ \text{cos}\ \phi} \right\rangle,0 \leq \theta \leq 2\pi,0 \leq \phi \leq \pi.$$

The idea of this parameterization is that as $\phi$ sweeps downward from the positive *z*-axis, a circle of radius $\rho\ \text{sin}\ \phi$ is traced out by letting $\theta$ run from 0 to $2\pi.$ To see this, let $\phi$ be fixed. Then

$$\begin{array}{cl} {x^{2} + y^{2}} & {= \left( {\rho\ \text{cos}\ \theta\ \text{sin}\ \phi} \right)^{2} + \left( {\rho\ \text{sin}\ \theta\ \text{sin}\ \phi} \right)^{2}} \\ & {= \rho^{2}\text{sin}^{2}\phi\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right)} \\ & {= \rho^{2}\text{sin}^{2}\phi} \\ & {= \left( {\rho\ \text{sin}\ \phi} \right)^{2}.} \end{array}$$

This results in the desired circle (Figure 6.61).

Finally, to parameterize the graph of a two-variable function, we first let $z = f\left( {x,y} \right)$ be a function of two variables. The simplest parameterization of the graph of $f$ is $\mathbf{\text{r}}(x,y) = \left\langle {x,y,f\left( {x,y} \right)} \right\rangle,$ where *x* and *y* vary over the domain of $f$ (Figure 6.62). For example, the graph of $f(x,y) = x^{2}y$ can be parameterized by $\mathbf{\text{r}}(x,y) = \left\langle {x,y,x^{2}y} \right\rangle,$ where the parameters *x* and *y* vary over the domain of $f.$ If we only care about a piece of the graph of $f$—say, the piece of the graph over rectangle $\left\lbrack {1,3} \right\rbrack\ \times \ \left\lbrack {2,5} \right\rbrack$—then we can restrict the parameter domain to give this piece of the surface:

$$\mathbf{\text{r}}(x,y) = \left\langle {x,y,x^{2}y} \right\rangle,1 \leq x \leq 3,2 \leq y \leq 5.$$

Similarly, if *S* is a surface given by equation $x = g\left( {y,z} \right)$ or equation $y = h(x,z),$ then a parameterization of *S* is

$\mathbf{\text{r}}(y,z) = \left\langle {g\left( {y,z} \right),y,z} \right\rangle$ or $\mathbf{\text{r}}(x,z) = \left\langle {x,h\left( {x,z} \right),z} \right\rangle,$ respectively. For example, the graph of paraboloid $2y = x^{2} + z^{2}$ can be parameterized by $\mathbf{\text{r}}(x,z) = \left\langle {x,\frac{x^{2} + z^{2}}{2},z} \right\rangle,0 \leq x < \infty,0 \leq z < \infty.$ Notice that we do not need to vary over the entire domain of *y* because *x* and *z* are squared.

Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization $\mathbf{\text{r}}(t),a \leq t \leq b$ is regular if $\mathbf{\text{r}}\prime(t) \neq 0$ for all *t* in $\lbrack a,b\rbrack.$ For a curve, this condition ensures that the image of r really is a curve, and not just a point. For example, consider curve parameterization $\mathbf{\text{r}}(t) = \left\langle {1,2} \right\rangle,0 \leq t \leq 5.$ The image of this parameterization is simply point $(1,2),$ which is not a curve. Notice also that $\mathbf{\text{r}}\prime(t) = 0.$ The fact that the derivative is the zero vector indicates we are not actually looking at a curve.

Analogously, we would like a notion of regularity for surfaces so that a surface parameterization really does trace out a surface. To motivate the definition of regularity of a surface parameterization, consider parameterization

$$\mathbf{\text{r}}(u,v) = \left\langle {0,\text{cos}\ v,1} \right\rangle,0 \leq u \leq 1,0 \leq v \leq \pi.$$

Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure 6.63). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that $\mathbf{\text{r}}_{u} = \left\langle {0,0,0} \right\rangle$ and $\mathbf{\text{r}}_{v} = \left\langle {0,\text{−}\text{sin}\ v,0} \right\rangle,$ and the corresponding cross product is zero. The analog of the condition $\mathbf{\text{r}}\prime(t) = 0$ is that $\mathbf{\text{r}}_{u}\ \times \ \mathbf{\text{r}}_{v}$ is not zero for any point $(u,v)$ in the parameter domain, which is a regular parameterization.

Parameterization $\mathbf{\text{r}}(u,v) = \left\langle {x(u,v),y(u,v),z(u,v)} \right\rangle$ is a regular parameterization if $\mathbf{\text{r}}_{u}\ \times \ \mathbf{\text{r}}_{v}$ is not zero for any point $(u,v)$ in the parameter domain.

If parameterization r is regular, then the image of r is a two-dimensional object, as a surface should be. Throughout this chapter, parameterizations $\mathbf{\text{r}}(u,v) = \left\langle {x(u,v),y(u,v),z(u,v)} \right\rangle$ are assumed to be regular.

Recall that curve parameterization $\mathbf{\text{r}}(t),a \leq t \leq b$ is smooth if $\mathbf{\text{r}}\prime(t)$ is continuous and $\mathbf{\text{r}}\prime(t) \neq 0$ for all *t* in $\lbrack a,b\rbrack.$ Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is *smooth* if the resulting surface has no sharp corners.

A surface parameterization $\mathbf{\text{r}}(u,v) = \left\langle {x(u,v),y(u,v),z(u,v)} \right\rangle$ is *smooth* if vector $\mathbf{\text{r}}_{u}\ \times \ \mathbf{\text{r}}_{v}$ is not zero for any choice of *u* and *v* in the parameter domain.

A surface may also be *piecewise smooth* if it has smooth faces but also has locations where the directional derivatives do not exist.

Identifying Smooth and Nonsmooth Surfaces 识别光滑与不光滑曲面

Which of the figures in Figure 6.64 is smooth?

Solution 解

The surface in Figure 6.64(a) can be parameterized by

$$\mathbf{\text{r}}(u,v) = \left\langle {(2 + \text{cos}\ v)\text{cos}\ u,(2 + \text{cos}\ v)\text{sin}\ u,\text{sin}\ v} \right\rangle,0 \leq u < 2\pi,0 \leq v < 2\pi$$

(we can use technology to verify). Notice that vectors

$$\mathbf{\text{r}}_{u} = \left\langle {\text{−}(2 + \text{cos}\ v)\text{sin}\ u,(2 + \text{cos}\ v)\text{cos}\ u,0} \right\rangle\ \text{and}\ \mathbf{\text{r}}_{v} = \left\langle {\text{−}\text{sin}\ v\ \text{cos}\ u,\text{−}\text{sin}\ v\ \text{sin}\ u,\text{cos}\ v} \right\rangle$$

exist for any choice of *u* and *v* in the parameter domain, and

$$\begin{array}{cl} {\mathbf{\text{r}}_{u}\ \times \ \mathbf{\text{r}}_{v}} & {= \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ {\text{−}(2 + \text{cos}\ v)\text{sin}\ u} & {(2 + \text{cos}\ v)\text{cos}\ u} & 0 \\ {\text{−}\text{sin}\ v\ \text{cos}\ u} & {\text{−}\text{sin}\ v\ \text{sin}\ u} & {\text{cos}\ v} \end{matrix} \right|} \\ & {= \left\lbrack {(2 + \text{cos}\ v)\text{cos}\ u\ \text{cos}\ v} \right\rbrack\mathbf{\text{i}} + \left\lbrack {(2 + \text{cos}\ v)\text{sin}\ u\ \text{cos}\ v} \right\rbrack\mathbf{\text{j}}} \\ & {\mspace{9mu} + \left\lbrack {(2 + \text{cos}\ v)\text{sin}\ v\ \text{sin}^{2}u + (2 + \text{cos}\ v)\text{sin}\ v\ \text{cos}^{2}u} \right\rbrack\mathbf{\text{k}}} \\ & {= \left\lbrack {(2 + \text{cos}\ v)\text{cos}\ u\ \text{cos}\ v} \right\rbrack\mathbf{\text{i}} + \left\lbrack {(2 + \text{cos}\ v)\text{sin}\ u\ \text{cos}\ v} \right\rbrack\mathbf{\text{j}} + \left\lbrack {(2 + \text{cos}\ v)\text{sin}\ v} \right\rbrack\mathbf{\text{k}}.} \end{array}$$

The k component of this vector is zero only if $v = 0$ or $v = \pi.$ If $v = 0$ or $v = \pi,$ then the only choices for *u* that make the j component zero are $u = 0$ or $u = \pi.$ But, these choices of *u* do not make the i component zero. Therefore, $\mathbf{\text{r}}_{u}\ \times \ \mathbf{\text{r}}_{v}$ is not zero for any choice of *u* and *v* in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.

In the pyramid in Figure 6.64(b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of five smooth faces, and thus this surface is piecewise smooth.

Is the surface parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {u^{2v},v + 1,\text{sin}\ u} \right\rangle,0 \leq u \leq 2,0 \leq v \leq 3$ smooth?

Surface Area of a Parametric Surface 参数曲面的面积

Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. The second step is to define the surface area of a parametric surface. The notation needed to develop this definition is used throughout the rest of this chapter.

Let *S* be a surface with parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle$ over some parameter domain *D*. We assume here and throughout that the surface parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle$ is continuously differentiable—meaning, each component function has continuous partial derivatives. Assume for the sake of simplicity that *D* is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Divide rectangle *D* into subrectangles $D_{ij}$ with horizontal width $\text{Δ}u$ and vertical length $\text{Δ}v.$ Suppose that *i* ranges from 1 to *m* and *j* ranges from 1 to *n* so that *D* is subdivided into *mn* rectangles. This division of *D* into subrectangles gives a corresponding division of surface *S* into pieces $S_{ij}.$ Choose point $P_{ij}$ in each piece $S_{ij}.$ Point $P_{ij}$ corresponds to point $(u_{i},v_{j})$ in the parameter domain.

Note that we can form a grid with lines that are parallel to the *u*-axis and the *v*-axis in the *uv*-plane. These grid lines correspond to a set of grid curves on surface *S* that is parameterized by $\mathbf{\text{r}}\left( {u,v} \right).$ Without loss of generality, we assume that $P_{ij}$ is located at the corner of two grid curves, as in Figure 6.65. If we think of r as a mapping from the *uv*-plane to $\mathbb{R}^{3},$ the grid curves are the image of the grid lines under r. To be precise, consider the grid lines that go through point $(u_{i},v_{j}).$ One line is given by $x = u_{i},y = v;$ the other is given by $x = u,y = v_{j}.$ In the first grid line, the horizontal component is held constant, yielding a vertical line through $(u_{i},v_{j}).$ In the second grid line, the vertical component is held constant, yielding a horizontal line through $(u_{i},v_{j}).$ The corresponding grid curves are $\mathbf{\text{r}}(u_{i},v)$ and $\mathbf{\text{r}}(u,v_{j}),$ and these curves intersect at point $P_{ij}.$

Now consider the vectors that are tangent to these grid curves. For grid curve $\mathbf{\text{r}}(u_{i},v),$ the tangent vector at $P_{ij}$ is

$$\mathbf{\text{t}}_{v}\left( P_{ij} \right) = \mathbf{\text{r}}_{v}\left( {u_{i},v_{j}} \right) = \left\langle {x_{v}\left( {u_{i},v_{j}} \right),y_{v}\left( {u_{i},v_{j}} \right),z_{v}\left( {u_{i},v_{j}} \right)} \right\rangle.$$

For grid curve $\mathbf{\text{r}}(u,v_{j}),$ the tangent vector at $P_{ij}$ is

$$\mathbf{\text{t}}_{u}\left( P_{ij} \right) = \mathbf{\text{r}}_{u}\left( {u_{i},v_{j}} \right) = \left\langle {x_{u}\left( {u_{i},v_{j}} \right),y_{u}\left( {u_{i},v_{j}} \right),z_{u}\left( {u_{i},v_{j}} \right)} \right\rangle.$$

If vector $\mathbf{\text{N}} = \mathbf{\text{t}}_{u}\left( P_{ij} \right)\ \times \ \mathbf{\text{t}}_{v}\left( P_{ij} \right)$ exists and is not zero, then the tangent plane at $P_{ij}$ exists (Figure 6.66). If piece $S_{ij}$ is small enough, then the tangent plane at point $P_{ij}$ is a good approximation of piece $S_{ij}.$

The tangent plane at $P_{ij}$ contains vectors $\mathbf{\text{t}}_{u}\left( P_{ij} \right)$ and $\mathbf{\text{t}}_{v}\left( P_{ij} \right),$ and therefore the parallelogram spanned by $\mathbf{\text{t}}_{u}\left( P_{ij} \right)$ and $\mathbf{\text{t}}_{v}\left( P_{ij} \right)$ is in the tangent plane. Since the original rectangle in the *uv*-plane corresponding to $S_{ij}$ has width $\text{Δ}u$ and length $\text{Δ}v,$ the parallelogram that we use to approximate $S_{ij}$ is the parallelogram spanned by $\text{Δ}u\mathbf{\text{t}}_{u}\left( P_{ij} \right)$ and $\text{Δ}v\mathbf{\text{t}}_{v}\left( P_{ij} \right).$ In other words, we scale the tangent vectors by the constants $\text{Δ}u$ and $\text{Δ}v$ to match the scale of the original division of rectangles in the parameter domain. Therefore, the area of the parallelogram used to approximate the area of $S_{ij}$ is

$$\text{Δ}S_{ij} \approx \left\| {\left( {\text{Δ}u\mathbf{\text{t}}_{u}\left( P_{ij} \right)} \right)\ \times \ \left( {\text{Δ}v\mathbf{\text{t}}_{v}\left( P_{ij} \right)} \right)} \right\| = \left\| {\mathbf{\text{t}}_{u}\left( P_{ij} \right)\ \times \ \mathbf{\text{t}}_{v}\left( P_{ij} \right)} \right\|\text{Δ}u\text{Δ}v.$$

Varying point $P_{ij}$ over all pieces $S_{ij}$ and the previous approximation leads to the following definition of surface area of a parametric surface (Figure 6.67).

Let $\mathbf{\text{r}}(u,v) = \left\langle {x(u,v),y(u,v),z(u,v)} \right\rangle$ with parameter domain *D* be a smooth parameterization of surface *S*. Furthermore, assume that *S* is traced out only once as $(u,v)$ varies over *D*. The surface area of *S* is

$${\iint_{D}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| dA}},$$ (6.18)

where $\mathbf{\text{t}}_{u} = \left\langle {\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}} \right\rangle$ and all partial derivatives are continuous.

Calculating Surface Area 计算曲面面积

Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height *h* and radius *r*.

Solution 解

Before calculating the surface area of this cone using Equation 6.18, we need a parameterization. We assume this cone is in $\mathbb{R}^{3}$ with its vertex at the origin (Figure 6.68). To obtain a parameterization, let $\alpha$ be the angle that is swept out by starting at the positive *z*-axis and ending at the cone, and let $k = \text{tan}\ \alpha.$ For a height value *v* with $0 \leq v \leq h,$ the radius of the circle formed by intersecting the cone with plane $z = v$ is $kv.$ Therefore, a parameterization of this cone is

$$\mathbf{\text{s}}(u,v) = \left\langle {kv\ \text{cos}\ u,kv\ \text{sin}\ u,v} \right\rangle,0 \leq u < 2\pi,0 \leq v \leq h.$$

The idea behind this parameterization is that for a fixed *v* value, the circle swept out by letting *u* vary is the circle at height *v* and radius *kv*. As *v* increases, the parameterization sweeps out a “stack” of circles, resulting in the desired cone.

With a parameterization in hand, we can calculate the surface area of the cone using Equation 6.18. The tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {\text{−}kv\ \text{sin}\ u,kv\ \text{cos}\ u,0} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {k\ \text{cos}\ u,k\ \text{sin}\ u,1} \right\rangle.$ Therefore,

$$\begin{array}{cl}{\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} & {= \left| \begin{matrix}\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\{\text{−}kv\ \text{sin}\ u} & {kv\ \text{cos}\ u} & 0 \\{k\ \text{cos}\ u} & {k\ \text{sin}\ u} & 1\end{matrix} \right|} \\ & {= \left\langle {kv\ \text{cos}\ u,kv\ \text{sin}\ u,\text{−}k^{2}v\ \text{sin}^{2}u - k^{2}v\ \text{cos}^{2}u} \right\rangle} \\ & {= \left\langle {kv\ \text{cos}\ u,kv\ \text{sin}\ u,\text{−}k^{2}v} \right\rangle.}\end{array}$$

The magnitude of this vector is

$$\begin{array}{cl}\left\| \left\langle {kv\ \text{cos}\ u,kv\ \text{sin}\ u,\text{−}k^{2}v} \right\rangle \right\| & {= \sqrt{k^{2}v^{2}\text{cos}^{2}u + k^{2}v^{2}\text{sin}^{2}u + k^{4}v^{2}}} \\ & {= \sqrt{k^{2}v^{2} + k^{4}v^{2}}} \\ & {= kv\sqrt{1 + k^{2}}.}\end{array}$$

By Equation 6.18, the surface area of the cone is

$$\begin{array}{cl}{\iint_{D}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| dA}} & {= {\int_{0}^{h}{{\int_{0}^{2\pi}{kv\sqrt{1 + k^{2}}}}dudv}}} \\ & {= 2\pi k\sqrt{1 + k^{2}}{\int_{0}^{h}{vdv}}} \\ & {= 2\pi k\sqrt{1 + k^{2}}\left\lbrack \frac{v^{2}}{2} \right\rbrack_{0}^{h}} \\ & {= \pi kh^{2}\sqrt{1 + k^{2}}.}\end{array}$$

Since $k = \text{tan}\ \alpha = r\text{/}h,$

$$\begin{array}{cl}{\pi kh^{2}\sqrt{1 + k^{2}}} & {= \pi\frac{r}{h}h^{2}\sqrt{1 + \frac{r^{2}}{h^{2}}}} \\ & {= \pi rh\sqrt{1 + \frac{r^{2}}{h^{2}}}} \\ & {= \pi r\sqrt{h^{2} + h^{2}\left( \frac{r^{2}}{h^{2}} \right)}} \\ & {= \pi r\sqrt{h^{2} + r^{2}}.}\end{array}$$

Therefore, the lateral surface area of the cone is $\pi r\sqrt{h^{2} + r^{2}}.$

Analysis 分析

The surface area of a right circular cone with radius *r* and height *h* is usually given as $\pi r^{2} + \pi r\sqrt{h^{2} + r^{2}}.$ The reason for this is that the circular base is included as part of the cone, and therefore the area of the base $\pi r^{2}$ is added to the lateral surface area $\pi r\sqrt{h^{2} + r^{2}}$ that we found.

Find the surface area of the surface with parameterization $\mathbf{\text{r}}(u,v) = \left\langle {u + v,u^{2},2v} \right\rangle,0 \leq u \leq 3,0 \leq v \leq 2.$

Calculating Surface Area 计算曲面面积

Show that the surface area of the sphere $x^{2} + y^{2} + z^{2} = r^{2}$ is $4\pi r^{2}.$

Solution 解

The sphere has parameterization

$$\left\langle {r\ \text{cos}\ \theta\ \text{sin}\ \phi,r\ \text{sin}\ \theta\ \text{sin}\ \phi,r\ \text{cos}\ \phi} \right\rangle,0 \leq \theta < 2\pi,0 \leq \phi \leq \pi.$$

The tangent vectors are

$$\mathbf{\text{t}}_{\theta} = \left\langle {\text{−}r\ \text{sin}\ \theta\ \text{sin}\ \phi,r\ \text{cos}\ \theta\ \text{sin}\ \phi,0} \right\rangle\ \text{and}\ \mathbf{\text{t}}_{\phi} = \left\langle {r\ \text{cos}\ \theta\ \text{cos}\ \phi,r\ \text{sin}\ \theta\ \text{cos}\ \phi,\text{−}r\ \text{sin}\ \phi} \right\rangle.$$

Therefore,

$$\begin{array}{cl}{\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta}} & {= \left\langle {r^{2}\text{cos}\ \theta\ \text{sin}^{2}\phi,r^{2}\text{sin}\ \theta\ \text{sin}^{2}\phi,r^{2}\text{sin}^{2}\theta\ \text{sin}\ \phi\ \text{cos}\ \phi + r^{2}\text{cos}^{2}\theta\ \text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle} \\ & {= \left\langle {r^{2}\text{cos}\ \theta\ \text{sin}^{2}\phi,r^{2}\text{sin}\ \theta\ \text{sin}^{2}\phi,r^{2}\text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle.}\end{array}$$

Now,

$$\begin{array}{cl}\left. \left\| {\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta}} \right. \right\| & {= \sqrt{r^{4}\text{sin}^{4}\phi\ \text{cos}^{2}\theta + r^{4}\text{sin}^{4}\phi\ \text{sin}^{2}\theta + r^{4}\text{sin}^{2}\phi\ \text{cos}^{2}\phi}} \\ & {= \sqrt{r^{4}\text{sin}^{4}\phi + r^{4}\text{sin}^{2}\phi\ \text{cos}^{2}\phi}} \\ & {= r^{2}\sqrt{\text{sin}^{2}\phi}} \\ & {= r^{2}\ \text{sin}\ \phi.}\end{array}$$

Notice that $\text{sin}\ \phi \geq 0$ on the parameter domain because $0 \leq \phi < \pi,$ and this justifies equation $\sqrt{\text{sin}^{2}\phi} = \text{sin}\ \phi.$ The surface area of the sphere is

$${\int_{0}^{2\pi}{\int_{0}^{\pi}{r^{2}\text{sin}\ \phi d\phi d\theta}}} = r^{2}{\int_{0}^{2\pi}{2d\theta}} = 4\pi r^{2}.$$

We have derived the familiar formula for the surface area of a sphere using surface integrals.

Show that the surface area of cylinder $x^{2} + y^{2} = r^{2},0 \leq z \leq h$ is $2\pi rh.$ Notice that this cylinder does not include the top and bottom circles.

In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. Let $y = f(x) \geq 0$ be a positive single-variable function on the domain $a \leq x \leq b$ and let *S* be the surface obtained by rotating $f$ about the *x*-axis (Figure 6.69). Let $\theta$ be the angle of rotation. Then, *S* can be parameterized with parameters *x* and $\theta$ by

$$\mathbf{\text{r}}\left( {x,\theta} \right) = \left\langle {x,f(x)\text{cos}\ \theta,f(x)\text{sin}\ \theta} \right\rangle,a \leq x \leq b,0 \leq \theta < 2\pi.$$

Calculating Surface Area 计算曲面面积

Find the area of the surface of revolution obtained by rotating $y = x^{2},0 \leq x \leq b$ about the *x*-axis (Figure 6.70).

Solution 解

This surface has parameterization

$$\mathbf{\text{r}}\left( {x,\theta} \right) = \left\langle {x,x^{2}\text{cos}\ \theta,x^{2}\text{sin}\ \theta} \right\rangle,0 \leq x \leq b,0 \leq \theta < 2\pi.$$

The tangent vectors are $\mathbf{\text{t}}_{x} = \left\langle {1,2x\ \text{cos}\ \theta,2x\ \text{sin}\ \theta} \right\rangle\ \text{and}\ \mathbf{\text{t}}_{\theta} = \left\langle {0,\text{−}x^{2}\text{sin}\ \theta,x^{2}\text{cos}\ \theta} \right\rangle.$ Therefore,

$$\begin{array}{cl}{\mathbf{\text{t}}_{x}\ \times \ \mathbf{\text{t}}_{\theta}} & {= \left\langle {2x^{3}\text{cos}^{2}\theta + 2x^{3}\text{sin}^{2}\theta,\text{−}x^{2}\text{cos}\ \theta,\text{−}x^{2}\text{sin}\ \theta} \right\rangle} \\ & {= \left\langle {2x^{3},\text{−}x^{2}\text{cos}\ \theta,\text{−}x^{2}\text{sin}\ \theta} \right\rangle}\end{array}$$

and

$$\begin{matrix}\mathbf{\left. ||{\text{t}_{x}\ \times \ \text{t}_{\theta}} \right.||} & {= \sqrt{4x^{6} + x^{4}\text{cos}^{2}\theta + x^{4}\text{sin}^{2}\theta}} \\ & {= \sqrt{4x^{6} + x^{4}}} \\ & {= x^{2}\sqrt{4x^{2} + 1}.}\end{matrix}$$

The area of the surface of revolution is

$$\begin{array}{cl}{\int_{0}^{b}{\int_{0}^{\pi}{x^{2}\sqrt{4x^{2} + 1}d\theta dx}}} & {= 2\pi{\int_{0}^{b}{x^{2}\sqrt{4x^{2} + 1}}}dx} \\ & {= 2\pi\left\lbrack {\frac{1}{64}\left( {2\sqrt{4x^{2} + 1}\left( {8x^{3} + x} \right){- \text{sinh}}^{-1}\left( {2x} \right)} \right)} \right\rbrack_{0}^{b}} \\ & {= 2\pi\left\lbrack {\frac{1}{64}\left( {2\sqrt{4b^{2} + 1}\left( {8b^{3} + b} \right){- \text{sinh}}^{-1}\left( {2b} \right)} \right)} \right\rbrack.}\end{array}$$

Use Equation 6.18 to find the area of the surface of revolution obtained by rotating curve $y = \text{sin}\ x,0 \leq x \leq \pi$ about the *x*-axis.

Surface Integral of a Scalar-Valued Function 标量值函数的曲面积分

Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let’s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain *region* (no longer a curve) into tiny pieces and proceed in the same fashion.

Let *S* be a piecewise smooth surface with parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle$ with parameter domain *D* and let $f\left( {x,y,z} \right)$ be a function with a domain that contains *S.* For now, assume the parameter domain *D* is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Divide rectangle *D* into subrectangles $D_{ij}$ with horizontal width $\text{Δ}u$ and vertical length $\text{Δ}v.$ Suppose that *i* ranges from 1 to *m* and *j* ranges from 1 to *n* so that *D* is subdivided into *mn* rectangles. This division of *D* into subrectangles gives a corresponding division of *S* into pieces $S_{ij}.$ Choose point $P_{ij}$ in each piece $S_{ij},$ evaluate $P_{ij}$ at $f$, and multiply by area $\text{Δ}S_{ij}$ to form the Riemann sum

$${\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( P_{ij} \right)}}}\text{Δ}S_{ij}.$$

To define a surface integral of a scalar-valued function, we let the areas of the pieces of *S* shrink to zero by taking a limit.

The surface integral of a scalar-valued function of $f$ over a piecewise smooth surface *S* is

$${\iint\limits_{S}{f\left( {x,y,z} \right)}}dS = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( P_{ij} \right)}}}\text{Δ}S_{ij}.$$

Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.

The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.

Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas $\text{Δ}S_{ij}$ with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors $\mathbf{\text{t}}_{u}$ and $\mathbf{\text{t}}_{v}\text{:}$

$$\mathbf{\text{t}}_{u} = \left\langle {\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}} \right\rangle\ \text{and}\ \mathbf{\text{t}}_{v} = \left\langle {\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}} \right\rangle.$$

From the material we have already studied, we know that

$$\text{Δ}S_{ij} \approx \left\| {\mathbf{\text{t}}_{u}\left( P_{ij} \right)\ \times \ \mathbf{\text{t}}_{v}\left( P_{ij} \right)} \right\|\text{Δ}u\text{Δ}v.$$

Therefore,

$${\iint_{S}{f(x,y,z)dS \approx \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( P_{ij} \right)\left\| {\mathbf{\text{t}}_{u}\left( P_{ij} \right)\ \times \ \mathbf{\text{t}}_{v}\left( P_{ij} \right)} \right\|\text{Δ}u\text{Δ}v}}}}}.$$

This approximation becomes arbitrarily close to $\underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( P_{ij} \right)\text{Δ}S_{ij}}}}$ as we increase the number of pieces $S_{ij}$ by letting *m* and *n* go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

$$\left. \iint{}_{S}{f(x,y,z)dS} \right. = {\iint\limits_{D}{f\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right)}}\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| dA.$$ (6.19)

Equation 6.19 allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to Equation 6.7 for line integrals:

$$\left. \int{}_{C}{f(x,y,z)ds} \right. = {\int_{a}^{b}{f\left( {\mathbf{\text{r}}(t)} \right)}}\left\| {\mathbf{r^{\prime}}(t)} \right\| dt.$$

In this case, vector $\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}$ is perpendicular to the surface, whereas vector $\mathbf{r^{\prime}}(t)$ is tangent to the curve.

Calculating a Surface Integral 计算一个曲面积分

Calculate surface integral ${\iint_{S}5}dS,$ where $S$ is the surface with parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {u,u^{2},v} \right\rangle$ for $0 \leq u \leq 2$ and $0 \leq v \leq u.$

Solution 解

Notice that this parameter domain *D* is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because Equation 6.19 does not place any restrictions on the shape of the parameter domain.

To use Equation 6.19 to calculate the surface integral, we first find vector $\mathbf{\text{t}}_{u}$ and $\mathbf{\text{t}}_{v}.$ Note that $\mathbf{\text{t}}_{u} = \left\langle {1,2u,0} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {0,0,1} \right\rangle.$ Therefore,

$$\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ 1 & {2u} & 0 \\ 0 & 0 & 1 \end{matrix} \right| = \left\langle {2u,-1,0} \right\rangle$$

and

$$\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| = \sqrt{1 + 4u^{2}}.$$

By Equation 6.19,

$$\begin{array}{cl} {{\iint_{S}5}dS} & {= 5{\iint_{D}\sqrt{1 + 4u^{2}}}dA} \\ & {= 5{\int_{0}^{2}{\int_{0}^{u}\sqrt{1 + 4u^{2}}}}dvdu = 5{\int_{0}^{2}{u\sqrt{1 + 4u^{2}}}}du} \\ & {= 5\left\lbrack \frac{\left( {1 + 4u^{2}} \right)^{3\text{/}2}}{3} \right\rbrack_{0}^{2} = \frac{5\left( {17^{3\text{/}2} - 1} \right)}{12} \approx 28.79.} \end{array}$$

Calculating the Surface Integral of a Cylinder 计算柱面的曲面积分

Calculate surface integral ${\iint_{S}{\left( {x + y^{2}} \right)dS}},$ where *S* is cylinder $x^{2} + y^{2} = 4,0 \leq z \leq 3$ (Figure 6.71).

Solution 解

To calculate the surface integral, we first need a parameterization of the cylinder. Following Example 6.58, a parameterization is

$$\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {2\text{cos}\ u,2\text{sin}\ u,v} \right\rangle,0 \leq u \leq 2\pi,0 \leq v \leq 3.$$

The tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {-2\text{sin}\ u,2\text{cos}\ u,0} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {0,0,1} \right\rangle.$ Then,

$$\textbf{t}_{u}\ \times \ \textbf{t}_{v} = \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ {- 2\text{sin}\ u} & {2\text{cos}\ u} & 0 \\ 0 & 0 & 1 \end{matrix} \right| = \left\langle 2\text{cos}\ u,2\text{sin}\ u,0 \right\rangle$$

and $\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| = \sqrt{{4\text{cos}}^{2}u + {4\text{sin}}^{2}u} = 2.$ By Equation 6.19,

$$\begin{matrix} \\ \\ \\ {\mspace{22mu}\iint_{S}{f(x,y,z)dS = \iint_{D}f\left( {\textbf{r}(u,v)} \right)\left\| \textbf{t}_{u}\ \times \ \textbf{t}_{v} \right\|}\ dA} \\ {= \int_{0}^{3}\int_{0}^{2\pi}\left( 2\text{cos}\ u + 4\text{sin}^{2}u \right)2dudv} \\ {= 2\int_{0}^{3}\left\lbrack 2\text{sin}\ u + 2u - \sin(2u) \right\rbrack_{0}^{2\pi}dv = 2\int_{0}^{3}4\pi dv = 24\pi.} \end{matrix}$$

Calculate ${\iint_{S}\left( {x^{2} - z} \right)}dS,$ where *S* is the surface with parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {v,u^{2} + v^{2},1} \right\rangle,0 \leq u \leq 2,0 \leq v \leq 3.$

Calculating the Surface Integral of a Piece of a Sphere 计算球面的一部分的曲面积分

Calculate surface integral ${\iint_{S}{f(x,y,z)dS}},$ where $f(x,y,z) = z^{2}$ and *S* is the surface that consists of the piece of sphere $x^{2} + y^{2} + z^{2} = 4$ that lies on or above plane $z = 1$ and the disk that is enclosed by intersection plane $z = 1$ and the given sphere (Figure 6.72).

Solution 解

Notice that *S* is not smooth but is piecewise smooth; *S* can be written as the union of its base $S_{1}$ and its spherical top $S_{2},$ and both $S_{1}$ and $S_{2}$ are smooth. Therefore, to calculate ${\iint_{S}{z^{2}dS}},$ we write this integral as ${\iint_{S_{1}}{z^{2}dS}} + {\iint_{S_{2}}{z^{2}dS}}$ and we calculate integrals $\iint_{S_{1}}{z^{2}dS}$ and ${\iint_{S_{2}}{z^{2}dS}}.$

First, we calculate ${\iint_{S_{1}}{z^{2}dS}}.$ To calculate this integral we need a parameterization of $S_{1}.$ This surface is a disk in plane $z = 1$ centered at $(0,0,1).$ To parameterize this disk, we need to know its radius. Since the disk is formed where plane $z = 1$ intersects sphere $x^{2} + y^{2} + z^{2} = 4,$ we can substitute $z = 1$ into equation $x^{2} + y^{2} + z^{2} = 4\text{:}$

$$x^{2} + y^{2} + 1 = 4\Rightarrow x^{2} + y^{2} = 3.$$

Therefore, the radius of the disk is $\sqrt{3}$ and a parameterization of $S_{1}$ is $\mathbf{\text{r}}(u,v) = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,1} \right\rangle,0 \leq u \leq \sqrt{3},0 \leq v \leq 2\pi.$ The tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {\text{cos}\ v,\text{sin}\ v,0} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {\text{−}u\ \text{sin}\ v,u{co}sv,0} \right\rangle,$ and thus

$$\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ {\text{cos}\ v} & {\text{sin}\ v} & 0 \\ {\text{−}u\ \text{sin}\ v} & {u\ \text{cos}\ v} & 0 \end{matrix} \right| = \left\langle {0,0,u\ \text{cos}^{2}v + u\ \text{sin}^{2}v} \right\rangle = \left\langle {0,0,u} \right\rangle.$$

The magnitude of this vector is *u*. Therefore,

$$\begin{array}{cl} {\iint_{S_{1}}{z^{2}dS}} & {= {\int_{0}^{\sqrt{3}}{\int_{0}^{2\pi}{f\left( {\mathbf{\text{r}}(u,v)} \right)}}}\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\|\ dv\ du} \\ & {= {\int_{0}^{\sqrt{3}}{\int_{0}^{2\pi}u}}\ dv\ du} \\ & {= 2\pi{\int_{0}^{\sqrt{3}}{udu}}} \\ & {= 3\pi.} \end{array}$$

Now we calculate ${\iint_{S_{2}}{dS}}.$ To calculate this integral, we need a parameterization of $S_{2}.$ The parameterization of full sphere $x^{2} + y^{2} + z^{2} = 4$ is

$$\mathbf{\text{r}}\left( {\phi,\theta} \right) = \left\langle {2\ \text{cos}\ \theta\ \text{sin}\ \phi,2\ \text{sin}\ \theta\ \text{sin}\ \phi,2\ \text{cos}\ \phi} \right\rangle,0 \leq \theta \leq 2\pi,0 \leq \phi \leq \pi.$$

Since we are only taking the piece of the sphere on or above plane $z = 1,$ we have to restrict the domain of $\phi.$ To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in Figure 6.73 (the $\sqrt{3}$ comes from the fact that the base of *S* is a disk with radius $\sqrt{3}).$ Therefore, the tangent of $\phi$ is $\sqrt{3},$ which implies that $\phi$ is $\pi\text{/}3.$ We now have a parameterization of $S_{2}\text{:}$

$$\mathbf{\text{r}}\left( {\phi,\theta} \right) = \left\langle {2\ \text{cos}\ \theta\ \text{sin}\ \phi,2\ \text{sin}\ \theta\ \text{sin}\ \phi,2\ \text{cos}\ \phi} \right\rangle,0 \leq \theta \leq 2\pi,0 \leq \phi \leq \pi\text{/}3.$$

The tangent vectors are

$$\mathbf{\text{t}}_{\phi} = \left\langle {2\ \text{cos}\ \theta\ \text{cos}\ \phi,2\ \text{sin}\ \theta\ \text{cos}\ \phi,-2\ \text{sin}\ \phi} \right\rangle\ \text{and}\ \mathbf{\text{t}}_{\theta} = \left\langle {-2\ \text{sin}\ \theta\ \text{sin}\ \phi,u\ \text{cos}\ \theta\ \text{sin}\ \phi,0} \right\rangle,$$

and thus

$$\begin{array}{cl} {\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta}} & {= \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ {2\ \text{cos}\ \theta\ \text{cos}\ \phi} & {2\ \text{sin}\ \theta\ \text{cos}\ \phi} & {-2\ \text{sin}\ \phi} \\ {-2\ \text{sin}\ \theta\ \text{sin}\ \phi} & {2\ \text{cos}\ \theta\ \text{sin}\ \phi} & 0 \end{matrix} \right|} \\ & {= \left\langle {4\ \text{cos}\ \theta\ \text{sin}^{2}\phi,4\ \text{sin}\ \theta\ \text{sin}^{2}\phi,4\ \text{cos}^{2}\theta\ \text{cos}\ \phi\ \text{sin}\ \phi + 4\ \text{sin}^{2}\theta\ \text{cos}\ \phi\ \text{sin}\ \phi} \right\rangle} \\ & {= \left\langle {4\ \text{cos}\ \theta\ \text{sin}^{2}\phi,4\ \text{sin}\ \theta\ \text{sin}^{2}\phi,4\ \text{cos}\ \phi\ \text{sin}\ \phi} \right\rangle.} \end{array}$$

The magnitude of this vector is

$$\begin{array}{cl} \left\| {\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta}} \right\| & {= \sqrt{16\ \text{cos}^{2}\theta\ \text{sin}^{4}\phi + 16\ \text{sin}^{2}\theta\ \text{sin}^{4}\phi + 16\ \text{cos}^{2}\phi\ \text{sin}^{2}\phi}} \\ & {= 4\sqrt{\text{sin}^{4}\phi + \text{cos}^{2}\phi\ \text{sin}^{2}\phi}.} \end{array}$$

Therefore,

$$\begin{matrix} {\iint_{S_{2}}zdS} & {= \int_{0}^{\pi\text{/3}}\int_{0}^{2\pi}f\left( {\textbf{r}(\phi,\theta)} \right)\left\| \textbf{t}_{\phi}\ \times \ \textbf{t}_{\theta} \right\|\ d\theta\ d\phi} \\ & {= \int_{0}^{\pi\text{/3}}\int_{0}^{2\pi}16\ \text{cos}^{2}\phi\sqrt{\text{sin}^{4}\phi + \text{cos}^{2}\phi\ \text{sin}^{2}\phi}d\theta\ d\phi} \\ & {= 32\pi\int_{0}^{\pi\text{/3}}\text{cos}^{2}\phi\sqrt{\text{sin}^{4}\phi + \text{cos}^{2}\phi\ \text{sin}^{2}\phi}\ d\phi} \\ & {= 32\pi\int_{0}^{\pi\text{/3}}\text{cos}^{2}\phi\ \text{sin}\ \phi\sqrt{\text{sin}^{2}\phi + \text{cos}^{2}\phi}\ d\phi} \\ & {= 32\pi\int_{0}^{\pi\text{/3}}\text{cos}^{2}\phi\ \text{sin}\ \phi\ d\phi} \\ & {= 32\pi\left\lbrack - \frac{\text{cos}^{3}\phi}{3} \right\rbrack_{0}^{\pi\text{/}3} = 32\pi\left\lbrack \frac{1}{3} - \frac{\sqrt{3}}{8} \right\rbrack = \frac{28\pi}{3}.} \end{matrix}$$

Since $\iint_{S}z^{2}dS = \iint_{S_{1}}z^{2}dS + \iint_{S_{2}}z^{2}dS = 3\pi + \frac{28\pi}{3} = \frac{37\pi}{3}$

Analysis 分析

In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.

Calculate surface integral ${\iint_{S}{\left( {x - y} \right)dS}},$ where *S* is cylinder $x^{2} + y^{2} = 1,0 \leq z \leq 2,$ including the circular top and bottom.

Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface *S* and the density of the sheet at point $\left( {x,y,z} \right)$ is $\rho\left( {x,y,z} \right),$ then mass *m* of the sheet is $m = {\iint_{S}{\rho\left( {x,y,z} \right)dS}}.$

Calculating the Mass of a Sheet 计算一片薄板的质量

A flat sheet of metal has the shape of surface $z = 1 + x + 2y$ that lies above rectangle $0 \leq x \leq 4$ and $0 \leq y \leq 2.$ If the density of the sheet is given by $\rho\left( {x,y,z} \right) = x^{2}yz,$ what is the mass of the sheet?

Solution 解

Let *S* be the surface that describes the sheet. Then, the mass of the sheet is given by $m = {\iint_{S}{x^{2}yzdS}}.$ To compute this surface integral, we first need a parameterization of *S*. Since *S* is given by the function $f\left( {x,y} \right) = 1 + x + 2y,$ a parameterization of *S* is $\mathbf{\text{r}}\left( {x,y} \right) = \left\langle {x,y,1 + x + 2y} \right\rangle,0 \leq x \leq 4,0 \leq y \leq 2.$

The tangent vectors are $\mathbf{\text{t}}_{x} = \left\langle {1,0,1} \right\rangle$ and $\mathbf{\text{t}}_{y} = \left\langle {1,0,2} \right\rangle.$ Therefore, $\mathbf{\text{t}}_{x}\ \times \ \mathbf{\text{t}}_{y} = \left\langle {-1,-2,1} \right\rangle$ and $\left\| \left. \mathbf{\text{t}}_{x}\ \times \ \mathbf{\text{t}}_{y} \right\| \right. = \sqrt{6}.$ By Equation 6.5,

$$\begin{array}{cl} m & {= {\iint_{S}{x^{2}yz^{}dS}}} \\ & {= \sqrt{6}{\int_{0}^{4}{\int_{0}^{2}{x^{2}y\left( {1 + x + 2y} \right)dydx}}}} \\ & {= \sqrt{6}{\int_{0}^{4}\frac{22x^{2}}{3}} + 2x^{3}dx} \\ & {= \frac{2560\sqrt{6}}{9}} \\ & {\approx 696.74.} \end{array}$$

A piece of metal has a shape that is modeled by paraboloid $z = x^{2} + y^{2},0 \leq z \leq 4,$ and the density of the metal is given by $\rho(x,y,z) = z + 1.$ Find the mass of the piece of metal.

Orientation of a Surface 曲面的定向

Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration.

On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation.

Let *S* be a smooth surface. For any point $\left( {x,y,z} \right)$ on *S,* we can identify two unit normal vectors $\mathbf{\text{N}}$ and $\text{−}\mathbf{\text{N}}.$ If it is possible to choose a unit normal vector N at every point $\left( {x,y,z} \right)$ on *S* so that N varies continuously over *S*, then *S* is “*orientable*.” Such a choice of unit normal vector at each point gives the orientation of a surface *S*. If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the “negative” side and the side of the surface at which the water flows away is the “positive” side. Informally, a choice of orientation gives *S* an “outer” side and an “inner” side (or an “upward” side and a “downward” side), just as a choice of orientation of a curve gives the curve “forward” and “backward” directions.

Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the *positive orientation of the closed surface* (Figure 6.74). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.

A portion of the graph of any smooth function $z = f(x,y)$ is also orientable. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points “below” the surface at each point. To get such an orientation, we parameterize the graph of $f$ in the standard way: $\mathbf{\text{r}}(x,y) = \left\langle {x,y,f(x,y)} \right\rangle,$ where *x* and *y* vary over the domain of $f.$ Then, $\mathbf{\text{t}}_{x} = \left\langle {1,0,f_{x}} \right\rangle$ and $\mathbf{\text{t}}_{y} = \left\langle {0,1,f_{y}} \right\rangle,$ and therefore the cross product $\mathbf{\text{t}}_{x}\ \times \ \mathbf{\text{t}}_{y}$ (which is normal to the surface at any point on the surface) is $\left\langle {\text{−}f_{x},\text{−}f_{y},1} \right\rangle.$ Since the *z* component of this vector is one, the corresponding unit normal vector points “upward,” and the upward side of the surface is chosen to be the “positive” side.

Let *S* be a smooth orientable surface with parameterization $\mathbf{\text{r}}\left( {u,v} \right).$ For each point $\mathbf{\text{r}}\left( {a,b} \right)$ on the surface, vectors $\mathbf{\text{t}}_{u}$ and $\mathbf{\text{t}}_{v}$ lie in the tangent plane at that point. Vector $\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}$ is normal to the tangent plane at $\mathbf{\text{r}}\left( {a,b} \right)$ and is therefore normal to *S* at that point. Therefore, the choice of unit normal vector

$$\mathbf{\text{N}} = \frac{\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\|}$$

gives an orientation of surface *S*.

Choosing an Orientation 选择定向

Give an orientation of cylinder $x^{2} + y^{2} = r^{2},0 \leq z \leq h.$

Solution 解

This surface has parameterization

$$\mathbf{\text{r}}(u,v) = \left\langle {r\ \text{cos}\ u,r\ \text{sin}\ u,v} \right\rangle,0 \leq u < 2\pi,0 \leq v \leq h.$$

The tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {\text{−}r\ \text{sin}\ u,r\ \text{cos}\ u,0} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {0,0,1} \right\rangle.$ To get an orientation of the surface, we compute the unit normal vector

$$\mathbf{\text{N}} = \frac{\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\|}.$$

In this case, $\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left\langle {r\ \text{cos}\ u,r\ \text{sin}\ u,0} \right\rangle$ and therefore

$$\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| = \sqrt{r^{2}\text{cos}^{2}u + r^{2}\text{sin}^{2}u} = r.$$

An orientation of the cylinder is

$$\mathbf{\text{N}}(u,v) = \frac{\left\langle {r\ \text{cos}\ u,r\ \text{sin}\ u,0} \right\rangle}{r} = \left\langle {\text{cos}\ u,\text{sin}\ u,0} \right\rangle.$$

Notice that all vectors are parallel to the *xy*-plane, which should be the case with vectors that are normal to the cylinder. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure 6.75).

Give the “upward” orientation of the graph of $f(x,y) = xy.$

Since every curve has a “forward” and “backward” direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. Hence, it is possible to think of every curve as an oriented curve. This is not the case with surfaces, however. Some surfaces cannot be oriented; such surfaces are called *nonorientable*. Essentially, a surface can be oriented if the surface has an “inner” side and an “outer” side, or an “upward” side and a “downward” side. Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side.

The classic example of a nonorientable surface is the Möbius strip. To create a Möbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure 6.76). Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Therefore, the strip really only has one side.

Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve.

Surface Integral of a Vector Field 向量场的曲面积分

With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field. The definition is analogous to the definition of the flux of a vector field along a plane curve. Recall that if F is a two-dimensional vector field and *C* is a plane curve, then the definition of the flux of F along *C* involved chopping *C* into small pieces, choosing a point inside each piece, and calculating $\mathbf{\text{F}} \cdot \mathbf{\text{N}}$ at the point (where N is the unit normal vector at the point). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface *S* into small pieces, choose a point in the small (two-dimensional) piece, and calculate $\mathbf{\text{F}} \cdot \mathbf{\text{N}}$ at the point.

To place this definition in a real-world setting, let *S* be an oriented surface with unit normal vector N. Let v be a velocity field of a fluid flowing through *S*, and suppose the fluid has density $\rho\left( {x,y,z} \right).$ Imagine the fluid flows through *S*, but *S* is completely permeable so that it does not impede the fluid flow (Figure 6.77). The mass flux of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area. How could we calculate the mass flux of the fluid across *S*?

The rate of flow, measured in mass per unit time per unit area, is $\rho\mathbf{\text{N}}.$ To calculate the mass flux across *S*, chop *S* into small pieces $S_{ij}.$ If $S_{ij}$ is small enough, then it can be approximated by a tangent plane at some point *P* in $S_{ij}.$ Therefore, the unit normal vector at *P* can be used to approximate $\mathbf{\text{N}}\left( {x,y,z} \right)$ across the entire piece $S_{ij},$ because the normal vector to a plane does not change as we move across the plane. The component of the vector $\rho\mathbf{\text{v}}$ at *P* in the direction of N is $\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}$ at *P*. Since $S_{ij}$ is small, the dot product $\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}$ changes very little as we vary across $S_{ij},$ and therefore $\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}$ can be taken as approximately constant across $S_{ij}.$ To approximate the mass of fluid per unit time flowing across $S_{ij}$ (and not just locally at point *P*), we need to multiply $\left( {\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}} \right)(P)$ by the area of $S_{ij}.$ Therefore, the mass of fluid per unit time flowing across $S_{ij}$ in the direction of N can be approximated by $\left( {\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}} \right)\text{Δ}S_{ij},$ where N, $\rho,$ and v are all evaluated at *P* (Figure 6.78). This is analogous to the flux of two-dimensional vector field F across plane curve *C*, in which we approximated flux across a small piece of *C* with the expression $\left( {\mathbf{\text{F}} \cdot \mathbf{\text{N}}} \right)\text{Δ}s.$ To approximate the mass flux across *S*, form the sum ${\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}\left( {\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}} \right)}}\text{Δ}\mathbf{\text{S}}_{ij}.$ As pieces $S_{ij}$ get smaller, the sum ${\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}\left( {\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}} \right)}}\text{Δ}\mathbf{\text{S}}_{ij}$ gets arbitrarily close to the mass flux. Therefore, the mass flux is

$${\iint_{s}{\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}dS =}}\underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}\left( {\rho\mathbf{\text{v}} \cdot \mathbf{\text{N}}} \right)}}\text{Δ}\mathbf{\text{S}}_{ij}.$$

This is a surface integral of a vector field. Letting the vector field $\rho\mathbf{\text{v}}$ be an arbitrary vector field F leads to the following definition.

Let F be a continuous vector field with a domain that contains oriented surface *S* with unit normal vector N. The surface integral of F over *S* is

$${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}} = {\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}}}}.$$ (6.20)

Notice the parallel between this definition and the definition of vector line integral ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}.$ A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Integral $\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}$ is called the *flux of F across S*, just as integral $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}$ is the flux of F across curve *C*. A surface integral over a vector field is also called a flux integral.

Just as with vector line integrals, surface integral $\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}$ is easier to compute after surface *S* has been parameterized. Let $\mathbf{\text{r}}\left( {u,v} \right)$ be a parameterization of *S* with parameter domain *D*. Then, the unit normal vector is given by $\mathbf{\text{N}} = \frac{\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\|}$ and, from Equation 6.20, we have

$$\begin{array}{cl} {\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} & {= {\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}}} \\ & {= {\iint_{S}{\mathbf{\text{F}} \cdot \frac{\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\|}dS}}} \\ & {= {\iint_{D}{\left( {\mathbf{\text{F}}\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right) \cdot \frac{\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}}{\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\|}} \right)\left\| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right\| dA}}} \\ & {= {\iint_{D}{\left( {\mathbf{\text{F}}\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right) \cdot \left( {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right)} \right)dA}}}. \end{array}$$

Therefore, to compute a surface integral over a vector field we can use the equation

$${\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}} = {\iint_{D}{\left( {\mathbf{\text{F}}\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right) \cdot \left( {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right)} \right)dA}}.$$ (6.21)

Calculating a Surface Integral 计算曲面积分

Calculate the surface integral ${\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}},$ where $\mathbf{\text{F}} = \left\langle {\text{−}y,x,0} \right\rangle$ and $S$ is the surface with parameterization $\mathbf{\text{r}}(u,v) = \left\langle {u,v^{2} - u,u + v} \right\rangle,0 \leq u < 3,0 \leq v \leq 4.$

Solution 解

The tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {1,-1,1} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {0,2v,1} \right\rangle.$ Therefore,

$$\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left\langle {-1 - 2v,-1,2v} \right\rangle.$$

By Equation 6.21,

$$\begin{array}{cl} {{\iint_{S}\mathbf{\text{F}}} \cdot d\mathbf{\text{S}}} & {= {\int_{0}^{4}{\int_{0}^{3}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}(u,v)} \right)}}} \cdot \left( {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right)dudv} \\ & {= {\int_{0}^{4}{\int_{0}^{3}{\left\langle {u - v^{2},u,0} \right\rangle \cdot \left\langle {-1 - 2v,-1,2v} \right\rangle}}}\ dudv} \\ & {= {\int_{0}^{4}{\int_{0}^{3}\left\lbrack {\left( {u - v^{2}} \right)\left( {-1 - 2v} \right) - u} \right\rbrack}}dudv} \\ & {= {\int_{0}^{4}{\int_{0}^{3}\left( {2v^{3} + v^{2} - 2uv - 2u} \right)}}dudv} \\ & {= {\int_{0}^{4}\left\lbrack {2v^{3}u + v^{2}u - vu^{2} - u^{2}} \right\rbrack}_{0}^{3}dv} \\ & {= {\int_{0}^{4}{\left( {6v^{3} + 3v^{2} - 9v - 9} \right)dv}}} \\ & {= \left\lbrack {\frac{3v^{4}}{2} + v^{3} - \frac{9v^{2}}{2} - 9v} \right\rbrack_{0}^{4}} \\ & {= 340.} \end{array}$$

Therefore, the flux of F across *S* is 340.

Calculate surface integral ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}} = \left\langle {0,\text{−}z,y} \right\rangle$ and *S* is the portion of the unit sphere in the first octant with outward orientation.

Calculating Mass Flow Rate 计算质量流率

Let $\mathbf{\text{v}}\left( {x,y,z} \right) = \left\langle {2x,2y,z} \right\rangle$ represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m³. Let *S* be hemisphere $x^{2} + y^{2} + z^{2} = 9$ with $z \geq 0$ such that *S* is oriented outward. Find the mass flow rate of the fluid across *S*.

Solution 解

A parameterization of the surface is

$$\mathbf{\text{r}}\left( {\phi,\theta} \right) = \left\langle {3\ \text{cos}\ \theta\ \text{sin}\ \phi,3\ \text{sin}\ \theta\ \text{sin}\ \phi,3\ \text{cos}\ \phi} \right\rangle,0 \leq \theta \leq 2\pi,0 \leq \phi \leq {\pi\text{/}2.}$$

As in Example 6.64, the tangent vectors are

$$\mathbf{\text{t}}_{\theta}{= \left\langle {-3\ \text{sin}\ \theta\ \text{sin}\ \phi,3\ \text{cos}\ \theta\ \text{sin}\ \phi,0} \right\rangle}\ \text{and}\ \mathbf{\text{t}}_{\phi}{= \left\langle {3\ \text{cos}\ \theta\ \text{cos}\ \phi,3\ \text{sin}\ \theta\ \text{cos}\ \phi,-3\ \text{sin}\ \phi} \right\rangle},$$

and their cross product is

$$\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta} = \left\langle {9\ \text{cos}\ \theta\ \text{sin}^{2}\phi,9\ \text{sin}\ \theta\ \text{sin}^{2}\phi,9\ \text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle.$$

Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation $\mathbf{\text{N}} = \left\langle {9\ \text{cos}\ \theta\ \text{sin}^{2}\phi,9\ \text{sin}\ \theta\ \text{sin}^{2}\phi,9\ \text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle$ for the sphere.

By [link],

$$','$$']] }, svg: { fontCache: 'global' } }; $$ \begin{array}{cl} {\iint_{S}{\rho\mathbf{\text{v}} \cdot d\mathbf{\text{S}}}} & {= 80{\int_{0}^{2\pi}{\int_{0}^{\pi\text{/}2}{\mathbf{\text{v}}\left( {\mathbf{\text{r}}\left( {\phi,\theta} \right)} \right)}}} \cdot \left( {\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta}} \right)d\phi d\theta} \\ & {= 80{\int_{0}^{2\pi}{\int_{0}^{\pi\text{/}2}{\begin{array}{l} \left\langle {6\ \text{cos}\ \theta\ \text{sin}\ \phi,6\ \text{sin}\ \theta\ \text{sin}\ \phi,3\ \text{cos}\ \phi} \right\rangle \\ {\cdot \left\langle {9\ \text{cos}\ \theta\ \text{sin}^{2}\phi,9\ \text{sin}\ \theta\ \text{sin}^{2}\phi,9\ \text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle d\phi d\theta} \end{array}}}} \\ & {= 80{\int_{0}^{2\pi}{\int_{0}^{\pi\text{/}2}{54\ \text{sin}^{3}\phi + 27\ \text{cos}^{2}\phi\ \text{sin}\ \phi d\phi d\theta}}}} \\ & {= 80{\int_{0}^{2\pi}{\int_{0}^{\pi\text{/}2}{54\left( {1 - \text{cos}^{2}\phi} \right)\text{sin}\ \phi + 27\ \text{cos}^{2}\phi\ \text{sin}\ \phi d\phi d\theta}}}} \\ & {= 80{\int_{0}^{2\pi}{\int_{0}^{\pi\text{/}2}{54\ \text{sin}\ \phi - 27\ \text{cos}^{2}\phi\ \text{sin}\ \phi d\phi d\theta}}}} \\ & {= 80{\int_{0}^{2\pi}{\left\lbrack {-54\ \text{cos}\ \phi + 9\ \text{cos}^{3}\phi} \right\rbrack_{\phi = 0}^{\phi = 2\pi}d\theta}}} \\ & {= 80{\int_{0}^{2\pi}{45d\theta}} = 7200\pi.} \end{array}$$

Therefore, the mass flow rate is $7200\pi\ \text{kg}\text{/}\text{sec}\text{/}\text{m}^{2}.$

Let $\mathbf{\text{v}}(x,y,z) = \left\langle {x^{2} + y^{2},z,4y} \right\rangle$ m/sec represent a velocity field of a fluid with constant density 100 kg/m³. Let *S* be the half-cylinder $\mathbf{\text{r}}(u,v) = \left\langle {\text{cos}\ u,\text{sin}\ u,v} \right\rangle,0 \leq u \leq \pi,0 \leq v \leq 2$ oriented outward. Calculate the mass flux of the fluid across *S*.

In Example 6.71, we computed the mass flux, which is the rate of mass flow per unit area. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral $\iint_{S}\textbf{v} \cdot \textbf{N}dS,$ which leaves out the density. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization of the flow rate of a fluid with velocity v across a surface *S*:

$$\text{Flow rate of fluid across}\ S = \iint_{S}v \cdot d\textbf{S}.$$

To compute the flow rate of the fluid in Example 6.71, we simply remove the density constant, which gives a flow rate of $90\pi{\ \text{m}}^{3}\text{/}\text{sec}.$

Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.

In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point $(x,y,z)$ in an object is $T(x,y,z).$ Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field $\mathbf{\text{F}} = \text{−}k\nabla T,$ where the constant *k* is the *thermal conductivity* of the substance from which the object is made (this constant is determined experimentally). The rate of heat flow across surface *S* in the object is given by the flux integral

$${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\iint_{S}{\text{−}k\nabla T \cdot d\mathbf{\text{S}}}}.$$

Calculating Heat Flow 计算热流

A cast-iron solid cylinder is given by inequalities $x^{2} + y^{2} \leq 1,$ $1 \leq z \leq 4.$ The temperature at point $(x,y,z)$ in a region containing the cylinder is $T(x,y,z) = \left( {x^{2} + y^{2}} \right)z.$ Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.

Solution 解

Let *S* denote the boundary of the object. To find the heat flow, we need to calculate flux integral ${\iint_{S}{\text{−}k\nabla T \cdot d\mathbf{\text{S}}}}.$ Notice that *S* is not a smooth surface but is piecewise smooth, since *S* is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of *S*. Before calculating any integrals, note that the gradient of the temperature is $\nabla T = \left\langle {2xz,2yz,x^{2} + y^{2}} \right\rangle.$

First we consider the circular bottom of the object, which we denote $S_{1}.$ We can see that $S_{1}$ is a circle of radius 1 centered at point $(0,0,1),$ sitting in plane $z = 1.$ This surface has parameterization $\mathbf{\text{r}}(u,v) = \left\langle {v\ \text{cos}\ u,v\ \text{sin}\ u,1} \right\rangle,0 \leq u < 2\pi,0 \leq v \leq 1.$ Therefore,

$$\mathbf{\text{t}}_{u} = \left\langle {\text{−}v\ \text{sin}\ u,v\ \text{cos}\ u,0} \right\rangle\ \text{and}\ \mathbf{\text{t}}_{v} = \left\langle {\text{cos}\ u,v\ \text{sin}\ u,0} \right\rangle,$$

and

$$\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left\langle {0,0,\text{−}v\ \text{sin}^{2}u - v\ \text{cos}^{2}u} \right\rangle = \left\langle {0,0,\text{−}v} \right\rangle.$$

Since the surface is oriented outward and $S_{1}$ is the bottom of the object, it makes sense that this vector points downward. By Equation 6.21, the heat flow across $S_{1}$ is

$$\begin{array}{cl} {\iint_{S_{1}}{\text{−}k\nabla T \cdot d\mathbf{\text{S}}}} & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{\nabla T(u,v) \cdot \left( {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right)dvdu}}}} \\ & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{\left\langle {2v\ \text{cos}\ u,2v\ \text{sin}\ u,v^{2}\text{cos}^{2}u + v^{2}\text{sin}^{2}u} \right\rangle \cdot \left\langle {0,0,\text{−}v} \right\rangle dvdu}}} \\ & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{\left\langle {2v\ \text{cos}\ u,2v\ \text{sin}\ u,v^{2}} \right\rangle \cdot \left\langle {0,0,\text{−}v} \right\rangle dvdu}}}} \\ & {= -55{\int_{0}^{2\pi}{{\int_{0}^{1}{\text{−}v^{3}}}dvdu}} = -55{\int_{0}^{2\pi}{- \frac{1}{4}du}} = \frac{55\pi}{2}.} \end{array}}$$}

Now let’s consider the circular top of the object, which we denote $S_{2}.$ We see that $S_{2}$ is a circle of radius 1 centered at point $(0,0,4),$ sitting in plane $z = 4.$ This surface has parameterization $\mathbf{\text{r}}(u,v) = \left\langle {v\ \text{cos}\ u,v\ \text{sin}\ u,4} \right\rangle,0 \leq u < 2\pi,0 \leq v \leq 1.$ Therefore,

$$\mathbf{\text{t}}_{u} = \left\langle {\text{−}v\ \text{sin}\ u,v\ \text{cos}\ u,0} \right\rangle\ \text{and}\ \mathbf{\text{t}}_{v} = \left\langle {\text{cos}\ u,v\ \text{sin}\ u,0} \right\rangle,$$

and

$$\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left\langle {0,0,\text{−}v\ \text{sin}^{2}u - v\ \text{cos}^{2}u} \right\rangle = \left\langle {0,0,\text{−}v} \right\rangle.$$

Since the surface is oriented outward and $S_{1}$ is the top of the object, we instead take vector $\mathbf{\text{t}}_{v}\ \times \ \mathbf{\text{t}}_{u} = \left\langle {0,0,v} \right\rangle.$ By Equation 6.21, the heat flow across $S_{1}$ is

$$\begin{array}{cl} {{\int{\int_{S_{2}}{\text{−}k}}}\nabla T \cdot d\mathbf{\text{S}}} & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{\nabla T\left( {u,v} \right) \cdot \left( {\mathbf{\text{t}}_{v}\ \times \ \mathbf{\text{t}}_{u}} \right)dvdu}}}} \\ & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{\left\langle {8v\ \text{cos}\ u,8v\ \text{sin}\ u,v^{2}\text{cos}^{2}u + v^{2}\text{sin}^{2}u} \right\rangle \cdot \left\langle {0,0,v} \right\rangle dvdu}}}} \\ & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{\left\langle {8v\ \text{cos}\ u,8v\ \text{sin}\ u,v^{2}} \right\rangle \cdot \left\langle {0,0,v} \right\rangle dvdu}}}} \\ & {= -55{\int_{0}^{2\pi}{\int_{0}^{1}{v^{3}dvdu}}} = - \frac{55\pi}{2}.} \end{array}$$

Last, let’s consider the cylindrical side of the object. This surface has parameterization $\mathbf{\text{r}}(u,v) = \left\langle {\text{cos}\ u,\text{sin}\ u,v} \right\rangle,0 \leq u < 2\pi,1 \leq v \leq 4.$ By Example 6.66, we know that $\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left\langle {\text{cos}\ u,\text{sin}\ u,0} \right\rangle.$ By Equation 6.21,

$$\begin{array}{cl} {\iint_{S_{3}}{\text{−}k\nabla T \cdot d\mathbf{\text{S}}}} & {= -55{\int_{0}^{2\pi}{\int_{1}^{4}{\nabla T(u,v) \cdot \left( {\mathbf{\text{t}}_{v}\ \times \ \mathbf{\text{t}}_{u}} \right)dvdu}}}} \\ & {= -55{\int_{0}^{2\pi}{{\int_{1}^{4}{\left\langle {2v\ \text{cos}\ u,2v\ \text{sin}\ u,\text{cos}^{2}u + \text{sin}^{2}u} \right\rangle \cdot \left\langle {\text{cos}\ u,\text{sin}\ u,0} \right\rangle}}dvdu}}} \\ & {= -55{\int_{0}^{2\pi}{{\int_{0}^{1}{\left\langle {2v\ \text{cos}\ u,2v\ \text{sin}\ u,1} \right\rangle \cdot \left\langle {\text{cos}\ u,\text{sin}\ u,0} \right\rangle}}dvdu}}} \\ & {= -55{\int_{0}^{2\pi}{{\int_{0}^{1}\left( {2v\ \text{cos}^{2}u + 2v\ \text{sin}^{2}u} \right)}dvdu}}} \\ & {= -55{\int_{0}^{2\pi}{{\int_{0}^{1}{2v}}dvdu}} = -55{\int_{0}^{2\pi}{du}} = -110\pi.} \end{array}$$

Therefore, the rate of heat flow across *S* is $\frac{55\pi}{2} - \frac{55\pi}{2} - 110\pi = -110\pi.$

A cast-iron solid ball is given by inequality $x^{2} + y^{2} + z^{2} \leq 1.$ The temperature at a point in a region containing the ball is $T(x,y,z) = \frac{1}{3}\left( {x^{2} + y^{2} + z^{2}} \right).$ Find the heat flow across the boundary of the solid if this boundary is oriented outward.

Section 6.6 Exercises 6.6 节习题

For the following exercises, determine whether the statements are *true or false*.

269.

If surface *S* is given by $\left\{ {\left( {x,y,z} \right):0 \leq x \leq 1,0 \leq y \leq 1,z = 10} \right\},$ then ${\iint_{S}{f\left( {x,y,z} \right)}}dS = {\int_{0}^{1}{\int_{0}^{1}{f\left( {x,y,10} \right)}}}dxdy.$

270\.

If surface *S* is given by $\left\{ {\left( {x,y,z} \right):0 \leq x \leq 1,0 \leq y \leq 1,z = x} \right\},$ then ${\iint_{S}{f\left( {x,y,z} \right)}}dS = {\int_{0}^{1}{\int_{0}^{1}{f\left( {x,y,x} \right)}}}dxdy.$

271.

Surface $\mathbf{\text{r}} = \left\langle {v\ \text{cos}\ u,v\ \text{sin}\ u,v^{2}} \right\rangle,\ \text{for}\ 0 \leq u \leq \pi,0 \leq v \leq 2,$ is the same as surface $\mathbf{\text{r}} = \left\langle {\sqrt{v}\ \text{cos}\ 2u,\sqrt{v}\ \text{sin}\ 2u,v} \right\rangle,$ for $0 \leq u \leq \frac{\pi}{2},0 \leq v \leq 4.$

272\.

Given the standard parameterization of a sphere, normal vectors $\text{t}_{u}^{}\ \times \ \text{t}_{v}$ are outward normal vectors.

For the following exercises, find parametric descriptions for the following surfaces.

273.

Plane $3x - 2y + z = 2$

274\.

Paraboloid $z = x^{2} + y^{2},$ for $0 \leq z \leq 9.$

275.

Plane $2x - 4y + 3z = 16$

276\.

The frustum of cone $z^{2} = x^{2} + y^{2},\ \text{for}\ 2 \leq z \leq 8$

277.

The portion of cylinder $x^{2} + y^{2} = 9$ in the first octant, for $0 \leq z \leq 3$

278\.

A cone with base radius *r* and height *h*, where *r* and *h* are positive constants

For the following exercises, use a computer algebra system to approximate the area of the following surfaces using a parametric description of the surface.

279.

\[T\] Half cylinder $\left\{ {\left( {r,\theta,z} \right):r = 4,0 \leq \theta \leq \pi,0 \leq z \leq 7} \right\}$

280\.

\[T\] Plane $z = 10 - x - y$ above square $|x| \leq 2,|y| \leq 2$

For the following exercises, let *S* be the hemisphere $x^{2} + y^{2} + z^{2} = 4,$ with $z \geq 0,$ and evaluate each surface integral.

281.

$\iint_{S}{zdS}$

282\.

$\iint_{S}{(x - 2y)dS}$

283.

$\iint_{S}{\left( {x^{2} + y^{2}} \right)zdS}$

For the following exercises, evaluate $\int{\int_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}}$ for vector field F, where N is an upward pointing normal vector to surface *S.*

284\.

$\mathbf{\text{F}}\left( {x,y,z} \right) = x\mathbf{\text{i}} + 2y\mathbf{\text{j}} - 3z\mathbf{\text{k}},$ and *S* is that part of plane $15x - 12y + 3z = 6$ that lies above unit square $0 \leq x \leq 1,0 \leq y \leq 1.$

285.

$\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}},$ and *S* is hemisphere $z = \sqrt{1 - x^{2} - y^{2}}.$

286\.

$\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + y^{2}\mathbf{\text{j}} + z^{2}\mathbf{\text{k}},$ and *S* is the portion of plane $z = y + 1$ that lies inside cylinder $x^{2} + y^{2} = 1.$

For the following exercises, approximate the mass of the lamina that has the shape of given surface *S.* Round to four decimal places.

287.

\[T\] *S* is surface $z = 4 - x - 2y,\ \text{with}\ z \geq 0\text{,}\ x \geq 0\text{,}\ y \geq 0\text{;}\ \rho = x.$

288\.

\[T\] *S* is surface $z = x^{2} + y^{2},\ \text{with}\ z \leq 1\text{;}\ \rho = z.$

289.

\[T\] *S* is surface $x^{2} + y^{2} + z^{2} = 5,\ \text{with}\ z \geq 1\text{;}\ \rho = \theta^{2}.$

290\.

Evaluate ${\iint_{S}{\left( {y^{2}z\mathbf{\text{i}} + y^{3}\mathbf{\text{j}} + xz\mathbf{\text{k}}} \right) \cdot dS}}\text{,}$ where *S* is the surface of cube $-1 \leq x \leq 1,-1 \leq y \leq 1,\text{and}\ 0 \leq z \leq 2.$ Assume an outward pointing normal.

291.

Evaluate surface integral ${\iint_{S}{gdS}},$ where $g(x,y,z) = xz + 2x^{2} - 3xy$ and *S* is the portion of plane $2x - 3y + z = 6$ that lies over unit square *R*: $0 \leq x \leq 1\text{,}\ 0 \leq y \leq 1.$

292\.

Evaluate ${\iint_{S}{(x^{2} + y - z)dS}}\text{,}$ where $S$ is the surface defined parametrically by $\mathbf{\text{r}}(u,v) = (2u + v)\mathbf{\text{i}} + (u - 2v)\mathbf{\text{j}} + (u + 3v)\mathbf{\text{k}}$ for $0 \leq u \leq 1,\ \text{and}\ 0 \leq v \leq 2.$

293.

\[T\] Evaluate ${\iint_{S}{(x - y^{2} + z)dS}}\text{,}$ where *S* is the surface defined by $\mathbf{\text{r}}(u,v) = u^{2}\mathbf{\text{i}} + v\mathbf{\text{j}} + u\mathbf{\text{k}}\text{,}\ 0 \leq u \leq 1\text{,}\ 0 \leq v \leq 1.$

294\.

\[T\] Evaluate $\int\int_{S}\left( x^{2} + y^{2} - z \right)dS$ where $S$ is the surface defined by $\mathbf{\text{r}}(u,v) = u\mathbf{\text{i}} - u^{2}\mathbf{\text{j}} + v\mathbf{\text{k}}\text{,}\ 0 \leq u \leq 2\text{,}\ 0 \leq v \leq 1.$

295.

Evaluate ${\iint_{S}{\left( {x^{2} + y^{2}} \right)dS}},$ where *S* is the surface of hemisphere $z = \sqrt{1 - x^{2} - y^{2}},$ and above the plane $z = 0.$

296\.

Evaluate ${\iint_{S}{\left( {x^{2} + y^{2} + z^{2}} \right)dS}},$ where *S* is the portion of plane $z = x + 1$ that lies inside cylinder $x^{2} + y^{2} = 1.$

297.

\[T\] Evaluate $\iint_{S}{x^{2}zdS,}$ where *S* is the portion of cone $z^{2} = x^{2} + y^{2}$ that lies between planes $z = 1$ and $z = 4.$

298\.

\[T\] Evaluate ${\iint_{S}{\left( {{xz}\text{/}y} \right)dS}},$ where *S* is the portion of cylinder $x = y^{2}$ that lies in the first octant between planes $z = 0,z = 5,y = 1,$ and $y = 4.$

299.

\[T\] Evaluate ${\iint_{S}{\left( {z + y} \right)dS}},$ where *S* is the part of the graph of $z = \sqrt{1 - x^{2}}$ in the first octant between the *xz*-plane and plane $y = 3.$

300\.

Evaluate $\iint_{S}{xyzdS}$ if *S* is the part of plane $z = x + y$ that lies over the triangular region in the *xy*-plane with vertices (0, 0, 0), (1, 0, 0), and (0, 2, 0).

301.

Find the mass of a lamina of density $\rho(x,y,z) = z$ in the shape of hemisphere $z = \left( {a^{2} - x^{2} - y^{2}} \right)^{1\text{/}2}.$

302\.

Compute $\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS,}$ where $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} - 5y\mathbf{\text{j}} + 4z\mathbf{\text{k}}$ and N is an outward normal vector of *S*, where *S* is the union of two squares $S_{1}:x = 0\text{,}\ 0 \leq y \leq 1\text{,}\ 0 \leq z \leq 1$ and $S_{2}:z = 1\text{,}\ 0 \leq x \leq 1\text{,}\ 0 \leq y \leq 1.$

303.

Compute $\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS,}$ where $\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} + z\mathbf{\text{j}} + (x + y)\mathbf{\text{k}}$ and N is an upward pointing normal vector $S$, where *S* is the triangular region of the plane $x + y + z = 1$ in the first octant.

304\.

Compute $\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS,}$ where $\mathbf{\text{F}}\left( {x,y,z} \right) = 2yz\mathbf{\text{i}} + \left( {\text{tan}^{-1}(xz)} \right)\mathbf{\text{j}} + e^{xy}\mathbf{\text{k}}$ and N is an outward normal vector of *S*, where *S* is the surface of sphere $x^{2} + y^{2} + z^{2} = 1.$

305.

Compute $\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS,}$ where $\mathbf{\text{F}}(x,y,z) = xyz\mathbf{\text{i}} + xyz\mathbf{\text{j}} + xyz\mathbf{\text{k}}$ and N is an outward normal vector *S*, where *S* is the surface of the five faces of the unit cube $0 \leq x \leq 1\text{,}\ 0 \leq y \leq 1\text{,}\ 0 \leq z \leq 1$ missing $z = 0.$

For the following exercises, express the surface integral as an iterated double integral by using a projection on *S* on the *yz*-plane.

306\.

${\iint_{S}{xy^{2}z^{3}dS}};$ *S* is the first-octant portion of plane $2x + 3y + 4z = 12.$

307.

$\iint_{S}{\left( {x^{2} - 2y + z} \right)d{S;}}$ *S* is the portion of the graph of $4x + y = 8$ bounded by the coordinate planes and plane $z = 6.$

For the following exercises, express the surface integral as an iterated double integral by using a projection on *S* on the *xz*-plane

308\.

${\iint_{S}{xy^{2}z^{3}dS}};$ *S* is the first-octant portion of plane $2x + 3y + 4z = 12.$

309.

$\iint_{S}{\left( {x^{2} - 2y + z} \right)d{S;}}$ *S* is the portion of the graph of $4x + y = 8$ bounded by the coordinate planes and plane $z = 6.$

310\.

Evaluate surface integral ${\iint_{S}{yzdS}},$ where *S* is the first-octant part of plane $x + y + z = \lambda,$ where $\lambda$ is a positive constant.

311.

Evaluate surface integral ${\iint_{S}{\left( {x^{2}z + y^{2}z} \right)dS}},$ where *S* is hemisphere $x^{2} + y^{2} + z^{2} = a^{2},z \geq 0.$

312\.

Evaluate surface integral ${\iint_{S}{zdS}},$ where *S* is surface $z = \sqrt{x^{2} + y^{2}},0 \leq z \leq 2.$

313.

Evaluate surface integral ${\iint_{S}{x^{2}yzdS}},$ where *S* is the part of plane $z = 1 + 2x + 3y$ that lies above rectangle $0 \leq x \leq 3\ \text{and}\ 0 \leq y \leq 2.$

314\.

Evaluate surface integral $\iint_{S}{yzdS,}$ where *S* is plane $x + y + z = 1$ that lies in the first octant.

315.

Evaluate surface integral ${\iint_{S}{yzdS}},$ where *S* is the part of plane $z = y + 3$ that lies inside cylinder $x^{2} + y^{2} = 1.$

For the following exercises, use geometric reasoning to evaluate the given surface integrals.

316\.

${\iint_{S}{\sqrt{x^{2} + y^{2} + z^{2}}dS}},$ where *S* is surface $x^{2} + y^{2} + z^{2} = 4,z \geq 0$

317.

${\iint_{S}{(x\mathbf{\text{i}} + y\mathbf{\text{j}}) \cdot dS}},$ where *S* is surface $x^{2} + y^{2} = 4,1 \leq z \leq 3,$ oriented with unit normal vectors pointing outward

318\.

${\iint_{S}{(z\mathbf{\text{k}}) \cdot dS}},$ where *S* is disc $x^{2} + y^{2} \leq 9$ on plane $z = 4,$ oriented with unit normal vectors pointing upward

319.

A lamina has the shape of a portion of sphere $x^{2} + y^{2} + z^{2} = a^{2}$ that lies within cone $z = \sqrt{x^{2} + y^{2}}.$ Let *S* be the spherical shell centered at the origin with radius *a*, and let *C* be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the *z*-axis. Determine the mass of the lamina if $\rho(x,y,z) = x^{2}y^{2}z.$

320\.

A lamina has the shape of a portion of sphere $x^{2} + y^{2} + z^{2} = a^{2}$ that lies within cone $z = \text{cot}\varphi_{0}\sqrt{x^{2} + y^{2}}$ Let *S* be the spherical shell centered at the origin with radius *a*, and let *C* be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the *z*-axis. Suppose the angle between the sides of the cone and the *z*-axis is $\phi_{0},\ \text{with}\ 0 \leq \phi_{0} < \frac{\pi}{2}.$ Determine the mass of that portion of the shape enclosed in the intersection of *S* and *C*. Assume $\rho(x,y,z) = x^{2}y^{2}z.$

321.

A paper cup has the shape of an inverted right circular cone of height 6 in. and radius of top 3 in. If the cup is full of water weighing ${62.5\ \text{lb}}\text{/}{\text{ft}^{3},}$ find the total force exerted by the water on the inside surface of the cup.

For the following exercises, the heat flow vector field for conducting objects is $\mathbf{\text{F}} = \text{−}k\nabla T,\ \text{where}\ T(x,y,z)$ is the temperature in the object and $k > 0$ is a constant that depends on the material. Find the outward flux of F across the following surfaces *S* for the given temperature distributions and assume $k = 1.$

322\.

$T(x,y,z) = 100e^{\text{−}x - y};$ *S* consists of the faces of cube $|x| \leq 1,|y| \leq 1,|z| \leq 1.$

323.

$T(x,y,z) = \text{−}\text{ln}\left( {x^{2} + y^{2} + z^{2}} \right);$ *S* is sphere $x^{2} + y^{2} + z^{2} = a^{2}.$

For the following exercises, consider the radial fields $\textbf{F} = \frac{\left\langle x,y,z \right\rangle}{\left( x^{2} + y^{2} + z^{2} \right)^{\frac{p}{2}}} = \frac{\mathbf{\text{r}}}{\left. ||\textbf{r} \right.||^{p}},$ where *p* is a real number. Let *S* consist of spheres *A* and *B* centered at the origin with radii $0 < a < b.$ The total outward flux across *S* consists of the outward flux across the outer sphere *B* less the flux into *S* across inner sphere *A*.

324\.

Find the total flux across *S* with $p = 0.$

325.

Show that for $p = 3$ the flux across *S* is independent of *a* and *b*.

6.7 Stokes’ Theorem 6.7 斯托克斯定理

• 6.7.1 Explain the meaning of Stokes’ theorem.
• 6.7.2 Use Stokes’ theorem to evaluate a line integral.
• 6.7.3 Use Stokes’ theorem to calculate a surface integral.
• 6.7.4 Use Stokes’ theorem to calculate a curl.

• 6.7.1 解释斯托克斯定理的含义。
• 6.7.2 使用斯托克斯定理计算线积分。
• 6.7.3 使用斯托克斯定理计算曲面积分。
• 6.7.4 使用斯托克斯定理计算旋度。

In this section, we study Stokes’ theorem, a higher-dimensional generalization of Green’s theorem. This theorem, like the Fundamental Theorem for Line Integrals and Green’s theorem, is a generalization of the Fundamental Theorem of Calculus to higher dimensions. Stokes’ theorem relates a vector surface integral over surface S in space to a line integral around the boundary of S. Therefore, just as the theorems before it, Stokes’ theorem can be used to reduce an integral over a geometric object S to an integral over the boundary of S.

In addition to allowing us to translate between line integrals and surface integrals, Stokes’ theorem connects the concepts of curl and circulation. Furthermore, the theorem has applications in fluid mechanics and electromagnetism. We use Stokes’ theorem to derive Faraday’s law, an important result involving electric fields.

Stokes’ Theorem 斯托克斯定理

Stokes’ theorem says we can calculate the flux of curl F across surface S by knowing information only about the values of F along the boundary of S. Conversely, we can calculate the line integral of vector field F along the boundary of surface S by translating to a double integral of the curl of F over S.

Let S be an oriented smooth surface with unit normal vector N. Furthermore, suppose the boundary of S is a simple closed curve C. The orientation of S induces the positive orientation of C if, as you walk in the positive direction around C with your head pointing in the direction of N, the surface is always on your left. With this definition in place, we can state Stokes’ theorem.

Stokes’ Theorem 斯托克斯定理

Let S be a piecewise smooth oriented surface with a boundary that is a simple closed curve C with positive orientation (Figure 6.79). If F is a vector field with component functions that have continuous partial derivatives on an open region containing S, then

$${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$$

Suppose surface S is a flat region in the xy-plane with upward orientation. Then the unit normal vector is k and surface integral ${\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}}}} \cdot d\mathbf{\text{S}}$ is actually the double integral ${\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}}}} \cdot \mathbf{\text{k}}dA.$ In this special case, Stokes’ theorem gives ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{k}}dA.}}$ However, this is the circulation form of Green’s theorem, which shows us that Green’s theorem is a special case of Stokes’ theorem. Green’s theorem can only handle surfaces in a plane, but Stokes’ theorem can handle surfaces in a plane or in space.

The complete proof of Stokes’ theorem is beyond the scope of this text. We look at an intuitive explanation for the truth of the theorem and then see proof of the theorem in the special case that surface S is a portion of a graph of a function, and S, the boundary of S, and F are all fairly tame.

Proof 证明

First, we look at an informal proof of the theorem. This proof is not rigorous, but it is meant to give a general feeling for why the theorem is true. Let S be a surface and let D be a small piece of the surface so that D does not share any points with the boundary of S. We choose D to be small enough so that it can be approximated by an oriented square E. Let D inherit its orientation from S, and give E the same orientation. This square has four sides; denote them $E_{l},$ $E_{r},$ $E_{u},$ and $E_{d}$ for the left, right, up, and down sides, respectively. On the square, we can use the flux form of Green’s theorem:

$${\int_{E_{l} + E_{d} + E_{r} + E_{u}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{E}{\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}} = {\iint_{E}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$$

To approximate the flux over the entire surface, we add the values of the flux on the small squares approximating small pieces of the surface (Figure 6.80). By Green’s theorem, the flux across each approximating square is a line integral over its boundary. Let F be an approximating square with an orientation inherited from S and with a right side $E_{l}$ (so F is to the left of E). Let $F_{r}$ denote the right side of $F$; then, $E_{l} = \text{−}F_{r}.$ In other words, the right side of $F$ is the same curve as the left side of E, just oriented in the opposite direction. Therefore,

$${\int_{E_{l}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = \text{−}{\int_{F_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$$

As we add up all the fluxes over all the squares approximating surface S, line integrals $\int_{E_{l}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ and $\int_{F_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ cancel each other out. The same goes for the line integrals over the other three sides of E. These three line integrals cancel out with the line integral of the lower side of the square above E, the line integral over the left side of the square to the right of E, and the line integral over the upper side of the square to the left of E (Figure 6.81). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of S. Therefore, the sum of all the fluxes (which, by Green’s theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated by a line integral over the boundary of S. In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.

Let’s now look at a rigorous proof of the theorem in the special case that S is the graph of function $z = g\left( {x,y} \right),$ where x and y vary over a bounded, simply connected region D of finite area (Figure 6.82). Furthermore, assume that $g$ has continuous second-order partial derivatives. Let C denote the boundary of S and let C′ denote the boundary of D. Then, D is the “shadow” of S in the plane and C′ is the “shadow” of C. Suppose that S is oriented upward. The counterclockwise orientation of C is positive, as is the counterclockwise orientation of $C^{\prime}.$ Let $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {P,Q,R} \right\rangle$ be a vector field with component functions that have continuous partial derivatives.

We take the standard parameterization of $S:x = x,y = y,z = g\left( {x,y} \right).$ The tangent vectors are $\mathbf{\text{t}}_{x} = \left\langle {1,0,g_{x}} \right\rangle$ and $\mathbf{\text{t}}_{y} = \left\langle {0,1,g_{y}} \right\rangle,$ and therefore, $\mathbf{\text{t}}_{x} \times \mathbf{\text{t}}_{y} = \left\langle {\text{−}g_{x},\text{−}g_{y},1} \right\rangle.$ By Equation 6.19,

$${\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}} =}}{\iint\limits_{D}\left\lbrack {\text{−}\left( {R_{y} - Q_{z}} \right)z_{x} - \left( {P_{z} - R_{x}} \right)z_{y} + \left( {Q_{x} - P_{y}} \right)} \right\rbrack}dA,$$

where the partial derivatives are all evaluated at $\left( {x,y,g\left( {x,y} \right)} \right),$ making the integrand depend on x and y only. Suppose $\left\langle {x(t),y(t)} \right\rangle,a \leq t \leq b$ is a parameterization of $C^{\prime}.$ Then, a parameterization of C is $\left\langle {x(t),y(t),g\left( {x(t),y(t)} \right)} \right\rangle,a \leq t \leq b.$ Armed with these parameterizations, the Chain rule, and Green’s theorem, and keeping in mind that P, Q, and R are all functions of x and y, we can evaluate line integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}\text{:}$

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{a}^{b}\left( {Px^{\prime}(t) + Qy^{\prime}(t) + Rz^{\prime}(t)} \right)}dt} \\ & {= {\int_{a}^{b}\left\lbrack {Px^{\prime}(t) + Qy^{\prime}(t) + R\left( {\frac{\partial z}{\partial x}\ \frac{dx}{dt} + \frac{\partial z}{\partial y}\ \frac{dy}{dt}} \right)} \right\rbrack}dt} \\ & {= {\int_{a}^{b}\left\lbrack {\left( {P + R\frac{\partial z}{\partial x}} \right)x^{\prime}(t) + \left( {Q + R\frac{\partial z}{\partial y}} \right)y^{\prime}(t)} \right\rbrack}dt} \\ & {= {\int\limits_{C^{\prime}}{\left( {P + R\frac{\partial z}{\partial x}} \right)dx + \left( {Q + R\frac{\partial z}{\partial y}} \right)dy}}} \\ & {= {\iint\limits_{D}\left\lbrack {\frac{\partial}{\partial x}\left( {Q + R\frac{\partial z}{\partial y}} \right) - \frac{\partial}{\partial y}\left( {P + R\frac{\partial z}{\partial x}} \right)} \right\rbrack}dA} \\ & {= \begin{array}{l} {\iint\limits_{D}\begin{matrix} \left( {\frac{\partial Q}{\partial x} + \frac{\partial Q}{\partial z}\ \frac{\partial z}{\partial x} + \frac{\partial R}{\partial x}\ \frac{\partial z}{\partial y} + \frac{\partial R}{\partial z}\ \frac{\partial z}{\partial x}\ \frac{\partial z}{\partial y} + R\frac{\partial^{2}z}{\partial x\partial y}} \right) \\ {\text{−}\left( \frac{\partial P}{\partial y} + \frac{\partial P}{\partial z}\frac{\partial z}{\partial y} + \frac{\partial R}{\partial y}\frac{\partial z}{\partial x} + \frac{\partial R}{\partial z}\frac{\partial z}{\partial y}\frac{\partial z}{\partial x} + R\frac{\partial^{2}z}{\partial y\partial x} \right)} \end{matrix}} \end{array}dA.} \end{array}$$

By Clairaut’s theorem, $\frac{\partial^{2}z}{\partial x\partial y} = \frac{\partial^{2}z}{\partial y\partial x}.$ Therefore, four of the terms disappear from this double integral, and we are left with

$${\iint\limits_{D}\left\lbrack {\text{−}\left( {R_{y} - Q_{z}} \right)z_{x} - \left( {P_{z} - R_{x}} \right)z_{y} + \left( {Q_{x} - P_{y}} \right)} \right\rbrack}dA,$$

which equals $\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}.}$

□

We have shown that Stokes’ theorem is true in the case of a function with a domain that is a simply connected region of finite area. We can quickly confirm this theorem for another important case: when vector field F is conservative. If F is conservative, the curl of F is zero, so $\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}} = 0.}$ Since the boundary of S is a closed curve, $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ is also zero.

Verifying Stokes’ Theorem for a Specific Case 验证特定情形下的斯托克斯定理

Verify that Stokes’ theorem is true for vector field $\mathbf{\text{F}}\left( {x,y,z} \right) = \left\langle {y,2z,x^{2}} \right\rangle$ and surface S, where S is the paraboloid $z = 4 - x^{2} - y^{2}$. Assume the surface is outward oriented and $z \geq 0$.

Solution 解

As a surface integral, you have $g(x,y) = 4 - x^{2} - y^{2},g_{x} = -2x$ and $g_{y} = -2y$

$$\begin{aligned} {\text{curl}\mathbf{F}\ } & {= \left| \begin{matrix} \mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & {2z} & x^{2} \end{matrix} \right|} \end{aligned} = \left\langle -2,-2x,-1 \right\rangle\text{.}$$

By Equation 6.19,

$$\begin{array}{cl} {\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} & {= {\iint_{D}{\text{curl}\ \mathbf{\text{F}}\left( {\mathbf{\text{r}}\left( {\phi,\theta} \right)} \right)}} \cdot \left( {\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta}} \right)dA} \\ & {= {\iint_{D}{\left\langle {-2,-2x,-1} \right\rangle \cdot \left\langle 2x,2y,1 \right\rangle dA}}} \\ & {= {\int_{-2}^{2}{\int_{\sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}}\left( {-4x-4xy-1} \right)}}dydx} \\ & {= \int_{-2}^{2}{\left( -8x\sqrt{4 - x^{2}}-2\sqrt{4 - x^{2}} \right)dx}} \\ & {{= -4}\pi} \end{array}$$

As a line integral, you can parameterize C by $\mathbf{\text{r}}(t) = \left\langle {2\ \text{cos}\ t,2\ \text{sin}\ t,0} \right\rangle\ 0 \leq t \leq 2\pi$. By Equation 6.19,

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\int_{0}^{2\pi}\left\langle \left. 2\text{sin}\ t,0,4\text{cos}^{2}t \right\rangle \cdot \left\langle -2\text{sin}\ t,2\text{cos}\ t,0 \right\rangle dt \right.}} \\ & {= {\int_{0}^{2\pi}{-4\text{sin}^{2}tdt = -4\pi}}} \end{array}$$ (6.22)

Therefore, we have verified Stokes' theorem for this example.

Verify that Stokes’ theorem is true for vector field $\mathbf{\text{F}}(x,y,z) = \left\langle {y,x,\text{−}z} \right\rangle$ and surface S, where S is the upwardly oriented portion of the graph of $f(x,y) = x^{2}y$ over a triangle in the xy-plane with vertices $(0,0),$ $(2,0),$ and $(0,2).$

Applying Stokes’ Theorem 应用斯托克斯定理

Stokes’ theorem translates between the flux integral of surface S to a line integral around the boundary of S. Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa. We now study some examples of each kind of translation.

Calculating a Surface Integral 计算曲面积分

Calculate surface integral ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where S is the surface, oriented outward, in Figure 6.84 and $\mathbf{\text{F}} = \left\langle {z,2xy,x + y} \right\rangle.$

Solution 解

Note that to calculate $\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ without using Stokes’ theorem, we would need to use Equation 6.19. Use of this equation requires a parameterization of *S*. Surface *S* is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes’ theorem, noting that the boundary *C* of the surface is merely a single circle with radius 1.

By Stokes’ theorem,

$${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}},}}$$

where *C* has parameterization $\mathbf{r}(t) = \left\langle \left. \text{sin}\ t,0,1 - \text{cos}\ t \right\rangle \right.,0 \leq t < 2\pi.$ By Equation 6.9,

$$\begin{matrix} {\iint_{S}\text{curl}\ \textbf{F} \cdot d\textbf{S}} & {= \int_{C}\textbf{F} \cdot d\textbf{r}} \\ & {= \int_{0}^{2\pi}\left\langle {\left. 1 - \text{cos}\ t\ ,0, + \text{sin}\ t \right\rangle \cdot \ \left\langle - \text{sin}\ t,0,\text{cos}\ t \right\rangle dt} \right.\ } \\ & {= \int_{0}^{2\pi}\left( - \text{sin}\ t + 2\text{sin}\ t~\text{cos}\ t \right)~dt} \\ & {= {\int_{0}^{2\pi}\left( - \text{sin}\ t + \text{sin~2t)~}\textit{dt} \right.}} \\ & {= \left\lbrack {\text{+cos}~t + \frac{1}{2}\text{cos}2t} \right\rbrack_{0}^{2\pi}} \\ & {= \left\lbrack {\text{+cos}~2\pi + \frac{1}{2}\text{cos}4\pi} \right\rbrack - \left\lbrack {\text{+cos}0 + \frac{1}{2}\text{cos}0} \right\rbrack} \\ & {= 0} \end{matrix}$$

An amazing consequence of Stokes’ theorem is that if *S*′ is any other smooth surface with boundary *C* and the same orientation as *S*, then ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = 0$ because Stokes’ theorem says the surface integral depends on the line integral around the boundary only.

In Example 6.74, we calculated a surface integral simply by using information about the boundary of the surface. In general, let $S_{1}$ and $S_{2}$ be smooth surfaces with the same boundary *C* and the same orientation. By Stokes’ theorem,

$${\iint_{S_{1}}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{S_{2}}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$$ (6.23)

Therefore, if $\iint_{S_{1}}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ is difficult to calculate but $\iint_{S_{2}}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ is easy to calculate, Stokes’ theorem allows us to calculate the easier surface integral. In Example 6.74, we could have calculated $\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ by calculating ${\iint_{S^{\prime}}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $S^{\prime}$ is the disk enclosed by boundary curve *C* (a much more simple surface with which to work).

Equation 6.23 shows that flux integrals of curl vector fields are surface independent in the same way that line integrals of gradient fields are path independent. Recall that if F is a two-dimensional conservative vector field defined on a simply connected domain, $f$ is a potential function for F, and *C* is a curve in the domain of F, then $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ depends only on the endpoints of *C*. Therefore if *C*′ is any other curve with the same starting point and endpoint as *C* (that is, *C*′ has the same orientation as *C*), then ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{C\text{'}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$ In other words, the value of the integral depends on the boundary of the path only; it does not really depend on the path itself.

Analogously, suppose that *S* and *S*′ are surfaces with the same boundary and same orientation, and suppose that G is a three-dimensional vector field that can be written as the curl of another vector field F (so that F is like a “potential field” of G). By Equation 6.23,

$${\iint_{S}{\mathbf{\text{G}} \cdot d\mathbf{\text{S}}}} = {\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{S\text{'}}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\iint_{S\text{'}}{\mathbf{\text{G}} \cdot d\mathbf{\text{S}}}}.$$

Therefore, the flux integral of G does not depend on the surface, only on the boundary of the surface. Flux integrals of vector fields that can be written as the curl of a vector field are surface independent in the same way that line integrals of vector fields that can be written as the gradient of a scalar function are path independent.

Use Stokes’ theorem to calculate surface integral ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}} = \left\langle {z,x,y} \right\rangle$ and *S* is the surface as shown in the following figure. The boundary curve, *C*, is oriented clockwise when looking along the positive y-axis.

Calculating a Line Integral 计算线积分

Calculate the line integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}} = \left\langle {xy,x^{2} + y^{2} + z^{2},yz} \right\rangle$ and *C* is the boundary of the parallelogram with vertices $(0,0,1),(0,1,0),(2,1, - 2),$ and $(2,0, - 1).$

Solution 解

To calculate the line integral directly, we need to parameterize each side of the parallelogram separately, calculate four separate line integrals, and add the result. This is not overly complicated, but it is time-consuming.

By contrast, let’s calculate the line integral using Stokes’ theorem. Let *S* denote the surface of the parallelogram. Note that *S* is the portion of the graph of $z = 1 - x - y$ for $\left( {x,y} \right)$ varying over the rectangular region with vertices $\left( {0,0} \right),$ $\left( {0,1} \right),$ $\left( {2,0} \right),$ and $\left( {2,1} \right)$ in the *xy*-plane. Therefore, a parameterization of *S* is $\left\langle {x,y,1 - x - y} \right\rangle,0 \leq x \leq 2,0 \leq y \leq 1.$ The curl of F is $\left\langle {{-z},0,x} \right\rangle,$ and Stokes’ theorem and Equation 6.19 give

$$\begin{array}{cl} {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} & {= {\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}} \\ & {= {\int_{0}^{2}{\int_{0}^{1}{\text{curl}\ \mathbf{\text{F}}\left( {x,y} \right) \cdot \left( {\mathbf{\text{t}}_{x} \times \mathbf{\text{t}}_{y}} \right)}}}dydx} \\ & {= {\int_{0}^{2}{\int_{0}^{1}{\left\langle {\text{−}\left( {1 - x - y} \right)\text{,0,x}} \right\rangle \cdot \left( {\left\langle {1,0,-1} \right\rangle \times \left\langle {0,1,-1} \right\rangle} \right)}}}dydx} \\ & {= {\int_{0}^{2}{\int_{0}^{1}{\left\langle {x + y - 1,0,x} \right\rangle \cdot \left\langle {1,1,1} \right\rangle}}}dydx} \\ & {\mspace{22mu}{\int_{0}^{2}{\int_{0}^{1}{2x + y - 1}}}dydx} \\ & {= 3.} \end{array}$$

Use Stokes’ theorem to calculate line integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}},$ where $\mathbf{\text{F}} = \left\langle {z,x,y} \right\rangle$ and *C* is oriented clockwise and is the boundary of a triangle with vertices $\left( {0,0,1} \right),\left( {3,0,-2} \right),$ and $\left( {0,1,2} \right).$

Interpretation of Curl 旋度的解释

In addition to translating between line integrals and flux integrals, Stokes’ theorem can be used to justify the physical interpretation of curl that we have learned. Here we investigate the relationship between curl and circulation, and we use Stokes’ theorem to state Faraday’s law—an important law in electricity and magnetism that relates the curl of an electric field to the rate of change of a magnetic field.

Recall that if *C* is a closed curve and F is a vector field defined on *C*, then the circulation of F around *C* is line integral ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$ If F represents the velocity field of a fluid in space, then the circulation measures the tendency of the fluid to move in the direction of *C*.

Let F be a continuous vector field and let $D_{r}$ be a small disk of radius *r* with center $P_{0}$ (Figure 6.85). If $D_{r}$ is small enough, then $(\text{curl}\ \mathbf{\text{F}})(P) \approx (\text{curl}\ \mathbf{\text{F}})(P_{0})$ for all points *P* in $D_{r}$ because the curl is continuous. Let $C_{r}$ be the boundary circle of $D_{r}.$ By Stokes’ theorem,

$${\int_{C_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}} = {\iint_{D_{r}}{\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}}}} \approx {\iint_{D_{r}}{\left( {\text{curl}\ \mathbf{\text{F}}} \right)\left( P_{0} \right) \cdot \mathbf{\text{N}}\left( P_{0} \right)dS}}.$$

The quantity $\left( {\text{curl}\ \mathbf{\text{F}}} \right)\left( P_{0} \right) \cdot \mathbf{\text{N}}\left( P_{0} \right)$ is constant, and therefore

$${\iint_{D_{r}}{\left( {\text{curl}\ \mathbf{\text{F}}} \right)\left( P_{0} \right) \cdot \mathbf{\text{N}}\left( P_{0} \right)dS}} = \pi r^{2}\left\lbrack {\left( {\text{curl}\ \mathbf{\text{F}}} \right)\left( P_{0} \right) \cdot \mathbf{\text{N}}\left( P_{0} \right)} \right\rbrack.$$

Thus

$${\int_{C_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} \approx \pi r^{2}\left\lbrack {\left( {\text{curl}\ \mathbf{\text{F}}} \right)\left( P_{0} \right) \cdot \mathbf{\text{N}}\left( P_{0} \right)} \right\rbrack,$$

and the approximation gets arbitrarily close as the radius shrinks to zero. Therefore Stokes’ theorem implies that

$$\left( {\text{curl}\ \mathbf{\text{F}}} \right)\left( P_{0} \right) \cdot \mathbf{\text{N}}\left( P_{0} \right) = \underset{r\rightarrow 0^{+}}{\text{lim}}\frac{1}{\pi r^{2}}{\int_{C_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$$

This equation relates the curl of a vector field to the circulation. Since the area of the disk is $\pi r^{2},$ this equation says we can view the curl (in the limit) as the circulation per unit area. Recall that if F is the velocity field of a fluid, then circulation ${\int_{C_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{C_{r}}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}}$ is a measure of the tendency of the fluid to move around $C_{r}.$ The reason for this is that $\mathbf{\text{F}} \cdot \mathbf{\text{T}}$ is a component of F in the direction of T, and the closer the direction of F is to T, the larger the value of $\mathbf{\text{F}} \cdot \mathbf{\text{T}}$ (remember that if a and b are vectors and b is fixed, then the dot product $\mathbf{\text{a}} \cdot \mathbf{\text{b}}$ is maximal when a points in the same direction as b). Therefore, if F is the velocity field of a fluid, then $\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}}$ is a measure of how the fluid rotates about axis N. The effect of the curl is largest about the axis that points in the direction of N, because in this case $\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}}$ is as large as possible.

To see this effect in a more concrete fashion, imagine placing a tiny paddlewheel at point $P_{0}$ (Figure 6.86). The paddlewheel achieves its maximum speed when the axis of the wheel points in the direction of curlF. This justifies the interpretation of the curl we have learned: curl is a measure of the rotation in the vector field about the axis that points in the direction of the normal vector N, and Stokes’ theorem justifies this interpretation.

Now that we have learned about Stokes’ theorem, we can discuss applications in the area of electromagnetism. In particular, we examine how we can use Stokes’ theorem to translate between two equivalent forms of Faraday’s law. Before stating the two forms of Faraday’s law, we need some background terminology.

Let *C* be a closed curve that models a thin wire. In the context of electric fields, the wire may be moving over time, so we write $C(t)$ to represent the wire. At a given time *t*, curve $C(t)$ may be different from original curve *C* because of the movement of the wire, but we assume that $C(t)$ is a closed curve for all times *t*. Let $D(t)$ be a surface with $C(t)$ as its boundary, and orient $C(t)$ so that $D(t)$ has positive orientation. Suppose that $C(t)$ is in a magnetic field $\mathbf{\text{B}}(t)$ that can also change over time. In other words, B has the form

$$\mathbf{\text{B}}(x,y,z) = \left\langle {P(x,y,z),Q(x,y,z),R(x,y,z)} \right\rangle,$$

where *P, Q,* and *R* can all vary continuously over time. We can produce current along the wire by changing field $\mathbf{\text{B}}(t)$ (this is a consequence of Ampere’s law). Flux $\phi(t) = {\iint_{D(t)}{\mathbf{\text{B}}(t) \cdot d\mathbf{\text{S}}}}$ creates electric field $\mathbf{\text{E}}(t)$ that does work. The integral form of Faraday’s law states that

$$\text{Work} = {\int_{C(t)}{\mathbf{\text{E}}(t)}} \cdot d\mathbf{\text{r}} = - \frac{\partial\phi}{\partial t}.$$

In other words, the work done by E is the line integral around the boundary, which is also equal to the rate of change of the flux with respect to time. The differential form of Faraday’s law states that

$$\text{curl}\ \mathbf{\text{E}} = - \frac{\partial\mathbf{\text{B}}}{\partial t}.$$

Using Stokes’ theorem, we can show that the differential form of Faraday’s law is a consequence of the integral form. By Stokes’ theorem, we can convert the line integral in the integral form into surface integral

$$- \frac{\partial\phi}{\partial t} = {\int_{C(t)}{\mathbf{\text{E}}(t)}} \cdot d\mathbf{\text{r}} = {\iint_{D(t)}{\text{curl}\ \mathbf{\text{E}}(t)}} \cdot d\mathbf{\text{S}}.$$

Since $\phi(t) = {\iint_{D(t)}{\mathbf{\text{B}}(t) \cdot d\mathbf{\text{S}}}},$ then as long as the integration of the surface does not vary with time we also have

$$- \frac{\partial\phi}{\partial t} = {\iint_{D(t)}{- \frac{\partial\mathbf{\text{B}}}{\partial t} \cdot d\mathbf{\text{S}}}}.$$

Therefore,

$${\iint_{D{(t)}}{- \frac{\partial\mathbf{\text{B}}}{\partial t}}} \cdot d\mathbf{\text{S}} = {\iint_{D{(t)}}{\text{curl}\ \mathbf{\text{E}} \cdot d\mathbf{\text{S}}}}.$$

To derive the differential form of Faraday’s law, we would like to conclude that $\text{curl}\ \mathbf{\text{E}} = - \frac{\partial\mathbf{\text{B}}}{\partial t}.$ In general, the equation

$${\iint_{D{(t)}}{- \frac{\partial\mathbf{\text{B}}}{\partial t}}} \cdot d\mathbf{\text{S}} = {\iint_{D{(t)}}{\text{curl}\ \mathbf{\text{E}} \cdot d\mathbf{\text{S}}}}$$

is not enough to conclude that $\text{curl}\ \mathbf{\text{E}} = - \frac{\partial\mathbf{\text{B}}}{\partial t}.$ The integral symbols do not simply “cancel out,” leaving equality of the integrands. To see why the integral symbol does not just cancel out in general, consider the two single-variable integrals $\int_{0}^{1}{xdx}$ and ${\int_{0}^{1}{f(x)dx}},$ where

$$f(x) = \left\{ {}_{0,\mspace{11mu} 1\text{/}2 \leq x \leq 1.}^{1,\mspace{11mu} 0 \leq x \leq 1\text{/}2} \right.$$

Both of these integrals equal $\frac{1}{2},$ so ${\int_{0}^{1}{xdx}} = {\int_{0}^{1}{f(x)dx}}.$ However, $x \neq f(x).$ Analogously, with our equation ${\iint_{D{(t)}}{- \frac{\partial\mathbf{\text{B}}}{\partial t}}} \cdot d\mathbf{\text{S}} = {\iint_{D{(t)}}{\text{curl}\ \mathbf{\text{E}} \cdot d\mathbf{\text{S}}}},$ we cannot simply conclude that $\text{curl}\ \mathbf{\text{E}} = - \frac{\partial\mathbf{\text{B}}}{\partial t}$ just because their integrals are equal. However, in our context, equation ${\iint_{D{(t)}}{- \frac{\partial\mathbf{\text{B}}}{\partial t}}} \cdot d\mathbf{\text{S}} = {\iint_{D{(t)}}{\text{curl}\ \mathbf{\text{E}} \cdot d\mathbf{\text{S}}}}$ is true for *any* region, however small (this is in contrast to the single-variable integrals just discussed). If F and G are three-dimensional vector fields such that ${\iint_{s}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}} =}}{\iint_{s}{\mathbf{\text{G}} \cdot d\mathbf{\text{S}}}}$ for any surface *S*, then it is possible to show that $\mathbf{\text{F}} = \mathbf{\text{G}}$ by shrinking the area of *S* to zero by taking a limit (the smaller the area of *S*, the closer the value of $\iint_{s}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ to the value of F at a point inside *S*). Therefore, we can let area $D(t)$ shrink to zero by taking a limit and obtain the differential form of Faraday’s law:

$$\text{curl}\ \mathbf{\text{E}} = - \frac{\partial\mathbf{\text{B}}}{\partial t}.$$

In the context of electric fields, the curl of the electric field can be interpreted as the negative of the rate of change of the corresponding magnetic field with respect to time.

Using Faraday’s Law 应用法拉第定律

Calculate the curl of electric field E if the corresponding magnetic field is constant field $\mathbf{\text{B}}(t) = \left\langle {1,-4,2} \right\rangle.$

Solution 解

Since the magnetic field does not change with respect to time, $- \frac{\partial\mathbf{\text{B}}}{\partial t} = \mathbf{0}.$ By Faraday’s law, the curl of the electric field is therefore also zero.

Analysis 分析

A consequence of Faraday’s law is that the curl of the electric field corresponding to a constant magnetic field is always zero.

Calculate the curl of electric field E if the corresponding magnetic field is $\mathbf{\text{B}}(t) = \left\langle {tx,ty,-2tz} \right\rangle,0 \leq t < \infty.$

Notice that the curl of the electric field does not change over time, although the magnetic field does change over time.

Section 6.7 Exercises 6.7 节习题

For the following exercises, without using Stokes’ theorem, calculate directly both the flux of $\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}}$ over the given surface and the circulation integral around its boundary, assuming all boundaries have positive orientation.

326\.

$\mathbf{\text{F}}(x,y,z) = y^{2}\mathbf{\text{i}} + z^{2}\mathbf{\text{j}} + x^{2}\mathbf{\text{k}}\text{;}$ *S* is the first-octant portion of plane $x + y + z = 1.$

327.

$\mathbf{\text{F}}(x,y,z) = z\mathbf{\text{i}} + x\mathbf{\text{j}} + y\mathbf{\text{k}}\text{;}$ *S* is hemisphere $z = \left( {a^{2} - x^{2} - y^{2}} \right)^{1\text{/}2}.$

328\.

$\mathbf{\text{F}}(x,y,z) = y^{2}\mathbf{\text{i}} + 2x\mathbf{\text{j}} + 5\mathbf{\text{k}}\text{;}$ *S* is hemisphere $z = \left( {4 - x^{2} - y^{2}} \right)^{1\text{/}2}.$

329.

$\mathbf{\text{F}}(x,y,z) = z\mathbf{\text{i}} + 2x\mathbf{\text{j}} + 3y\mathbf{\text{k}}\text{;}$ *S* is upper hemisphere $z = \sqrt{9 - x^{2} - y^{2}}.$

330\.

$\mathbf{\text{F}}(x,y,z) = \left( {x + 2z} \right)\mathbf{\text{i}} + \left( {y - x} \right)\mathbf{\text{j}} + \left( {z - y} \right)\mathbf{\text{k}}\text{;}$ *S* is a triangular region with vertices (3, 0, 0), (0, 3/2, 0), and (0, 0, 3).

331.

$\mathbf{\text{F}}(x,y,z) = 2y\mathbf{\text{i}} - 6z\mathbf{\text{j}} + 3x\mathbf{\text{k}}\text{;}$ *S* is a portion of paraboloid $z = 4 - x^{2} - y^{2}$ and is above the *xy*-plane.

For the following exercises, use Stokes’ theorem to evaluate $\iint_{S}{(\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}})dS}$ for the vector fields and surface.

332\.

$\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} - z\mathbf{\text{j}}$ and *S* is the surface of the cube $0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1,$ except for the face where $z = 0,$ and using the outward unit normal vector.

333.

$\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} + x^{2}\mathbf{\text{j}} + z^{2}\mathbf{\text{k}}\text{;}$ and *S* is the part of paraboloid $z = x^{2} + y^{2}$ below plane $z = y,$ and using the outward normal vector.

334\.

$\mathbf{\text{F}}(x,y,z) = 4y\mathbf{\text{i}} + z\mathbf{\text{j}} + 2y\mathbf{\text{k}}$ and *S* is the part of sphere $x^{2} + y^{2} + z^{2} = 4$ above plane $z = 0,$ and using the outward normal vector

335.

Use Stokes’ theorem to evaluate ${\int\limits_{C}\left\lbrack {2xy^{2}zdx + 2x^{2}yzdy + \left( {x^{2}y^{2} - 2z} \right)dz} \right\rbrack},$ where *C* is the curve given by $x = \text{cos}\ t,y = \text{sin}\ t,z = \text{sin}\ t,0 \leq t \leq 2\pi,$ traversed in the direction of increasing *t*.

336\.

\[T\] Use a computer algebraic system (CAS) and Stokes’ theorem to approximate line integral ${\int\limits_{C}\left( {ydx + zdy + xdz} \right)},$ where *C* is the intersection of plane $x + y = 2$ and surface $x^{2} + y^{2} + z^{2} = 2\left( {x + y} \right),$ traversed counterclockwise viewed from the origin.

337.

\[T\] Use a CAS and Stokes’ theorem to approximate line integral ${\int\limits_{C}\left( {3ydx + 2zdy - 5xdz} \right)},$ where *C* is the intersection of the *xy*-plane and hemisphere $z = \sqrt{1 - x^{2} - y^{2}},$ traversed counterclockwise viewed from the top—that is, from the positive *z*-axis toward the *xy*-plane.

338\.

\[T\] Use a CAS and Stokes’ theorem to approximate line integral ${\int\limits_{C}\left\lbrack {\left( {1 + y} \right)zdx + \left( {1 + z} \right)xdy + \left( {1 + x} \right)ydz} \right\rbrack},$ where *C* is a triangle with vertices $\left( {1,0,0} \right),$ $\left( {0,1,0} \right),$ and $\left( {0,0,1} \right)$ oriented counterclockwise.

339.

Use Stokes’ theorem to evaluate ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = e^{xy}\text{cos}\ z\mathbf{\text{i}} + x^{2}z\mathbf{\text{j}} + xy\mathbf{\text{k}},$ and *S* is half of sphere $x = \sqrt{1 - y^{2} - z^{2}},$ oriented out toward the positive *x*-axis.

340\.

\[T\] Use a CAS and Stokes’ theorem to evaluate ${\iint_{S}{(\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}})dS}},$ where $\mathbf{\text{F}}(x,y,z) = x^{2}y\mathbf{\text{i}} + xy^{2}\mathbf{\text{j}} + z^{3}\mathbf{\text{k}}$ and *S* is the curve of the part of plane $3x + 2y + z = 6$ above cylinder $x^{2} + y^{2} = 4,$ oriented clockwise when viewed from above.

341.

\[T\] Use a CAS and Stokes’ theorem to evaluate $\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}},}$ where $\mathbf{\text{F}}(x,y,z) = \left( {\text{sin}\left( {y + z} \right) - yx^{2} - \frac{y^{3}}{3}} \right)\mathbf{\text{i}} + x\ \text{cos}\left( {y + z} \right)\mathbf{\text{j}} + \text{cos}\left( {2y} \right)\mathbf{\text{k}}$ and *S* consists of the top and the four sides but not the bottom of the cube with vertices $\left( {\pm 1,\pm 1,\pm 1} \right),$ oriented outward.

342\.

\[T\] Use a CAS and Stokes’ theorem to evaluate $\iint\limits_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}},}$ where $\mathbf{\text{F}}(x,y,z) = z^{2}\mathbf{\text{i}} - 3xy\mathbf{\text{j}} + x^{3}y^{3}\mathbf{\text{k}}$ and *S* is the top part of $z = 5 - x^{2} - y^{2}$ above plane $z = 1,$ and *S* is oriented upward.

343.

Use Stokes’ theorem to evaluate ${\iint_{S}{(\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}})dS}},$ where $\mathbf{\text{F}}(x,y,z) = z^{2}\mathbf{\text{i}} + y^{2}\mathbf{\text{j}} + x\mathbf{\text{k}}$ and *S* is a triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) with upward orientation.

344\.

Use Stokes’ theorem to evaluate line integral ${\int\limits_{C}\left( {zdx + xdy + ydz} \right)},$ where *C* is a triangle with vertices (3, 0, 0), (0, 0, 2), and (0, 6, 0) traversed in the given order.

345.

Use Stokes’ theorem to evaluate ${\int\limits_{C}\left( {\frac{1}{2}y^{2}dx + zdy + xdz} \right)},$ where *C* is the curve of intersection of plane $x + z = 1$ and ellipsoid $x^{2} + 2y^{2} + z^{2} = 1,$ oriented clockwise from the origin.

346\.

Use Stokes’ theorem to evaluate ${\iint_{S}{(\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}})dS}},$ where $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y^{2}\mathbf{\text{j}} + ze^{xy}\mathbf{\text{k}}$ and *S* is the part of surface $z = 1 - x^{2} - 2y^{2}$ with $z \geq 0\text{,}\ $ oriented upward.

347.

Use Stokes’ theorem to evaluate ${\iint_{S}{(\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}})dS}},$ for vector field $\mathbf{\text{F}}(x,y,z) = z\mathbf{\text{i}} + 3x\mathbf{\text{j}} + 2z\mathbf{\text{k}}$ where *S* is surface $z = 1 - x^{2} - y^{2},z \geq 0,$ *C* is boundary circle $x^{2} + y^{2} = 1,$ and *S* is oriented in the positive *z*-direction.

348\.

Use Stokes’ theorem to evaluate ${\iint_{S}{(\text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{N}})dS}},$ for vector field $\mathbf{\text{F}}(x,y,z) = - \frac{3}{2}y^{2}\mathbf{\text{i}} - 2xy\mathbf{\text{j}} + yz\mathbf{\text{k}}\text{,}$ where *S* is that part of the surface of plane $x + y + z = 1$ contained within triangle *C* with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), traversed counterclockwise as viewed from above.

349.

A certain closed path *C* in plane $2x + 2y + z = 1$ is known to project onto unit circle $x^{2} + y^{2} = 1$ in the *xy*-plane. Let *c* be a constant and let $\mathbf{\text{R}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}.$ Use Stokes’ theorem to evaluate ${\int_{C}^{}{(c\mathbf{\text{k}}\ \times \ \mathbf{\text{R}}) \cdot d\mathbf{\text{r}}}}.$

350\.

Use Stokes’ theorem and let *C* be the boundary of surface $z = x^{2} + y^{2}$ with $0 \leq x \leq 2$ and $0 \leq y \leq 1,$ oriented with upward facing normal. Define

$$\mathbf{\text{F}}(x,y,z) = \left\lbrack {\text{sin}\left( x^{3} \right) + xz} \right\rbrack\mathbf{\text{i}} + (x - yz)\mathbf{\text{j}} + \text{cos}\left( z^{4} \right)\mathbf{\text{k}}\ \text{and evaluate}\ {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$$ 351.

Let *S* be hemisphere $x^{2} + y^{2} + z^{2} = 4$ with $z \geq 0,$ oriented upward. Let $\mathbf{\text{F}}(x,y,z) = x^{2}e^{yz}\mathbf{\text{i}} + y^{2}e^{xz}\mathbf{\text{j}} + z^{2}e^{xy}\mathbf{\text{k}}$ be a vector field. Use Stokes’ theorem to evaluate ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$

352\.

Let $\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} + \left( {e^{z^{2}} + y} \right)\mathbf{\text{j}} + (x + y)\mathbf{\text{k}}$ and let *S* be the graph of function $y = \frac{x^{2}}{9} + \frac{z^{2}}{9} - 1$ with $y \leq 0$ oriented so that the normal vector of *S* has a positive *j* component. Use Stokes’ theorem to compute integral ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$

353.

Use Stokes’ theorem to evaluate $\int_{C}\mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{\text{F}}(x,y,z) = y\mathbf{\text{i}} + z\mathbf{\text{j}} + x\mathbf{\text{k}}$ and *C* is a triangle with vertices (0, 0, 0), (2, 0, 0) and $(0,-2,2)$ oriented counterclockwise when viewed from above.

354\.

Use the surface integral in Stokes’ theorem to calculate the circulation of field F, $\mathbf{\text{F}}(x,y,z) = x^{2}y^{3}\mathbf{\text{i}} + \mathbf{\text{j}} + z\mathbf{\text{k}}$ around *C*, which is the intersection of cylinder $x^{2} + y^{2} = 4$ and hemisphere $x^{2} + y^{2} + z^{2} = 16,z \geq 0,$ oriented counterclockwise when viewed from above.

355.

Use Stokes’ theorem to compute ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = \mathbf{\text{i}} + xy^{2}\mathbf{\text{j}} + xy^{2}\mathbf{\text{k}}$ and *S* is a part of plane $y + z = 2$ inside cylinder $x^{2} + y^{2} = 1$ and oriented upward.

356\.

Use Stokes’ theorem to evaluate ${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = \text{−}y^{2}\mathbf{\text{i}} + x\mathbf{\text{j}} + z^{2}\mathbf{\text{k}}$ and *S* is the part of plane $x + y + z = 1$ in the first octant and oriented upward $x \geq 0\text{,}\ y \geq 0\text{,}\ z \geq 0.$

357.

Let $\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} + 2z\mathbf{\text{j}} - 2y\mathbf{\text{k}}$ and let *C* be the intersection of plane $x + z = 5$ and cylinder $x^{2} + y^{2} = 9,$ which is oriented counterclockwise when viewed from the top. Compute the line integral of F over *C* using Stokes’ theorem.

358\.

\[T\] Use a CAS and let $\mathbf{\text{F}}(x,y,z) = xy^{2}\mathbf{\text{i}} + (yz - x)\mathbf{\text{j}} + e^{yxz}\mathbf{\text{k}}.$ Use Stokes’ theorem to compute the surface integral of curl F over surface *S* with inward orientation consisting of cube $\lbrack 0,1\rbrack\ \times \ \lbrack 0,1\rbrack\ \times \ \lbrack 0,1\rbrack$ with the right side missing.

359.

Let *S* be ellipsoid $\frac{x^{2}}{4} + \frac{y^{2}}{9} + z^{2} = 1$ oriented outward and let F be a vector field with component functions that have continuous partial derivatives. Compute $\iint_{s}\text{curl~}\mathbf{F} \cdot d\mathbf{S}$

360\.

Let *S* be the part of paraboloid $z = 9 - x^{2} - y^{2}$ with $z \geq 0$ oriented upward. Verify Stokes’ theorem for vector field $\mathbf{\text{F}}(x,y,z) = 3z\mathbf{\text{i}} + 4x\mathbf{\text{j}} + 2y\mathbf{\text{k}}.$

361.

\[T\] Use a CAS and Stokes’ theorem to evaluate $\int_{C}\mathbf{F} \cdot d\mathbf{r}$ if $\mathbf{\text{F}}(x,y,z) = \left( {3z - \text{sin}\ x} \right)\mathbf{\text{i}} + \left( {x^{2} + e^{y}} \right)\mathbf{\text{j}} + \left( {y^{3} - \text{cos}\ z} \right)\mathbf{\text{k}}\text{,}$ where *C* is the curve given by $x = \text{cos}\ t,y = \text{sin}\ t,z = 1;0 \leq t \leq 2\pi.$

362\.

\[T\] Use a CAS and Stokes’ theorem to evaluate $\iint_{S}\text{curl}\textbf{F} \cdot d\textbf{S}\text{,}$ where $\mathbf{\text{F}}(x,y,z) = 2y\mathbf{\text{i}} + e^{z}\mathbf{\text{j}} - \text{arctan}\ x\mathbf{\text{k}}$ with *S* as a portion of paraboloid $z = 4 - x^{2} - y^{2}$ cut off by the *xy*-plane oriented upward.

363.

\[T\] Use a CAS to evaluate $\iint_{S}{\text{curl}\mathbf{\text{F}} \cdot d\mathbf{\text{S}}\text{,}}$ where $\mathbf{\text{F}}(x,y,z) = 2z\mathbf{\text{i}} + 3x\mathbf{\text{j}} + 5y\mathbf{\text{k}}$ and *S* is the surface parametrically by $\mathbf{\text{r}}(r,\theta) = r\ \text{cos}\ \theta\mathbf{\text{i}} + r\ \text{sin}\ \theta\mathbf{\text{j}} + \left( {4 - r^{2}} \right)\mathbf{\text{k}}$ $\left( {0 \leq \theta \leq 2\pi\text{,}\ 0 \leq r \leq 3} \right).$

364\.

Let *S* be paraboloid $z = a\left( {1 - x^{2} - y^{2}} \right),$ for $z \geq 0,$ where $a > 0$ is a real number. Let $\mathbf{\text{F}} = \left\langle {x - y,y + z,z - x} \right\rangle.$ For what value(s) of *a* (if any) does $\iint_{S}\text{curl}~\textbf{F} \cdot d\mathbf{S}$ have its maximum value?

For the following application exercises, the goal is to evaluate $A = \iint_{S}\text{curl}~\textbf{F} \cdot d\mathbf{S}$ where $\mathbf{\text{F}} = \left\langle {xz,\text{−}xz,xy} \right\rangle$ and *S* is the upper half of ellipsoid $x^{2} + y^{2} + 8z^{2} = 1,\ \text{where}\ z \geq 0.$

365.

Evaluate a surface integral over a more convenient surface to find the value of *A*.

366\.

Evaluate *A* using a line integral.

367.

Take paraboloid $z = x^{2} + y^{2},$ for $0 \leq z \leq 4,$ and slice it with plane $y = 0.$ Let *S* be the surface that remains for $y \geq 0,$ including the planar surface in the *xz*-plane. Let *C* be the semicircle and line segment that bounded the cap of *S* in plane $z = 4$ with counterclockwise orientation. Let $\mathbf{\text{F}} = \left\langle {2z + y,2x + z,2y + x} \right\rangle.$ Evaluate $\iint_{S}\text{curl}~\textbf{F} \cdot d\mathbf{S}$

For the following exercises, let *S* be the disk enclosed by curve

$C:\mathbf{\text{r}}(t) = \left\langle {\text{cos}\ \varphi\ \text{cos}\ t,\text{sin}\ t,\text{sin}\ \varphi\ \text{cos}\ t} \right\rangle,$ for $0 \leq t \leq 2\pi,$ where $0 \leq \varphi \leq \frac{\pi}{2}$ is a fixed angle.

368\.

What is the length of *C* in terms of $\varphi?$

369.

What is the circulation of *C* of vector field $\mathbf{\text{F}} = \left\langle {\text{−}y,\text{−}z,x} \right\rangle$ as a function of $\varphi?$

370\.

For what value of $\varphi$ is the circulation a maximum?

371.

Circle *C* in plane $x + y + z = 8$ has radius 4 and center (2, 3, 3). Evaluate $\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}$ for $\mathbf{F} = \left\langle {0,\text{−}z,2y} \right\rangle,$ where *C* has a counterclockwise orientation when viewed from above.

372\.

Velocity field $\mathbf{\text{v}} = \left\langle {0,1 - x^{2},0} \right\rangle,$ for $|x| \leq 1\ \text{and}\ |z| \leq 1,$ represents a horizontal flow in the *y*-direction. Compute the curl of v in a clockwise rotation.

373.

Evaluate integral $\iint_{S}\text{curl}~\textbf{F} \cdot d\mathbf{S}$ where $\mathbf{\text{F}} = \text{−}xz\mathbf{\text{i}} + yz\mathbf{\text{j}} + xye^{z}\mathbf{\text{k}}$ and *S* is the cap of paraboloid $z = 5 - x^{2} - y^{2}$ above plane $z = 3,$ and n points in the positive *z*-direction on *S*.

For the following exercises, use Stokes’ theorem to find the circulation of the following vector fields around any smooth, simple closed curve C.

374\.

$\mathbf{\text{F}} = \nabla\left( {x\ \text{sin}\ ye^{z}} \right)$

375.

$\mathbf{\text{F}} = \left\langle {y^{2}z^{3},2xyz^{3},3xy^{2}z^{2}} \right\rangle$

6.8 The Divergence Theorem 6.8 散度定理

• 6.8.1 Explain the meaning of the divergence theorem.
• 6.8.2 Use the divergence theorem to calculate the flux of a vector field.
• 6.8.3 Apply the divergence theorem to an electrostatic field.

• 6.8.1 解释散度定理的含义。
• 6.8.2 使用散度定理计算向量场的通量。
• 6.8.3 将散度定理应用于静电场。

We have examined several versions of the Fundamental Theorem of Calculus in higher dimensions that relate the integral around an oriented boundary of a domain to a “derivative” of that entity on the oriented domain. In this section, we state the divergence theorem, which is the final theorem of this type that we will study. The divergence theorem has many uses in physics; in particular, the divergence theorem is used in the field of partial differential equations to derive equations modeling heat flow and conservation of mass. We use the theorem to calculate flux integrals and apply it to electrostatic fields.

Overview of Theorems 定理总览

Before examining the divergence theorem, it is helpful to begin with an overview of the versions of the Fundamental Theorem of Calculus we have discussed:

1. The Fundamental Theorem of Calculus:

$${\int_{a}^{b}{f^{\prime}(x)dx = f(b) - f(a)}}.$$

This theorem relates the integral of derivative $f^{\prime}$ over line segment $\left\lbrack {a,b} \right\rbrack$ along the *x*-axis to a difference of $f$ evaluated on the boundary.

2. The Fundamental Theorem for Line Integrals:

$${\int_{C}{\nabla f \cdot d\mathbf{\text{r}} = f\left( P_{1} \right) - f\left( P_{0} \right)}},$$

where $P_{0}$ is the initial point of *C* and $P_{1}$ is the terminal point of *C*. The Fundamental Theorem for Line Integrals allows path *C* to be a path in a plane or in space, not just a line segment on the *x*-axis. If we think of the gradient as a derivative, then this theorem relates an integral of derivative $\nabla f$ over path *C* to a difference of $f$ evaluated on the boundary of *C*.

3. Green’s theorem, circulation form:

$${\iint_{D}{(Q_{x} - P_{y})dA}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$$

Since $Q_{x} - P_{y} = \text{curl}\ \mathbf{\text{F}} \cdot \mathbf{\text{k}}$ and curl is a derivative of sorts, Green’s theorem relates the integral of derivative curlF over planar region *D* to an integral of F over the boundary of *D*.

4. Green’s theorem, flux form:

$${\iint_{D}{(P_{x} + Q_{y})dA}} = {\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}.$$

Since $P_{x} + Q_{y} = \text{div}\ \mathbf{\text{F}}$ and divergence is a derivative of sorts, the flux form of Green’s theorem relates the integral of derivative divF over planar region *D* to an integral of F over the boundary of *D*.

5. Stokes’ theorem:

$${\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}.$$

If we think of the curl as a derivative of sorts, then Stokes’ theorem relates the integral of derivative curlF over surface *S* (not necessarily planar) to an integral of F over the boundary of *S*.

Stating the Divergence Theorem 散度定理的表述

The divergence theorem follows the general pattern of these other theorems. If we think of divergence as a derivative of sorts, then the divergence theorem relates a triple integral of derivative divF over a solid to a flux integral of F over the boundary of the solid. More specifically, the divergence theorem relates a flux integral of vector field F over a closed surface *S* to a triple integral of the divergence of F over the solid enclosed by *S*.

The Divergence Theorem 散度定理

Let *S* be a piecewise, smooth closed surface that encloses solid *E* in space. Assume that *S* is oriented outward, and let F be a vector field with continuous partial derivatives on an open region containing *E* (Figure 6.87). Then

$$\left. \iiint{}_{E}{\text{div}\ \mathbf{\text{F}}dV =} \right.{\iint\limits_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$$ (6.24)

Recall that the flux form of Green’s theorem states that ${\iint_{D}{\text{div}\ \mathbf{\text{F}}dA = {\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}}}}.$ Therefore, the divergence theorem is a version of Green’s theorem in one higher dimension.

The proof of the divergence theorem is beyond the scope of this text. However, we look at an informal proof that gives a general feel for why the theorem is true, but does not prove the theorem with full rigor. This explanation follows the informal explanation given for why Stokes’ theorem is true.

Proof 证明

Let *B* be a small box with sides parallel to the coordinate planes inside *E* (Figure 6.88). Let the center of *B* have coordinates $\left( {x,y,z} \right)$ and suppose the edge lengths are $\text{Δ}x,\text{Δ}y,$ and $\text{Δ}z$ (Figure 6.88(b)). The normal vector out of the top of the box is k and the normal vector out of the bottom of the box is $\text{−}\mathbf{\text{k}}.$ The dot product of $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle$ with k is *R* and the dot product with $\text{−}\mathbf{\text{k}}$ is $\text{−}R.$ The area of the top of the box (and the bottom of the box) $\text{Δ}S$ is $\text{Δ}x\text{Δ}y.$

The flux out of the top of the box can be approximated by $R\left( {x,y,z + \frac{\text{Δ}z}{2}} \right)\text{Δ}x\text{Δ}y$ (Figure 6.88(c)) and the flux out of the bottom of the box is $\text{−}R\left( {x,y,z - \frac{\text{Δ}z}{2}} \right)\text{Δ}x\text{Δ}y.$ If we denote the difference between these values as $\text{Δ}R,$ then the net flux in the vertical direction can be approximated by $\text{Δ}R\text{Δ}x\text{Δ}y.$ However,

$$\text{Δ}R\text{Δ}x\text{Δ}y = \left( \frac{\text{Δ}R}{\text{Δ}z} \right)\text{Δ}x\text{Δ}y\text{Δ}z \approx \left( \frac{\partial R}{\partial z} \right)\text{Δ}V.$$

Therefore, the net flux in the vertical direction can be approximated by $\left( \frac{\partial R}{\partial z} \right)\text{Δ}V.$ Similarly, the net flux in the *x*-direction can be approximated by $\left( \frac{\partial P}{\partial x} \right)\text{Δ}V$ and the net flux in the *y*-direction can be approximated by $\left( \frac{\partial Q}{\partial y} \right)\text{Δ}V.$ Adding the fluxes in all three directions gives an approximation of the total flux out of the box:

$$\text{Total flux} \approx \left( {\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}} \right)\text{Δ}V = \text{div}\ \mathbf{\text{F}}\text{Δ}V.$$

This approximation becomes arbitrarily close to the value of the total flux as the volume of the box shrinks to zero.

The sum of $\text{div}\ \mathbf{\text{F}}\text{Δ}V$ over all the small boxes approximating *E* is approximately $\left. \iiint{}_{E}{\text{div}\ \mathbf{\text{F}}dV} \right..$ On the other hand, the sum of $\text{div}\ \mathbf{\text{F}}\text{Δ}V$ over all the small boxes approximating *E* is the sum of the fluxes over all these boxes. Just as in the informal proof of Stokes’ theorem, adding these fluxes over all the boxes results in the cancelation of a lot of the terms. If an approximating box shares a face with another approximating box, then the flux over one face is the negative of the flux over the shared face of the adjacent box. These two integrals cancel out. When adding up all the fluxes, the only flux integrals that survive are the integrals over the faces approximating the boundary of *E*. As the volumes of the approximating boxes shrink to zero, this approximation becomes arbitrarily close to the flux over *S*.

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Verifying the Divergence Theorem 验证散度定理

Verify the divergence theorem for vector field $\mathbf{\text{F}} = \left\langle {x - y,x + z,z - y} \right\rangle$ and surface *S* that consists of cone $x^{2} + y^{2} = z^{2},0 \leq z \leq 1,$ and the circular top of the cone (see the following figure). Assume this surface is oriented outward.

Solution 解

Let *E* be the solid cone enclosed by *S*. To verify the theorem for this example, we show that $\left. \iiint{}_{E}{\text{div}\ \mathbf{\text{F}}dV =} \right.{\iint\limits_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}$ by calculating each integral separately.

To compute the triple integral, note that $\text{div}\ \mathbf{\text{F}} = P_{x} + Q_{y} + R_{z} = 2,$ and therefore the triple integral is

$$\begin{array}{cl} {\iiint_{E}{\text{div}\ \mathbf{\text{F}}dV}} & {= 2{\iiint_{E}{dV}}} \\ & {= 2\left( {\text{volume of}\ E} \right).} \end{array}$$

The volume of a right circular cone is given by $\pi r^{2}\frac{h}{3}.$ In this case, $h = r = 1.$ Therefore,

$${\iiint_{E}{\text{div}\ \mathbf{\text{F}}dV}} = 2\left( {\text{volume of}\ E} \right) = \frac{2\pi}{3}.$$

To compute the flux integral, first note that *S* is piecewise smooth; *S* can be written as a union of smooth surfaces. Therefore, we break the flux integral into two pieces: one flux integral across the circular top of the cone and one flux integral across the remaining portion of the cone. Call the circular top $S_{1}$ and the portion under the top $S_{2}.$ We start by calculating the flux across the circular top of the cone. Notice that $S_{1}$ has parameterization

$$\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,1} \right\rangle,0 \leq u \leq 1,0 \leq v \leq 2\pi.$$

Then, the tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {\text{cos}\ v,\text{sin}\ v,0} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {\text{−}u\ \text{cos}\ v,u\ \text{sin}\ v,0} \right\rangle.$ Therefore, the flux across $S_{1}$ is

$$\begin{array}{cl} {\iint_{S_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} & {= {\int_{0}^{1}{\int_{0}^{2\pi}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right)}}} \cdot \left( {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right)dA} \\ & {= {\int_{0}^{1}{\int_{0}^{2\pi}{\left\langle {u\ \text{cos}\ v - u\ \text{sin}\ v,u\ \text{cos}\ v + 1,1 - u\ \text{sin}\ v} \right\rangle \cdot \left\langle {0,0,u} \right\rangle}}}dvdu} \\ & {= {\int_{0}^{1}{\int_{0}^{2\pi}{u - u^{2}\text{sin}\ v}}}\ dvdu = \pi.} \end{array}$$

We now calculate the flux over $S_{2}.$ A parameterization of this surface is

$$\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,u} \right\rangle,0 \leq u \leq 1,0 \leq v \leq 2\pi.$$

The tangent vectors are $\mathbf{\text{t}}_{u} = \left\langle {\text{cos}\ v,\text{sin}\ v,1} \right\rangle$ and $\mathbf{\text{t}}_{v} = \left\langle {\text{−}u\ \text{sin}\ v,u\ \text{cos}\ v,0} \right\rangle,$ so the cross product is

$$\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v} = \left\langle {\text{−}u\ \text{cos}\ v,\text{−}u\ \text{sin}\ v,u} \right\rangle.$$

Notice that the negative signs on the *x* and *y* components induce the inward orientation of the cone. Since the surface is oriented outward, we use vector $\mathbf{\text{t}}_{v}\ \times \ \mathbf{\text{t}}_{u} = \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,\text{−}u} \right\rangle$ in the flux integral. The flux across $S_{2}$ is then

$$\begin{array}{cl} {\iint_{S_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} & {= {\int_{0}^{1}{\int_{0}^{2\pi}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right)}}} \cdot \left( {\mathbf{\text{t}}_{v}\ \times \ \mathbf{\text{t}}_{u}} \right)dA} \\ & {= {\int_{0}^{1}{\int_{0}^{2\pi}{\left\langle {u\ \text{cos}\ v - u\ \text{sin}\ v,u\ \text{cos}\ v + u,u - \text{sin}\ v} \right\rangle \cdot \left\langle {u\ \text{cos}\ v,u\ \text{sin}\ v,\text{−}u} \right\rangle}}}} \\ & {= {\int_{0}^{1}{\int_{0}^{2\pi}u^{2}}}\text{cos}^{2}v + 2u^{2}\text{sin}\ v - u^{2}dvdu = - \frac{\pi}{3}.} \end{array}$$

The total flux across *S* is

$${\iint_{S_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = {\iint_{S_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} + {\iint_{S_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = \frac{2\pi}{3} = \left. \iiint{}_{E}\text{div} \right.\ \mathbf{\text{F}}dV,$$

and we have verified the divergence theorem for this example.

Verify the divergence theorem for vector field $\mathbf{\text{F}}(x,y,z) = \left\langle {x + y + z,y,2x - y} \right\rangle$ and surface *S* given by the cylinder $x^{2} + y^{2} = 1,0 \leq z \leq 3$ plus the circular top and bottom of the cylinder. Assume that *S* is oriented outward.

Recall that the divergence of continuous field F at point *P* is a measure of the “outflowing-ness” of the field at *P*. If F represents the velocity field of a fluid, then the divergence can be thought of as the rate per unit volume of the fluid flowing out less the rate per unit volume flowing in. The divergence theorem confirms this interpretation. To see this, let *P* be a point and let $B_{r}$ be a ball of small radius *r* centered at *P* (Figure 6.89). Let $S_{r}$ be the boundary sphere of $B_{r}.$ Since the radius is small and F is continuous, $\text{div}\ \mathbf{\text{F}}(Q) \approx \text{div}\ \mathbf{\text{F}}(P)$ for all other points *Q* in the ball. Therefore, the flux across $S_{r}$ can be approximated using the divergence theorem:

$${\iint_{S_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} = \left. \iiint{}_{B_{r}}\text{div} \right.\ \mathbf{\text{F}}dV \approx \left. \iiint{}_{B_{r}}\text{div} \right.\ \mathbf{\text{F}}(P)dV.$$

Since $\text{div}\ \mathbf{\text{F}}(P)$ is a constant,

$$\left. \iiint{}_{B_{r}}\text{div} \right.\ \mathbf{\text{F}}(P)dV = \text{div}\ \mathbf{\text{F}}(P)V\left( B_{r} \right).$$

Therefore, flux $\iint_{S_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ can be approximated by $\text{div}\ \mathbf{\text{F}}(P)V\left( B_{r} \right).$ This approximation gets better as the radius shrinks to zero, and therefore

$$\text{div}\ \mathbf{\text{F}}(P) = \underset{r\rightarrow 0}{\text{lim}}\frac{1}{V\left( B_{r} \right)}{\iint_{S_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}.$$

This equation says that the divergence at *P* is the net rate of outward flux of the fluid per unit volume.

Using the Divergence Theorem 散度定理的应用

The divergence theorem translates between the flux integral of closed surface *S* and a triple integral over the solid enclosed by *S*. Therefore, the theorem allows us to compute flux integrals or triple integrals that would ordinarily be difficult to compute by translating the flux integral into a triple integral and vice versa.

Applying the Divergence Theorem 应用散度定理

Calculate the surface integral ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where *S* is cylinder $x^{2} + y^{2} = 1,0 \leq z \leq 2,$ including the circular top and bottom, and $\mathbf{\text{F}} = \left\langle {\frac{x^{3}}{3} + yz,\frac{y^{3}}{3} - \text{sin}\left( {xz} \right),z - x - y} \right\rangle.$

Solution 解

We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using Equation 6.19.

By contrast, the divergence theorem allows us to calculate the single triple integral $\left. \iiint{}_{E}\text{div} \right.\ \mathbf{\text{F}}dV,$ where *E* is the solid enclosed by the cylinder. Using the divergence theorem and converting to cylindrical coordinates, we have

$$\begin{array}{cl} \left. \iint{}_{s}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}} \right. & {= \left. \iiint{}_{E}\text{div} \right.\ \mathbf{\text{F}}\ dV} \\ & {= \left. \iiint{}_{E}\left( {x^{2} + y^{2} + 1} \right) \right.dV} \\ & {= {\int_{0}^{2\pi}{\int_{0}^{1}{\int_{0}^{2}{\left( {r^{2} + 1} \right)r\ dz}}}}\ dr\ d\theta} \\ & {= \frac{3}{2}{\int_{0}^{2\pi}{d\theta}} = 3\pi.} \end{array}$$

Use the divergence theorem to calculate flux integral ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where *S* is the boundary of the box given by $0 \leq x \leq 2,1 \leq y \leq 4,0 \leq z \leq 1,$ and $\mathbf{\text{F}} = \left\langle {x^{2} + yz,y - z,2x + 2y + 2z} \right\rangle$ (see the following figure).

Applying the Divergence Theorem 应用散度定理

Let $\mathbf{\text{v}} = \left\langle {- \frac{y}{z},\frac{x}{z},0} \right\rangle$ be the velocity field of a fluid. Let *C* be the solid cube given by $1 \leq x \leq 4,2 \leq y \leq 5,1 \leq z \leq 4,$ and let *S* be the boundary of this cube (see the following figure). Find the flow rate of the fluid across *S*.

Solution 解

The flow rate of the fluid across *S* is ${\iint_{S}{\mathbf{\text{v}} \cdot d\mathbf{\text{S}}}}.$ Before calculating this flux integral, let’s discuss what the value of the integral should be. Based on Figure 6.90, we see that if we place this cube in the fluid (as long as the cube doesn’t encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. The field is rotational in nature and, for a given circle parallel to the *xy*-plane that has a center on the *z*-axis, the vectors along that circle are all the same magnitude. That is how we can see that the flow rate is the same entering and exiting the cube. The flow into the cube cancels with the flow out of the cube, and therefore the flow rate of the fluid across the cube should be zero.

To verify this intuition, we need to calculate the flux integral. Calculating the flux integral directly requires breaking the flux integral into six separate flux integrals, one for each face of the cube. We also need to find tangent vectors, compute their cross product, and use Equation 6.19. However, using the divergence theorem makes this calculation go much more quickly:

$$\begin{array}{cl} {\iint_{S}{\mathbf{\text{v}} \cdot d\mathbf{\text{S}}}} & {= \left. \iiint{}_{C}{\text{div}\left( \mathbf{\text{v}} \right)} \right.dV} \\ & {= \left. \iiint{}_{C}0 \right.\ dV = 0.} \end{array}$$

Therefore the flux is zero, as expected.

Let $\mathbf{\text{v}} = \left\langle {\frac{x}{z},\frac{y}{z},0} \right\rangle$ be the velocity field of a fluid. Let *C* be the solid cube given by $1 \leq x \leq 4,2 \leq y \leq 5,1 \leq z \leq 4,$ and let *S* be the boundary of this cube (see the following figure). Find the flow rate of the fluid across *S*.

Example 6.79 illustrates a remarkable consequence of the divergence theorem. Let *S* be a piecewise, smooth closed surface and let F be a vector field defined on an open region containing the surface enclosed by *S*. If F has the form $\mathbf{\text{F}} = \left\langle {f\left( {y,z} \right),g\left( {x,z} \right),h\left( {x,y} \right)} \right\rangle,$ then the divergence of F is zero. By the divergence theorem, the flux of F across *S* is also zero. This makes certain flux integrals incredibly easy to calculate. For example, suppose we wanted to calculate the flux integral $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ where *S* is a cube and

$$\mathbf{\text{F}} = \left\langle {\text{sin}(y)e^{yz},x^{2}z^{2},\text{cos}\left( {xy} \right)e^{\text{sin}\ x}} \right\rangle.$$

Calculating the flux integral directly would be difficult, if not impossible, using techniques we studied previously. At the very least, we would have to break the flux integral into six integrals, one for each face of the cube. But, because the divergence of this field is zero, the divergence theorem immediately shows that the flux integral is zero.

We can now use the divergence theorem to justify the physical interpretation of divergence that we discussed earlier. Recall that if F is a continuous three-dimensional vector field and *P* is a point in the domain of F, then the divergence of F at *P* is a measure of the “outflowing-ness” of F at *P*. If F represents the velocity field of a fluid, then the divergence of F at *P* is a measure of the net flow rate out of point *P* (the flow of fluid out of *P* less the flow of fluid in to *P*). To see how the divergence theorem justifies this interpretation, let $B_{r}$ be a ball of very small radius *r* with center *P*, and assume that $B_{r}$ is in the domain of F. Furthermore, assume that $B_{r}$ has a positive, outward orientation. Since the radius of $B_{r}$ is small and F is continuous, the divergence of F is approximately constant on $B_{r}.$ That is, if $P^{\prime}$ is any point in $B_{r},$ then $\text{div}\ \mathbf{\text{F}}(P) \approx \text{div}\ \mathbf{\text{F}}(P^{\prime}).$ Let $S_{r}$ denote the boundary sphere of $B_{r}.$ We can approximate the flux across $S_{r}$ using the divergence theorem as follows:

$$\begin{array}{cl} {\iint_{S_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}} & {= {\iiint_{B_{r}}{\text{div}\ \mathbf{\text{F}}}}\ dV} \\ & {\approx {\iiint_{B_{r}}{\text{div}\ \mathbf{\text{F}}}}(P)dV} \\ & {= \text{div}\ \mathbf{\text{F}}(P)V(B_{r}).} \end{array}$$

As we shrink the radius *r* to zero via a limit, the quantity $\text{div}\ \mathbf{\text{F}}(P)V(B_{r})$ gets arbitrarily close to the flux. Therefore,

$$\text{div}\ \mathbf{\text{F}}(P) = \underset{r\rightarrow 0}{\text{lim}}\frac{1}{V(B_{r})}{\iint_{S_{r}}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}$$

and we can consider the divergence at *P* as measuring the net rate of outward flux per unit volume at *P*. Since “outflowing-ness” is an informal term for the net rate of outward flux per unit volume, we have justified the physical interpretation of divergence we discussed earlier, and we have used the divergence theorem to give this justification.

Application to Electrostatic Fields 静电场中的应用

The divergence theorem has many applications in physics and engineering. It allows us to write many physical laws in both an integral form and a differential form (in much the same way that Stokes’ theorem allowed us to translate between an integral and differential form of Faraday’s law). Areas of study such as fluid dynamics, electromagnetism, and quantum mechanics have equations that describe the conservation of mass, momentum, or energy, and the divergence theorem allows us to give these equations in both integral and differential forms.

One of the most common applications of the divergence theorem is to electrostatic fields. An important result in this subject is Gauss’ law. This law states that if *S* is a closed surface in electrostatic field E, then the flux of E across *S* is the total charge enclosed by *S* (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.

If $(x,y,z)$ is a point in space, then the distance from the point to the origin is $r = \sqrt{x^{2} + y^{2} + z^{2}}.$ Let $\mathbf{\text{F}}_{r}$ denote radial vector field $\mathbf{\text{F}}_{r} = \frac{1}{r^{2}}\left\langle {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right\rangle.$ The vector at a given position in space points in the direction of unit radial vector $\left\langle {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right\rangle$ and is scaled by the quantity $1\text{/}r^{2}.$ Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector’s distance from the origin. Suppose we have a stationary charge of *q* Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field E given by

$$\mathbf{\text{E}} = \frac{q}{4\pi\varepsilon_{0}}\mathbf{\text{F}}_{r},$$

where the approximation $\varepsilon_{0} = 8.854\ \times \ 10^{-12}$ farad (F)/m is an electric constant. (The constant $\varepsilon_{0}$ is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that E is a radial vector field similar to the gravitational field described in Example 6.6. The difference is that this field points outward whereas the gravitational field points inward. Because

$$\mathbf{\text{E}} = \frac{q}{4\pi\varepsilon_{0}}\mathbf{\text{F}}_{r} = \frac{q}{4\pi\varepsilon_{0}}\left( {\frac{1}{r^{2}}\left\langle {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right\rangle} \right),$$

we say that electrostatic fields obey an inverse-square law. That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface *S* is zero if the charge is outside of *S*, and that the flux is $q\text{/}\varepsilon_{0}$ if the charge is inside of *S*. In other words, the flux across *S* is the charge inside the surface divided by constant $\varepsilon_{0}.$ This is a special case of Gauss’ law, and here we use the divergence theorem to justify this special case.

To show that the flux across *S* is the charge inside the surface divided by constant $\varepsilon_{0},$ we need two intermediate steps. First we show that the divergence of $\mathbf{\text{F}}_{r}$ is zero and then we show that the flux of $\mathbf{\text{F}}_{r}$ across any smooth surface *S* is either zero or $4\pi.$ We can then justify this special case of Gauss’ law.

The Divergence of $\mathbf{\text{F}}_{r}$ Is Zero $\mathbf{\text{F}}_{r}$ 的散度为零

Verify that the divergence of $\mathbf{\text{F}}_{r}$ is zero where $\mathbf{\text{F}}_{r}$ is defined (away from the origin).

Solution 解

Since $r = \sqrt{x^{2} + y^{2} + z^{2}},$ the quotient rule gives us

$$\begin{array}{cl} {\frac{\partial}{\partial x}\left( \frac{x}{r^{3}} \right)} & {= \frac{\partial}{\partial x}\left( \frac{x}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2}} \right)} \\ & {= \frac{\left( {x^{2} + y^{2} + z^{2}} \right)^{3\text{/}2} - x\left\lbrack {\frac{3}{2}\left( {x^{2} + y^{2} + z^{2}} \right)^{1\text{/}2}2x} \right\rbrack}{\left( {x^{2} + y^{2} + z^{2}} \right)^{3}}} \\ & {= \frac{r^{3} - 3x^{2}r}{r^{6}} = \frac{r^{2} - 3x^{2}}{r^{5}}.} \end{array}$$

Similarly,

$$\frac{\partial}{\partial y}\left( \frac{y}{r^{3}} \right) = \frac{r^{2} - 3y^{2}}{r^{5}}\ \text{and}\ \frac{\partial}{\partial z}\left( \frac{z}{r^{3}} \right) = \frac{r^{2} - 3z^{2}}{r^{5}}.$$

Therefore,

$$\begin{array}{cl} {\text{div}\ \mathbf{\text{F}}_{r}} & {= \frac{r^{2} - 3x^{2}}{r^{5}} + \frac{r^{2} - 3y^{2}}{r^{5}} + \frac{r^{2} - 3z^{2}}{r^{5}}} \\ & {= \frac{3r^{2} - 3\left( {x^{2} + y^{2} + z^{2}} \right)}{r^{5}}} \\ & {= \frac{3r^{2} - 3r^{2}}{r^{5}} = 0.} \end{array}$$

Notice that since the divergence of $\mathbf{\text{F}}_{r}$ is zero and E is $\mathbf{\text{F}}_{r}$ scaled by a constant, the divergence of electrostatic field E is also zero (except at the origin).

Flux across a Smooth Surface 穿过光滑曲面的通量

Let *S* be a connected, piecewise smooth closed surface and let $\mathbf{\text{F}}_{r} = \frac{1}{r^{2}}\left\langle {\frac{x}{r},\frac{y}{r},\frac{z}{r}} \right\rangle.$ Then,

$${\iint_{S}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}} = \left\{ \begin{array}{lccl} 0 & & & {\text{if}\ S\ \text{does not encompass the origin}} \\ {4\pi} & & & {\text{if}\ S\ \text{encompasses the origin.}} \end{array} \right.$$

In other words, this theorem says that the flux of $\mathbf{\text{F}}_{r}$ across any piecewise smooth closed surface *S* depends only on whether the origin is inside of *S*.

Proof 证明

The logic of this proof follows the logic of Example 6.46, only we use the divergence theorem rather than Green’s theorem.

First, suppose that *S* does not encompass the origin. In this case, the solid enclosed by *S* is in the domain of $\mathbf{\text{F}}_{r},$ and since the divergence of $\mathbf{\text{F}}_{r}$ is zero, we can immediately apply the divergence theorem and find that $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ is zero.

Now suppose that *S* does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let $S_{a}$ be a sphere of radius *a* inside of *S* centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is

$$\mathbf{\text{t}}_{\phi}\ \times \ \mathbf{\text{t}}_{\theta} = \left\langle {a^{2}\text{cos}\ \theta\ \text{sin}^{2}\phi,a^{2}\text{sin}\ \theta\ \text{sin}^{2}\phi,a^{2}\text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle$$

(see Example 6.64). Therefore, on the surface of the sphere, the dot product $\mathbf{\text{F}}_{r} \cdot \mathbf{\text{N}}$ (in spherical coordinates) is

$$\begin{array}{cl} {\mathbf{\text{F}}_{r} \cdot \mathbf{\text{N}}} & {= \left\langle {\frac{\text{sin}\ \phi\ \text{cos}\ \theta}{a^{2}},\frac{\text{sin}\ \phi\ \text{sin}\ \theta}{a^{2}},\frac{\text{cos}\ \phi}{a^{2}}} \right\rangle \cdot \left\langle {a^{2}\text{cos}\ \theta\ \text{sin}^{2}\phi,a^{2}\text{sin}\ \theta\ \text{sin}^{2}\phi,a^{2}\text{sin}\ \phi\ \text{cos}\ \phi} \right\rangle} \\ & {= \text{sin}\ \phi\left( {\left\langle {\text{sin}\ \phi\ \text{cos}\ \theta,\text{sin}\ \phi\ \text{sin}\ \theta,\text{cos}\ \phi} \right\rangle \cdot \left\langle {\text{sin}\ \phi\ \text{cos}\ \theta,\text{sin}\ \phi\ \text{sin}\ \theta,\text{cos}\ \phi} \right\rangle} \right)} \\ & {= \text{sin}\ \phi.} \end{array}$$

The flux of $\mathbf{\text{F}}_{r}$ across $S_{a}$ is

$${\iint_{S_{a}}{\mathbf{\text{F}}_{r} \cdot \mathbf{\text{N}}dS =}}{\int_{0}^{2\pi}{{\int_{0}^{\pi}{\text{sin}\ \phi d\phi d\theta}} =}}4\pi.$$

Now, remember that we are interested in the flux across *S*, not necessarily the flux across $S_{a}.$ To calculate the flux across *S*, let *E* be the solid between surfaces $S_{a}$ and *S*. Then, the boundary of *E* consists of $S_{a}$ and *S*. Denote this boundary by $S - S_{a}$ to indicate that *S* is oriented outward but now $S_{a}$ is oriented inward. We would like to apply the divergence theorem to solid *E.* Notice that the divergence theorem, as stated, can’t handle a solid such as *E* because *E* has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green’s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,

$$\begin{array}{cl} {\iint_{S - S_{a}}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}} & {= {\iint_{S}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}} - {\iint_{S_{a}}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}}} \\ & {= {\iiint_{E}\text{div}}\mathbf{\text{F}}_{r}\ dV} \\ & {= {\iiint_{E}{0dV}} = 0.} \end{array}$$

Therefore,

$${\iint_{S}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}} = {\iint_{S_{a}}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}} = 4\pi,$$

and we have our desired result.

□

Now we return to calculating the flux across a smooth surface in the context of electrostatic field $\mathbf{\text{E}} = \frac{q}{4\pi\varepsilon_{0}}\mathbf{\text{F}}_{r}$ of a point charge at the origin. Let *S* be a piecewise smooth closed surface that encompasses the origin. Then

$$\begin{array}{cl} {\iint_{S}{\mathbf{\text{E}} \cdot d\mathbf{\text{S}}}} & {= {\iint_{S}{\frac{q}{4\pi\varepsilon_{0}}\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}}} \\ & {= \frac{q}{4\pi\varepsilon_{0}}{\iint_{S}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}}} \\ & {= \frac{q}{\varepsilon_{0}}.} \end{array}$$

If *S* does not encompass the origin, then

$${\iint_{S}{\mathbf{\text{E}} \cdot d\mathbf{\text{S}}}} = \frac{q}{4\pi\varepsilon_{0}}{\iint_{S}{\mathbf{\text{F}}_{r} \cdot d\mathbf{\text{S}}}} = 0.$$

Therefore, we have justified the claim that we set out to justify: the flux across closed surface *S* is zero if the charge is outside of *S*, and the flux is $q\text{/}\varepsilon_{0}$ if the charge is inside of *S*.

This analysis works only if there is a single point charge at the origin. In this case, Gauss’ law says that the flux of E across *S* is the total charge enclosed by *S*. Gauss’ law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss’ law states that if *S* is a piecewise smooth closed surface and *Q* is the total amount of charge inside of *S*, then the flux of E across *S* is $Q\text{/}\varepsilon_{0}.$

Using Gauss’ law 应用高斯定律

Suppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at $(0,1,1),(1,1,4),(-1,0,0),\ \text{and}\ (-2,-2,2).$ Let E denote the electrostatic field generated by these point charges. If *S* is the sphere of radius 2 oriented outward and centered at the origin, then find ${\iint_{S}{\mathbf{\text{E}} \cdot d\mathbf{\text{S}}}}.$

Solution 解

According to Gauss’ law, the flux of E across *S* is the total charge inside of *S* divided by the electric constant. Since *S* has radius 2, notice that only two of the charges are inside of *S*: the charge at $(0,1,1)$ and the charge at $(-1,0,0).$ Therefore, the total charge encompassed by *S* is 0.004 and, by Gauss’ law,

$${\iint_{S}{\mathbf{\text{E}} \cdot d\mathbf{\text{S}}}} = \frac{0.004}{8.854\ \times \ 10^{-12}} \approx 4.518\ \times \ 10^{9}\ \text{V-m}.$$

Work the previous example for surface *S* that is a sphere of radius 4 centered at the origin, oriented outward.

Section 6.8 Exercises 6.8 节习题

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral $\iint_{S}^{}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}ds}$ for the given choice of F and the boundary surface *S.* For each closed surface, assume N is the outward unit normal vector.

376\.

\[T\] $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}};$ *S* is the surface of cube $0 \leq x \leq 1,0 \leq y \leq 1,0 < z \leq 1.$

377.

\[T\] $\mathbf{\text{F}}(x,y,z) = (\text{cos}\ yz)\mathbf{\text{i}} + e^{xz}\mathbf{\text{j}} + 3z^{2}\mathbf{\text{k}}\text{;}$ *S* is the surface of hemisphere $z = \sqrt{4 - x^{2} - y^{2}}$ together with disk $x^{2} + y^{2} \leq 4$ in the *xy*-plane.

378\.

\[T\] ${\mathbf{\text{F}}(x,y,z) = \left( {x^{2} + y^{2} - x^{2}} \right)\mathbf{\text{i}} + x^{2}y\mathbf{\text{j}} + 3z\mathbf{\text{k}};}S$ is the surface of the unit cube $0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1$ excluding the face $z = 0$.

379.

\[T\] $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}\text{;}$ *S* is the surface of the solid bounded by the parabola $z = x^{2} + y^{2}\ $ and the plane $z = 9$.

380\.

\[T\] $\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + y^{2}\mathbf{\text{j}} + z^{2}\mathbf{\text{k}}\text{;}$ *S* is the surface of sphere $x^{2} + y^{2} + z^{2} = 4.$

381.

\[T\] $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + \left( {z^{2} - 1} \right)\mathbf{\text{k}}\text{;}$ *S* is the surface of the solid bounded by cylinder $x^{2} + y^{2} = 4$ and planes $z = 0\ \text{and}\ z = 1.$

382\.

\[T\] $\mathbf{\text{F}}(x,y,z) = xy^{2}\mathbf{\text{i}} + yz^{2}\mathbf{\text{j}} + x^{2}z\mathbf{\text{k}}\text{;}$ *S* is the surface of the solid bounded above by sphere $\rho = 2$ and below by cone $\varphi = \frac{\pi}{4}$ in spherical coordinates. (Think of *S* as the surface of an “ice cream cone.”)

383.

\[T\] $\mathbf{\text{F}}(x,y,z) = x^{3}\mathbf{\text{i}} + y^{3}\mathbf{\text{j}} + 3a^{2}z\mathbf{\text{k}}\ \text{(constant}\ a > 0)\text{;}$ *S* is the surface of the solid bounded by cylinder $x^{2} + y^{2} = a^{2}$ and planes $z = 0\ \text{and}\ z = 1.$

384\.

\[T\] Surface integral ${\iint_{S}{\mathbf{\text{F}} \cdot}}d\mathbf{\text{S}},$ where *S* is the surface of the solid bounded by paraboloid $z = x^{2} + y^{2}$ and plane $z = 4,$ and $\mathbf{\text{F}}(x,y,z) = \left( {x + y^{2}z^{2}} \right)\mathbf{\text{i}} + \left( {y + z^{2}x^{2}} \right)\mathbf{\text{j}} + \left( {z + x^{2}y^{2}} \right)\mathbf{\text{k}}$

385.

Use the divergence theorem to calculate surface integral ${\iint_{S}{\mathbf{\text{F}} \cdot}}d\mathbf{\text{S}},$ where $\mathbf{\text{F}}(x,y,z) = \left( e^{y^{2}} \right)\mathbf{\text{i}} + \left( {y + \text{sin}\left( z^{2} \right)} \right)\mathbf{\text{j}} + \left( {z - 1} \right)\mathbf{\text{k}}$ and *S* is the surface of the solid bounded by the sphere $x^{2} + y^{2} + z^{2} = 1\text{,}$ and below by the plane $z = 0$.

386\.

Use the divergence theorem to calculate surface integral ${\iint_{S}{\mathbf{\text{F}} \cdot}}ds,$ where $\mathbf{\text{F}}(x,y,z) = x^{4}\mathbf{\text{i}} - x^{3}z^{2}\mathbf{\text{j}} + 4xy^{2}z\mathbf{\text{k}}$ and $S$ is the surface bounded by cylinder $x^{2} + y^{2} = 1$ and planes $z = x + 2$ and $z = 0.$

387.

Use the divergence theorem to calculate surface integral $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ when $\mathbf{\text{F}}(x,y,z) = x^{2}z^{3}\mathbf{\text{i}} + 2xyz^{3}\mathbf{\text{j}} + xz^{4}\mathbf{\text{k}}$ and *S* is the surface of the box with vertices $(\pm 1,\pm 2,\pm 3).$

388\.

Use the divergence theorem to calculate surface integral $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ when $\mathbf{\text{F}}(x,y,z) = z\ \text{tan}^{-1}\left( y^{2} \right)\mathbf{\text{i}} + z^{3}\text{ln}\left( {x^{2} + 1} \right)\mathbf{\text{j}} + z\mathbf{\text{k}}$ and *S* is the surface of the solid bounded by the paraboloid $x^{2} + y^{2} + z = 2$ and the plane $z = 1$.

389.

\[T\] Use a CAS and the divergence theorem to calculate flux ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = \left( {x^{3} + y^{3}} \right)\mathbf{\text{i}} + \left( {y^{3} + z^{3}} \right)\mathbf{\text{j}} + \left( {z^{3} + x^{3}} \right)\mathbf{\text{k}}$ and *S* is a sphere with center (0, 0, 0) and radius 2.

390\.

Use the divergence theorem to compute the value of flux integral ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = \left( {y^{3} + 3x} \right)\mathbf{\text{i}} + \left( {xz + y} \right)\mathbf{\text{j}} + \left\lbrack {z + x^{4}\text{cos}\left( {x^{2}y} \right)} \right\rbrack\mathbf{\text{k}}$ and *S* is the surface of the solid bounded by $x^{2} + y^{2} = 1,x \geq 0,y \geq 0,\ \text{and}\ 0 \leq z \leq 1.$

391.

Use the divergence theorem to compute flux integral ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = y\mathbf{\text{j}} - z\mathbf{\text{k}}$ and *S* consists of the union of paraboloid $y = x^{2} + z^{2},0 \leq y \leq 1,$ and disk $x^{2} + z^{2} \leq 1,y = 1,$ oriented outward. What is the flux through just the paraboloid?

392\.

Use the divergence theorem to compute flux integral ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = x + y\mathbf{\text{j}} + z^{4}\mathbf{\text{k}}$ and *S* is a part of cone $z = \sqrt{x^{2} + y^{2}}$ beneath top plane $z = 1,$ oriented downward.

393.

Use the divergence theorem to calculate surface integral $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ for $\mathbf{\text{F}}(x,y,z) = x^{4}\mathbf{\text{i}} - x^{3}z^{2}\mathbf{\text{j}} + 4xy^{2}z\mathbf{\text{k}},$ where *S* is the surface inside the cylinder $x^{2} + y^{2} = 1$ between the planes $z = x + 2\ \text{and}\ z = 0.$

394\.

Consider $\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + xy\mathbf{\text{j}} + (z + 1)\mathbf{\text{k}}.$ Let *E* be the solid enclosed by paraboloid $z = 4 - x^{2} - y^{2}$ and plane $z = 0$ with normal vectors pointing outside *E*. Compute flux *F* across the boundary of *E* using the divergence theorem.

For the following exercises, use a CAS along with the divergence theorem to compute the net outward flux for the fields across the given surfaces *S*.

395.

\[T\] $\mathbf{\text{F}} = \left\langle {x,-2y,3z} \right\rangle;$ *S* is sphere $\left\{ {\left( {x,y,z} \right):x^{2} + y^{2} + z^{2} = 6} \right\}.$

396\.

\[T\] $\mathbf{\text{F}} = \left\langle {x,2y,z} \right\rangle;$ *S* is the boundary of the tetrahedron in the first octant formed by plane $x + y + z = 1.$

397.

\[T\] $\mathbf{\text{F}} = \left\langle {y - 2x,x^{3} - y,y^{2} - z} \right\rangle;$ *S* is sphere $\left\{ {\left( {x,y,z} \right):x^{2} + y^{2} + z^{2} = 4} \right\}.$

398\.

\[T\] $\mathbf{\text{F}} = \left\langle {x,y,z} \right\rangle;$ *S* is the surface of paraboloid $z = 4 - x^{2} - y^{2},$ for $z \geq 0,$ plus its base in the *xy*-plane.

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions *D*.

399.

\[T\] $\mathbf{\text{F}} = \left\langle {z - x,x - y,2y - z} \right\rangle;$ *D* is the region between spheres of radius 2 and 4 centered at the origin.

400\.

\[T\] $\mathbf{\text{F}} = \frac{\mathbf{\text{r}}}{\left. ||\mathbf{\text{r}} \right.||} = \frac{\left\langle {x,y,z} \right\rangle}{\sqrt{x^{2} + y^{2} + z^{2}}};$ *D* is the region between spheres of radius 1 and 2 centered at the origin.

401.

\[T\] $\mathbf{\text{F}} = \left\langle {x^{2},\text{−}y^{2},z^{2}} \right\rangle;$ *D* is the region in the first octant between planes $z = 4 - x - y$ and $z = 2 - x - y.$

402\.

Let $\mathbf{\text{F}}(x,y,z) = 2x\mathbf{\text{i}} - 3xy\mathbf{\text{j}} + xz^{2}\mathbf{\text{k}}.$ Use the divergence theorem to calculate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where *S* is the surface of the cube with corners at $(0,0,0),(1,0,0),(0,1,0),$ $(1,1,0),(0,0,1),(1,0,1),(0,1,1),\ \text{and}\ (1,1,1),$ oriented outward.

403.

Use the divergence theorem to find the outward flux of field $\mathbf{\text{F}}(x,y,z) = \left( {x^{3} - 3y} \right)\mathbf{\text{i}} + \left( {2yz + 1} \right)\mathbf{\text{j}} + xyz\mathbf{\text{k}}$ through the cube bounded by planes $x = \pm 1,y = \pm 1,\ \text{and}\ z = \pm 1.$

404\.

Let $\mathbf{\text{F}}(x,y,z) = 2x\mathbf{\text{i}} - 3y\mathbf{\text{j}} + 5z\mathbf{\text{k}}$ and let *S* be hemisphere $z = \sqrt{9 - x^{2} - y^{2}}$ together with disk $x^{2} + y^{2} \leq 9$ in the *xy*-plane. Use the divergence theorem to calculate $\iint_{S}~\mathbf{F}~ \cdot d\mathbf{S}$.

405.

Evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}},$ where $\mathbf{\text{F}}(x,y,z) = x^{2}\mathbf{\text{i}} + xy\mathbf{\text{j}} + x^{3}y^{3}\mathbf{\text{k}}$ and *S* is the surface consisting of all faces of the tetrahedron bounded by plane $x + y + z = 1$ and the coordinate planes, with outward unit normal vector N.

406\.

Find the net outward flux of field $\mathbf{\text{F}} = \left\langle {bz - cy,cx - az,ay - bx} \right\rangle$ across any smooth closed surface in $\mathbf{\text{R}}^{3},$ where *a*, *b*, and *c* are constants.

407.

Use the divergence theorem to evaluate $\iint_{S}\left\| \textbf{R} \right\|\textbf{R} \cdot \mathbf{n}dS,$ where $\mathbf{\text{R}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}$ and *S* is sphere $x^{2} + y^{2} + z^{2} = a^{2},$ with constant $a > 0.$

408\.

Use the divergence theorem to evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = y^{2}z\mathbf{\text{i}} + y^{3}\mathbf{\text{j}} + xz\mathbf{\text{k}}$ and *S* is the boundary of the cube defined by $-1 \leq x \leq 1,-1 \leq y \leq 1,\ \text{and}\ 0 \leq z \leq 2.$

409.

Let *R* be the region defined by $x^{2} + y^{2} + z^{2} \leq 1.$ Use the divergence theorem to find ${\iiint_{R}{z^{2}dV}}.$

410\.

Let *E* be the solid bounded by the *xy*-plane and paraboloid $z = 4 - x^{2} - y^{2}$ so that *S* is the surface of the paraboloid piece together with the disk in the *xy*-plane that forms its bottom. If $\mathbf{\text{F}}(x,y,z) = \left( {xz\ \text{sin}(yz) + x^{3}} \right)\mathbf{\text{i}} + \text{cos}\left( {yz} \right)\mathbf{\text{j}} + \left( {3zy^{2} - e^{x^{2} + y^{2}}} \right)\mathbf{\text{k}},$ find $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ using the divergence theorem.

411.

Let *E* be the solid unit cube with diagonally opposite corners at the origin and (1, 1, 1), and faces parallel to the coordinate planes. Let *S* be the surface of *E*, oriented with the outward-pointing normal. Use a CAS to find $\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}$ using the divergence theorem if $\mathbf{\text{F}}(x,y,z) = 2xy\mathbf{\text{i}} + 3ye^{z}\mathbf{\text{j}} + x\ \text{sin}\ z\mathbf{\text{k}}.$

412\.

Use the divergence theorem to calculate the flux of $\mathbf{\text{F}}(x,y,z) = x^{3}\mathbf{\text{i}} + y^{3}\mathbf{\text{j}} + z^{3}\mathbf{\text{k}}$ through sphere $x^{2} + y^{2} + z^{2} = 1.$

413.

Find ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}$ and *S* is the outwardly oriented surface obtained by removing cube $\left\lbrack {1,2} \right\rbrack\ \times \ \left\lbrack {1,2} \right\rbrack\ \times \ \left\lbrack {1,2} \right\rbrack$ from cube $\left\lbrack {0,2} \right\rbrack\ \times \ \left\lbrack {0,2} \right\rbrack\ \times \ \left\lbrack {0,2} \right\rbrack.$

414\.

Consider radial vector field $\mathbf{\text{F}} = \frac{\mathbf{\text{r}}}{\left| \mathbf{\text{r}} \right|} = \frac{\left\langle {x,y,z} \right\rangle}{\left( {x^{2} + y^{2} + z^{2}} \right)^{1\text{/}2}}.$ Compute the surface integral, where *S* is the surface of a sphere of radius *a* centered at the origin.

415.

Compute the flux of water through parabolic cylinder $S:y = x^{2},$ from $0 \leq x \leq 2,0 \leq z \leq 3,$ if the velocity vector is $\mathbf{\text{F}}(x,y,z) = 3z^{2}\mathbf{\text{i}} + 6\mathbf{\text{j}} + 6xz\mathbf{\text{k}}.$

416\.

\[T\] Use a CAS to find the flux of vector field $\mathbf{\text{F}}(x,y,z) = z\mathbf{\text{i}} + z\mathbf{\text{j}} + \sqrt{x^{2} + y^{2}}\mathbf{\text{k}}$ across the portion of hyperboloid $x^{2} + y^{2} = z^{2} + 1$ between planes $z = 0$ and $z = \frac{\sqrt{3}}{3},$ oriented so the unit normal vector points away from the *z*-axis.

417.

\[T\] Use a CAS to find the flux of vector field $\mathbf{\text{F}}(x,y,z) = \left( {e^{y} + x} \right)\mathbf{\text{i}} + \left( {3\ \text{cos}(xz) - y} \right)\mathbf{\text{j}} + z\mathbf{\text{k}}$ through surface *S*, where *S* is given by $z^{2} = 4x^{2} + 4y^{2}$ from $0 \leq z \leq 4,$ oriented so the unit normal vector points downward.

418\.

\[T\] Use a CAS to compute ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + 2z\mathbf{\text{k}}$ and *S* is the boundary of a part of solid sphere $x^{2} + y^{2} + z^{2} \leq 2$ with $0 \leq z \leq 1.$

419.

Evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = bxy^{2}\mathbf{\text{i}} + bx^{2}y\mathbf{\text{j}} + \left( {x^{2} + y^{2}} \right)z^{2}\mathbf{\text{k}}$ and *S* is the boundary of the solid cylinder $x^{2} + y^{2} \leq a^{2}$ and $0 \leq z \leq b.$

420\.

\[T\] Use a CAS to calculate the flux of $\mathbf{\text{F}}(x,y,z) = \left( {x^{3} + y\ \text{sin}\ z} \right)\mathbf{\text{i}} + \left( {y^{3} + z\ \text{sin}\ x} \right)\mathbf{\text{j}} + 3z\mathbf{\text{k}}$ across surface *S*, where *S* is the boundary of the solid bounded by hemispheres $z = \sqrt{4 - x^{2} - y^{2}}$ and $z = \sqrt{1 - x^{2} - y^{2}},$ and plane $z = 0.$

421.

Use the divergence theorem to evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = xy\mathbf{\text{i}} - \frac{1}{2}y^{2}\mathbf{\text{j}} + z\mathbf{\text{k}}$ and *S* is the surface consisting of three pieces: $z = 4 - 3x^{2} - 3y^{2},1 \leq z \leq 4$ on the top; $x^{2} + y^{2} = 1,0 \leq z \leq 1$ on the sides; and $z = 0$ on the bottom.

422\.

\[T\] Use a CAS and the divergence theorem to evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = \left( {2x + y\ \text{cos}\ z} \right)\mathbf{\text{i}} + \left( {x^{2} - y} \right)\mathbf{\text{j}} + y^{2}z\mathbf{\text{k}}$ and *S* is sphere $x^{2} + y^{2} + z^{2} = 4$ orientated outward.

423.

Use the divergence theorem to evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}},$ where $\mathbf{\text{F}}(x,y,z) = x\mathbf{\text{i}} + y\mathbf{\text{j}} + z\mathbf{\text{k}}$ and *S* is the boundary of the solid enclosed by paraboloid $y = x^{2} + z^{2} - 2,$ cylinder $x^{2} + z^{2} = 1,$ and plane $x + y = 2,$ and *S* is oriented outward.

For the following exercises, Fourier’s law of heat transfer states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, $\mathbf{\text{F}} = \text{−}k\nabla T,$ which means that heat energy flows hot regions to cold regions. The constant $k > 0$ is called the *conductivity*, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region *D* is given. Use the divergence theorem to find net outward heat flux ${\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS}} = \text{−}k{\iint_{S}{\nabla T \cdot \mathbf{\text{N}}dS}}$ across the boundary *S* of *D*, where $k = 1.$

424\.

$T(x,y,z) = 100 + x + 2y + z;$ $D = \left\{ {\left( {x,y,z} \right):0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1} \right\}$

425.

$T(x,y,z) = 100 + e^{\text{−}z};$ $D = \left\{ {\left( {x,y,z} \right):0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1} \right\}$

426\.

$T(x,y,z) = 100e^{\text{−}x^{2} - y^{2} - z^{2}};$ *D* is the sphere of radius *a* centered at the origin.

Key Terms 关键术语

circulation — the tendency of a fluid to move in the direction of curve *C*. If *C* is a closed curve, then the circulation of F along *C* is line integral $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds,}$ which we also denote $\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}$

closed curve — a curve for which there exists a parameterization $\mathbf{\text{r}}(t),$ $a \leq t \leq b,$ such that $\mathbf{\text{r}}(a) = \mathbf{\text{r}}(b),$ and the curve is traversed exactly once

closed curve — a curve that begins and ends at the same point

connected region — a region in which any two points can be connected by a path with a trace contained entirely inside the region

conservative field — a vector field for which there exists a scalar function $f$ such that $\text{∇}f = \mathbf{\text{F}}$

curl — the curl of vector field $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle,$ denoted $\nabla\ \times \ \mathbf{\text{F}},$ is the “determinant” of the matrix $\left| \begin{matrix}\mathbf{\text{i}} & \mathbf{\text{j}} & \mathbf{\text{k}} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\P & Q & R\end{matrix} \right|$ and is given by the expression $\left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}};$ it measures the tendency of particles at a point to rotate about the axis that points in the direction of the curl at the point

divergence — the divergence of a vector field $\mathbf{\text{F}} = \left\langle {P,Q,R} \right\rangle,$ denoted $\nabla\ \cdot \ \mathbf{\text{F}},$ is $P_{x} + Q_{y} + R_{z};$ it measures the “outflowing-ness” of a vector field

divergence theorem — a theorem used to transform a difficult flux integral into an easier triple integral and vice versa

flux — the rate of a fluid flowing across a curve in a vector field; the flux of vector field F across plane curve *C* is line integral ${\int_{C}{\mathbf{\text{F}} \cdot \frac{\mathbf{\text{n}}(t)}{\left\| {\mathbf{\text{n}}(t)} \right\|}}}\ ds$

flux integral — another name for a surface integral of a vector field; the preferred term in physics and engineering

Fundamental Theorem for Line Integrals — the value of line integral $\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}$ depends only on the value of $f$ at the endpoints of *C*: ${\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}} = f\left( {\mathbf{\text{r}}(b} \right)) - f\left( {\mathbf{\text{r}}(a)} \right)$

Gauss’ law — if *S* is a piecewise, smooth closed surface in a vacuum and *Q* is the total stationary charge inside of *S*, then the flux of electrostatic field E across *S* is $Q\text{/}\varepsilon_{0}$

gradient field — a vector field $\mathbf{\text{F}}$ for which there exists a scalar function $f$ such that $\text{∇}f = \mathbf{\text{F}};$ in other words, a vector field that is the gradient of a function; such vector fields are also called *conservative*

Green’s theorem — relates the integral over a connected region to an integral over the boundary of the region

grid curves — curves on a surface that are parallel to grid lines in a coordinate plane

heat flow — a vector field proportional to the negative temperature gradient in an object

independence of path — a vector field F has path independence if ${\int_{C_{1}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\int_{C_{2}}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}}$ for any curves $C_{1}$ and $C_{2}$ in the domain of F with the same initial points and terminal points

inverse-square law — the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge

line integral — the integral of a function along a curve in a plane or in space

mass flux — the rate of mass flow of a fluid per unit area, measured in mass per unit time per unit area

orientation of a curve — the orientation of a curve *C* is a specified direction of *C*

orientation of a surface — if a surface has an “inner” side and an “outer” side, then an orientation is a choice of the inner or the outer side; the surface could also have “upward” and “downward” orientations

parameter domain (parameter space) — the region of the *uv* plane over which the parameters *u* and *v* vary for parameterization $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle$

parameterized surface (parametric surface) — a surface given by a description of the form $\mathbf{\text{r}}\left( {u,v} \right) = \left\langle {x\left( {u,v} \right),y\left( {u,v} \right),z\left( {u,v} \right)} \right\rangle,$ where the parameters *u* and *v* vary over a parameter domain in the *uv*-plane

piecewise smooth curve — an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves

potential function — a scalar function $f$ such that $\text{∇}f = \mathbf{\text{F}}$

radial field — a vector field in which all vectors either point directly toward or directly away from the origin; the magnitude of any vector depends only on its distance from the origin

regular parameterization — parameterization $\mathbf{\text{r}}(u,v) = \left\langle {x(u,v),y(u,v),z(u,v)} \right\rangle$ such that $\mathbf{\text{r}}_{u}\ \times \ \mathbf{\text{r}}_{v}$ is not zero for any point $(u,v)$ in the parameter domain

rotational field — a vector field in which the vector at point $\left( {x,y} \right)$ is tangent to a circle with radius $r = \sqrt{x^{2} + y^{2}};$ in a rotational field, all vectors flow either clockwise or counterclockwise, and the magnitude of a vector depends only on its distance from the origin

scalar line integral — the scalar line integral of a function $f$ along a curve *C* with respect to arc length is the integral ${\int_{C}{fds}},$ it is the integral of a scalar function $f$ along a curve in a plane or in space; such an integral is defined in terms of a Riemann sum, as is a single-variable integral

simple curve — a curve that does not cross itself

simply connected region — a region that is connected and has the property that any closed curve that lies entirely inside the region encompasses points that are entirely inside the region

Stokes’ theorem — relates the flux integral over a surface *S* to a line integral around the boundary *C* of the surface *S*

stream function — if $\mathbf{\text{F}} = \left\langle {P,Q} \right\rangle$ is a source-free vector field, then stream function *g* is a function such that $P = g_{y}$ and $Q = \text{−}g_{x}$

surface area — the area of surface *S* given by the surface integral $\int{\int_{S}{dS}}$

surface independent — flux integrals of curl vector fields are surface independent if their evaluation does not depend on the surface but only on the boundary of the surface

surface integral — an integral of a function over a surface

surface integral of a scalar-valued function — a surface integral in which the integrand is a scalar function

surface integral of a vector field — a surface integral in which the integrand is a vector field

unit vector field — a vector field in which the magnitude of every vector is 1

vector field — measured in $\mathbb{R}^{2},$ an assignment of a vector $\mathbf{\text{F}}\left( {x,y} \right)$ to each point $\left( {x,y} \right)$ of a subset $D$ of $\mathbb{R}^{2};$ in $\mathbb{R}^{3},$ an assignment of a vector $\mathbf{\text{F}}\left( {x,y,z} \right)$ to each point $\left( {x,y,z} \right)$ of a subset $D$ of $\mathbb{R}^{3}$

vector line integral — the vector line integral of vector field F along curve *C* is the integral of the dot product of F with unit tangent vector T of *C* with respect to arc length, ${\int_{C}{\mathbf{\text{F}} \cdot \mathbf{\text{T}}ds}};$ such an integral is defined in terms of a Riemann sum, similar to a single-variable integral

Key Equations 关键公式

Vector field in ℝ² F(x,y) = ⟨P(x,y),Q(x,y)⟩
or
F(x,y) = P(x,y)i + Q(x,y)j

Vector field in ℝ³ F(x,y,z) = ⟨P(x,y,z),Q(x,y,z),R(x,y,z)⟩
or
F(x,y,z) = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k

Calculating a scalar line integral $\int_{C}{f\left( {x,y,z} \right)ds = {\int_{a}^{b}{f\left( {\mathbf{\text{r}}(t)} \right)\sqrt{\left( {x^{\prime}(t)} \right)^{2} + \left( {y^{\prime}(t)} \right)^{2} + \left( {z^{\prime}(t)} \right)^{2}}dt}}}$

Calculating a vector line integral ∫_CF ⋅ dr = ∫_CF ⋅ Tds = ∫_a^bF(r(t)) ⋅ r^′(t)dt
or
${\int_{C}{Pdx + Qdy + Rdz}} = {\int_{a}^{b}{\left( {P(\mathbf{\text{r}}(t))\frac{dx}{dt} + Q(\mathbf{\text{r}}(t))\frac{dy}{dt} + R(\mathbf{\text{r}}(t))\frac{dz}{dt}} \right)dt}}$

Calculating flux ${\int_{C}{\mathbf{\text{F}} \cdot \frac{\mathbf{\text{n}}(t)}{\left\| {\mathbf{\text{n}}(t)} \right\|}}}\ ds = {\int_{a}^{b}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}(t)} \right) \cdot \mathbf{\text{n}}(t)dt}}$

Fundamental Theorem for Line Integrals ${\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}} = f\left( {\mathbf{\text{r}}(b)} \right) - f\left( {\mathbf{\text{r}}(a)} \right)$

Circulation of a conservative field over curve C that encloses a simply connected region ${\int_{C}{\text{∇}f \cdot d\mathbf{\text{r}}}} = 0$

Green’s theorem, circulation form $\int_{C}{Pdx + Qdy = {\iint_{D}{Q_{x} - P_{y}dA,}}}$ where C is the boundary of D

Green’s theorem, flux form $\int_{C}\textbf{F} \cdot \textbf{N}ds = \iint_{D}P_{x} + Q_{y}dA$

Green’s theorem, extended version ${\int_{\partial D}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{D}{Q_{x} - P_{y}dA}}$

Curl $\nabla\ \times \ \mathbf{\text{F}} = \left( {R_{y} - Q_{z}} \right)\mathbf{\text{i}} + \left( {P_{z} - R_{x}} \right)\mathbf{\text{j}} + \left( {Q_{x} - P_{y}} \right)\mathbf{\text{k}}$

Divergence $\nabla \cdot \mathbf{\text{F}} = P_{x} + Q_{y} + R_{z}$

Divergence of curl is zero $\nabla \cdot \left( {\nabla\ \times \ \mathbf{\text{F}}} \right) = 0$

Curl of a gradient is the zero vector $\nabla\ \times \ \left( {\text{∇}f} \right) = 0$

Scalar surface integral $\int{\int_{S}{f\left( {x,y,z} \right)dS = {\int{\int_{D}{f\left( {\mathbf{\text{r}}\left( {u,v} \right)} \right)\left| \left| {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right| \right|dA}}}}}$

Flux integral ${\iint_{S}{\mathbf{\text{F}} \cdot \mathbf{\text{N}}dS =}}{\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}} =}}{\iint_{D}{\mathbf{\text{F}}\left( {\mathbf{\text{r}}(u,v)} \right) \cdot \left( {\mathbf{\text{t}}_{u}\ \times \ \mathbf{\text{t}}_{v}} \right)dA}}$

Stokes’ theorem ${\int_{C}{\mathbf{\text{F}} \cdot d\mathbf{\text{r}}}} = {\iint_{S}{\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}$

Divergence theorem $\left. \iiint{}_{E}{\text{div}\ \mathbf{\text{F}}dV =} \right.{\iint\limits_{S}{\mathbf{\text{F}} \cdot d\mathbf{\text{S}}}}$

Key Concepts 关键概念

6.1 Vector Fields 6.1 向量场

• A vector field assigns a vector $\mathbf{\text{F}}(x,y)$ to each point $(x,y)$ in a subset D of $\mathbb{R}^{2}\ \text{or}\ \mathbb{R}^{3}.$ $\mathbf{\text{F}}(x,y,z)$ to each point $(x,y,z)$ in a subset D of $\mathbb{R}^{3}.$
• Vector fields can describe the distribution of vector quantities such as forces or velocities over a region of the plane or of space. They are in common use in such areas as physics, engineering, meteorology, oceanography.
• We can sketch a vector field by examining its defining equation to determine relative magnitudes in various locations and then drawing enough vectors to determine a pattern.
• A vector field $\mathbf{\text{F}}$ is called conservative if there exists a scalar function $f$ such that $\text{∇}f = \mathbf{\text{F}}.$

• 向量场对 $\mathbb{R}^{2}$（或 $\mathbb{R}^{3}$）的子集 D 中每一点 $(x,y)$ 赋予一个向量 $\mathbf{\text{F}}(x,y)$；对 $\mathbb{R}^{3}$ 的子集 D 中每一点 $(x,y,z)$ 赋予一个向量 $\mathbf{\text{F}}(x,y,z)$。
• 向量场可用于描述平面或空间区域内向量量（如力或速度）的分布，广泛应用于物理学、工程学、气象学、海洋学等领域。
• 我们可以通过考察向量场的定义方程来确定各处的相对大小，再绘制足够多的向量以辨识其整体形态，从而勾画出该向量场。
• 若存在标量函数 $f$ 使得 $\text{∇}f = \mathbf{\text{F}}$，则称向量场 $\mathbf{\text{F}}$ 为保守场。

6.2 Line Integrals 6.2 线积分

• Line integrals generalize the notion of a single-variable integral to higher dimensions. The domain of integration in a single-variable integral is a line segment along the x-axis, but the domain of integration in a line integral is a curve in a plane or in space.
• If C is a curve, then the length of C is ${\int_{C}{ds}}.$
• There are two kinds of line integral: scalar line integrals and vector line integrals. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field.
• Scalar line integrals can be calculated using Equation 6.8; vector line integrals can be calculated using Equation 6.9.
• Two key concepts expressed in terms of line integrals are flux and circulation. Flux measures the rate that a field crosses a given line; circulation measures the tendency of a field to move in the same direction as a given closed curve.

• 线积分将单变量积分的概念推广到了高维。单变量积分的积分域是 x 轴上的一段线段，而线积分的积分域是平面或空间中的一条曲线。
• 若 C 为一条曲线，则其弧长等于 ${\int_{C}{ds}}$。
• 线积分有两种：标量线积分与向量线积分。标量线积分可用于计算金属丝的质量；向量线积分可用于计算粒子穿越场时所做的功。
• 标量线积分可用公式 6.8 计算；向量线积分可用公式 6.9 计算。
• 用线积分表达的两个核心概念是通量与环流量。通量度量场穿过给定曲线的速率；环流量度量场沿给定闭曲线同方向移动的趋势。

6.3 Conservative Vector Fields 6.3 保守向量场

• The theorems in this section require curves that are closed, simple, or both, and regions that are connected or simply connected.
• The line integral of a conservative vector field can be calculated using the Fundamental Theorem for Line Integrals. This theorem is a generalization of the Fundamental Theorem of Calculus in higher dimensions. Using this theorem usually makes the calculation of the line integral easier.
• Conservative fields are independent of path. The line integral of a conservative field depends only on the value of the potential function at the endpoints of the domain curve.
• Given vector field F, we can test whether F is conservative by using the cross-partial property. If F has the cross-partial property and the domain is simply connected, then F is conservative (and thus has a potential function). If F is conservative, we can find a potential function by using the Problem-Solving Strategy.
• The circulation of a conservative vector field on a simply connected domain over a closed curve is zero.

• 本节中的定理要求曲线为闭曲线、简单曲线或二者兼具，并要求区域为连通区域或单连通区域。
• 保守向量场的线积分可用线积分基本定理计算。该定理是微积分基本定理在高维的推广，通常能使线积分的计算更为简便。
• 保守场与路径无关。保守场的线积分仅取决于势函数在积分曲线端点处的取值。
• 给定向量场 F，可利用交叉偏导性质检验 F 是否为保守场。若 F 满足交叉偏导性质且定义域单连通，则 F 为保守场（从而存在势函数）。若 F 为保守场，则可借助问题求解策略求出势函数。
• 在单连通区域上，保守向量场沿闭曲线的环流量为零。

6.4 Green’s Theorem 6.4 格林定理

• Green’s theorem relates the integral over a connected region to an integral over the boundary of the region. Green’s theorem is a version of the Fundamental Theorem of Calculus in one higher dimension.
• Green’s Theorem comes in two forms: a circulation form and a flux form. In the circulation form, the integrand is $\mathbf{\text{F}} \cdot \mathbf{\text{T}}.$ In the flux form, the integrand is $\mathbf{\text{F}} \cdot \mathbf{\text{N}}.$
• Green’s theorem can be used to transform a difficult line integral into an easier double integral, or to transform a difficult double integral into an easier line integral.
• A vector field is source free if it has a stream function. The flux of a source-free vector field across a closed curve is zero, just as the circulation of a conservative vector field across a closed curve is zero.

• 格林定理将一个连通区域上的积分与该区域边界上的积分联系起来，它是微积分基本定理在高一维空间中的形式。
• 格林定理有两种形式：环流量形式与通量形式。在环流量形式中，被积函数为 $\mathbf{\text{F}} \cdot \mathbf{\text{T}}$；在通量形式中，被积函数为 $\mathbf{\text{F}} \cdot \mathbf{\text{N}}$。
• 格林定理可用于将复杂的线积分转化为较简单的二重积分，或将复杂的二重积分转化为较简单的线积分。
• 若向量场存在流函数，则称其为无源场。无源向量场穿过闭曲线的通量为零，正如保守向量场穿过闭曲线的环流量为零。

6.5 Divergence and Curl 6.5 散度与旋度

• The divergence of a vector field is a scalar function. Divergence measures the “outflowing-ness” of a vector field. If v is the velocity field of a fluid, then the divergence of v at a point is the outflow of the fluid less the inflow at the point.
• The curl of a vector field is a vector field. The curl of a vector field at point P measures the tendency of particles at P to rotate about the axis that points in the direction of the curl at P.
• A vector field with a simply connected domain is conservative if and only if its curl is zero.

• 向量场的散度是一个标量函数，它度量向量场的"向外发散程度"。若 v 为流体的速度场，则 v 在一点处的散度等于该点处流体的流出量减去流入量。
• 向量场的旋度仍是一个向量场。向量场在点 P 处的旋度度量了该点处粒子绕旋度方向的轴旋转的趋势。
• 定义域为单连通的向量场是保守场，当且仅当其旋度为零。

6.6 Surface Integrals 6.6 曲面积分

• Surfaces can be parameterized, just as curves can be parameterized. In general, surfaces must be parameterized with two parameters.
• Surfaces can sometimes be oriented, just as curves can be oriented. Some surfaces, such as a Möbius strip, cannot be oriented.
• A surface integral is like a line integral in one higher dimension. The domain of integration of a surface integral is a surface in a plane or space, rather than a curve in a plane or space.
• The integrand of a surface integral can be a scalar function or a vector field. To calculate a surface integral with an integrand that is a function, use Equation 6.19. To calculate a surface integral with an integrand that is a vector field, use Equation 6.20.
• If S is a surface, then the area of S is ${\int{\int_{S}{dS}}}.$

• 曲面可以像曲线一样被参数化。一般而言，曲面必须用两个参数来参数化。
• 曲面有时可以定向，正如曲线可以定向。但某些曲面（如莫比乌斯带）无法定向。
• 曲面积分类似于高一维空间中的线积分。曲面积分的积分域是平面或空间中的一张曲面，而非平面或空间中的一条曲线。
• 曲面积分的被积函数可以是标量函数或向量场。当被积函数为标量函数时，用公式 6.19 计算曲面积分；当被积函数为向量场时，用公式 6.20 计算。
• 若 S 为一张曲面，则其面积等于 ${\int{\int_{S}{dS}}}$。

6.7 Stokes’ Theorem 6.7 斯托克斯定理

• Stokes’ theorem relates a flux integral over a surface to a line integral around the boundary of the surface. Stokes’ theorem is a higher dimensional version of Green’s theorem, and therefore is another version of the Fundamental Theorem of Calculus in higher dimensions.
• Stokes’ theorem can be used to transform a difficult surface integral into an easier line integral, or a difficult line integral into an easier surface integral.
• Through Stokes’ theorem, line integrals can be evaluated using the simplest surface with boundary C.
• Faraday’s law relates the curl of an electric field to the rate of change of the corresponding magnetic field. Stokes’ theorem can be used to derive Faraday’s law.

• 斯托克斯定理将曲面上的通量积分与该曲面边界上的线积分联系起来。它是格林定理在高维的推广，因而也是微积分基本定理在高维的又一形式。
• 斯托克斯定理可用于将复杂的曲面积分转化为较简单的线积分，或将复杂的线积分转化为较简单的曲面积分。
• 借助斯托克斯定理，可用以 C 为边界的最简单曲面来计算线积分。
• 法拉第定律将电场的旋度与相应磁场的变化率联系起来。斯托克斯定理可用于推导法拉第定律。

6.8 The Divergence Theorem 6.8 散度定理

• The divergence theorem relates a surface integral across closed surface *S* to a triple integral over the solid enclosed by *S*. The divergence theorem is a higher dimensional version of the flux form of Green’s theorem, and is therefore a higher dimensional version of the Fundamental Theorem of Calculus.
• The divergence theorem can be used to transform a difficult flux integral into an easier triple integral and vice versa.
• The divergence theorem can be used to derive Gauss’ law, a fundamental law in electrostatics.

• 散度定理将闭曲面 *S* 上的曲面积分与 *S* 所围立体上的三重积分联系起来。散度定理是格林定理通量形式的高维版本，因而也是微积分基本定理的高维版本。
• 散度定理可把一个难算的通量积分转化为一个更易算的三重积分，反之亦然。
• 散度定理可用于推导高斯定律，这是静电学中的一条基本定律。

Review Exercises 复习题

*True or False?* Justify your answer with a proof or a counterexample.

427. Vector field $\mathbf{\text{F}}(x,y) = x^{2}y\mathbf{\text{i}} + y^{2}x\mathbf{\text{j}}$ is conservative.

428. For vector field $\mathbf{\text{F}}(x,y) = P(x,y)\mathbf{\text{i}} + Q(x,y)\mathbf{\text{j}},$ if $P_{y}(x,y) = Q_{x}(x,y)$ in open region $D,$ then ${\int_{C}{Pdx + Qdy = 0}}.$

429. The divergence of a vector field is a vector field.

430. If $\text{curl}\ \mathbf{\text{F}} = 0,$ then $\mathbf{\text{F}}$ is a conservative vector field.

Draw the following vector fields.

431. $\mathbf{\text{F}}\left( {x,y} \right) = \frac{1}{2}\mathbf{\text{i}} + 2x\mathbf{\text{j}}$

432. $\textbf{F}(x,y) = \frac{\mathbf{\left. <{\mathit{y},3\mathit{x}}> \right.}}{\sqrt{x^{2} + y^{2}}}$

Are the following the vector fields conservative? If so, find the potential function $f$ such that $\mathbf{\text{F}} = \nabla f.$

433. $\mathbf{\text{F}}\left( {x,y} \right) = y\mathbf{\text{i}} + \left( {x - 2e^{y}} \right)\mathbf{\text{j}}$

434. $\mathbf{\text{F}}\left( {x,y} \right) = \left( {6xy} \right)\mathbf{\text{i}} + \left( {3x^{2} - ye^{y}} \right)\mathbf{\text{j}}$

435. $\mathbf{\text{F}}\left( {x,y,z} \right) = \left( {2xy + z^{2}} \right)\mathbf{\text{i}} + \left( {x^{2} + 2yz} \right)\mathbf{\text{j}} + \left( {2xz + y^{2}} \right)\mathbf{\text{k}}$

436. $\mathbf{\text{F}}(x,y,z) = \left( {e^{x}y} \right)\mathbf{\text{i}} + \left( {e^{x} + z} \right)\mathbf{\text{j}} + \left( {e^{x} + y^{2}} \right)\mathbf{\text{k}}$

Evaluate the following integrals.

437. $\int\limits_{C}{x^{2}dy + \left( {2x - 3xy} \right)dx,}$ along $C:y = \frac{1}{2}x$ from (0, 0) to (4, 2)

438. $\int\limits_{C}{ydx + xy^{2}dy,}$ where $C:x = \sqrt{t},y = t - 1,0 \leq t \leq 1$

439. $\left. \iint{}_{S}{xy^{2}dS} \right.,$ where *S* is surface $z = x^{2} - y,0 \leq x \leq 1,0 \leq y \leq 4$

Find the divergence and curl for the following vector fields.

440. $\mathbf{\text{F}}(x,y,z) = 3xyz\mathbf{\text{i}} + xye^{z}\mathbf{\text{j}} - 3xy\mathbf{\text{k}}$

441. $\mathbf{\text{F}}(x,y,z) = e^{x}\mathbf{\text{i}} + e^{xy}\mathbf{\text{j}} + e^{xyz}\mathbf{\text{k}}$

Use Green’s theorem to evaluate the following integrals.

442. $\int\limits_{C}{3xydx + 2xy^{2}dy,}$ where *C* is a square with vertices (0, 0), (0, 2), (2, 2) and (2, 0) oriented counterclockwise.

443. $\left. \int{}_{C}{3ydx + \left( {x + e^{y}} \right)dy,} \right.$ where *C* is a circle centered at the origin with radius 3

Use Stokes’ theorem to evaluate ${\iint_{S}\text{curl}\ \mathbf{\text{F}} \cdot d\mathbf{S}}.$

444. $\mathbf{\text{F}}(x,y,z) = y\mathbf{\text{i}} - x\mathbf{\text{j}} + z\mathbf{\text{k}},$ where $S$ is the upper half of the unit sphere

445. $\mathbf{\text{F}}(x,y,z) = y\mathbf{\text{i}} + xyz\mathbf{\text{j}} - 2zx\mathbf{\text{k}},$ where $S$ is the upward-facing paraboloid $z = x^{2} + y^{2}$ lying in cylinder $x^{2} + y^{2} = 1$

Use the divergence theorem to evaluate ${\iint_{S}{\mathbf{\text{F}} \cdot d\mathbf{S}}}.$

446. $\mathbf{\text{F}}(x,y,z) = \left( {x^{3}y} \right)\mathbf{\text{i}} + \left( {3y - e^{x}} \right)\mathbf{\text{j}} + \left( {z + x} \right)\mathbf{\text{k}},$ over cube $S$ defined by $-1 \leq x \leq 1,$ $0 \leq y \leq 2,$ $0 \leq z \leq 2$

447. $\mathbf{\text{F}}(x,y,z) = \left( {2xy} \right)\mathbf{\text{i}} + \left( {\text{−}y^{2}} \right)\mathbf{\text{j}} + \left( {2z^{3}} \right)\mathbf{\text{k}},$ where $S$ is bounded by paraboloid $z = x^{2} + y^{2}$ and plane $z = 2$

448. Find the amount of work performed by a 50-kg woman ascending a helical staircase with radius 2 m and height 100 m. The woman completes five revolutions during the climb. Use 9.8 m/s² as the acceleration due to gravity.

449. Find the total mass of a thin wire in the shape of an upper semicircle with radius $\sqrt{2,}$ and a density function of $\rho\left( {x,y} \right) = y + x^{2}.$

450. Find the total mass of a thin sheet in the shape of a hemisphere with radius 2 for $z \geq 0$ with a density function $\rho\left( {x,y,z} \right) = x + y + z.$

451. Use the divergence theorem to compute the value of the flux integral over the unit sphere with $\mathbf{\text{F}}(x,y,z) = 3z\mathbf{\text{i}} + 2y\mathbf{\text{j}} + 2x\mathbf{\text{k}}.$