2 Applications of Integration 积分的应用

本页译自 OpenStax《Calculus Volume 2》第 2 章 Applications of Integration（积分的应用）：2.1–2.9 九节 + Key Terms/Key Equations/Key Concepts/Review Exercises 全译，段段对照。公式经本地 MathJax 渲染，自定义宏已注入。

2.1 Areas between Curves 2.1 曲线之间的面积

2.1.1 Determine the area of a region between two curves by integrating with respect to the independent variable.
2.1.2 Find the area of a compound region.
2.1.3 Determine the area of a region between two curves by integrating with respect to the dependent variable.

2.1.1 通过对自变量积分，确定两曲线之间区域的面积。
2.1.2 求复合区域的面积。
2.1.3 通过对因变量积分，确定两曲线之间区域的面积。

In Introduction to Integration, we developed the concept of the definite integral to calculate the area below a curve on a given interval. In this section, we expand that idea to calculate the area of more complex regions. We start by finding the area between two curves that are functions of $x,$ beginning with the simple case in which one function value is always greater than the other. We then look at cases when the graphs of the functions cross. Last, we consider how to calculate the area between two curves that are functions of $y.$

在《积分导论》中，我们发展了定积分的概念，用以计算给定区间上曲线下方的面积。本节中，我们将这一思想推广，用以计算更复杂区域的面积。我们首先求两曲线（均为 $x$ 的函数）之间的面积，从其中一个函数值始终大于另一个的简单情形开始。然后我们考察函数图像相交的情形。最后，我们考虑如何求两曲线（均为 $y$ 的函数）之间的面积。

Area of a Region between Two Curves 两曲线之间区域的面积

Let $f(x)$ and $g(x)$ be continuous functions over an interval $\left\lbrack {a,b} \right\rbrack$ such that $f(x) \geq g(x)$ on $\left\lbrack {a,b} \right\rbrack.$ We want to find the area between the graphs of the functions, as shown in the following figure.

设 $f(x)$ 与 $g(x)$ 是区间 $\left\lbrack {a,b} \right\rbrack$ 上的连续函数，且在 $\left\lbrack {a,b} \right\rbrack$ 上满足 $f(x) \geq g(x)$。我们想求这两个函数图像之间的面积，如下图所示。

As we did before, we are going to partition the interval on the $x\text{-axis}$ and approximate the area between the graphs of the functions with rectangles. So, for $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of $\left\lbrack {a,b} \right\rbrack.$ Then, for $i = 1,2\text{,…},n,$ choose a point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack,$ and on each interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ construct a rectangle that extends vertically from $g(x_{i}^{*})$ to $f(x_{i}^{*}).$ Figure 2.3(a) shows the rectangles when $x_{i}^{*}$ is selected to be the left endpoint of the interval and $n = 10.$ Figure 2.3(b) shows a representative rectangle in detail.

如同之前所做，我们将 $x\text{-轴}$ 上的区间进行分割，并用矩形逼近两函数图像之间的面积。于是，对于 $i = 0,1,2\text{,…},n,$ 令 $P = \left\{ x_{i} \right\}$ 为 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割。接着，对于 $i = 1,2\text{,…},n,$ 选取一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack,$ 并在每个区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 上构造一个竖直方向上从 $g(x_{i}^{*})$ 延伸到 $f(x_{i}^{*})$ 的矩形。图 2.3(a) 展示了当 $x_{i}^{*}$ 取为区间左端点且 $n = 10$ 时的矩形。图 2.3(b) 详细地展示了一个代表性的矩形。

Use this calculator to learn more about the areas between two curves.

使用这个计算器，进一步了解两曲线之间的面积。

The height of each individual rectangle is $f(x_{i}^{*}) - g(x_{i}^{*})$ and the width of each rectangle is $\text{Δ}x.$ Adding the areas of all the rectangles, we see that the area between the curves is approximated by

每个矩形的高度为 $f(x_{i}^{*}) - g(x_{i}^{*})$，每个矩形的宽度为 $\text{Δ}x$。把所有矩形的面积相加，我们看到两曲线之间的面积近似为

$$A \approx \sum\limits_{i = 1}^{n}\left\lbrack {f(x_{i}^{*}) - g(x_{i}^{*})} \right\rbrack\text{Δ}x.$$

This is a Riemann sum, so we take the limit as $n\rightarrow\infty$ and we get

这是一个黎曼和，因此取极限 $n\rightarrow\infty$，我们得到

$$A = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\left\lbrack {f(x_{i}^{*}) - g(x_{i}^{*})} \right\rbrack\text{Δ}x = {\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack}dx.$$

These findings are summarized in the following theorem.

这些结论总结于下列定理中。

Finding the Area between Two Curves 求两曲线之间的面积

Let $f(x)$ and $g(x)$ be continuous functions such that $f(x) \geq g(x)$ over an interval $\left\lbrack {a,b} \right\rbrack.$ Let $R$ denote the region bounded above by the graph of $f(x),$ below by the graph of $g(x),$ and on the left and right by the lines $x = a$ and $x = b,$ respectively. Then, the area of $R$ is given by

设 $f(x)$ 与 $g(x)$ 为连续函数，且在区间 $\left\lbrack {a,b} \right\rbrack$ 上满足 $f(x) \geq g(x)$。令 $R$ 表示由上方的 $f(x)$ 图像、下方的 $g(x)$ 图像、以及左右两侧的直线 $x = a$ 与 $x = b$ 所围成的区域。则 $R$ 的面积由下式给出

$$A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx.$$ (2.1)

We apply this theorem in the following example.

我们在下面的例子中应用这一定理。

Finding the Area of a Region between Two Curves 1 求两曲线之间区域的面积 1

If *R* is the region bounded above by the graph of the function $f(x) = x + 4$ and below by the graph of the function $g(x) = 3 - \frac{x}{2}$ over the interval $\left\lbrack {1,4} \right\rbrack,$ find the area of region $R.$

若 *R* 是区间 $\left\lbrack {1,4} \right\rbrack$ 上由函数 $f(x) = x + 4$ 的图像在上方、函数 $g(x) = 3 - \frac{x}{2}$ 的图像在下方所围成的区域，求区域 $R$ 的面积。

Solution 解答

The region is depicted in the following figure.

该区域如下图所示。

We have

我们有

$$\begin{array}{cl} A & {= \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx} \\ & {= {\int_{1}^{4}{\left\lbrack {\left( {x + 4} \right) - \left( {3 - \frac{x}{2}} \right)} \right\rbrack dx}} = {\int_{1}^{4}{\left\lbrack {\frac{3x}{2} + 1} \right\rbrack dx}}} \\ & {= \left. \left\lbrack {\frac{3x^{2}}{4} + x} \right\rbrack\ \right|_{1}^{4} = \left( {16 - \frac{7}{4}} \right) = \frac{57}{4}.} \end{array}$$

The area of the region is $\frac{57}{4}\ \text{units}^{2}.$

该区域的面积为 $\frac{57}{4}\ \text{units}^{2}.$

If $R$ is the region bounded by the graphs of the functions $f(x) = \frac{x}{2} + 5$ and $g(x) = x + \frac{1}{2}$ over the interval $\left\lbrack {1,5} \right\rbrack,$ find the area of region $R.$

若 $R$ 是区间 $\left\lbrack {1,5} \right\rbrack$ 上由函数 $f(x) = \frac{x}{2} + 5$ 与 $g(x) = x + \frac{1}{2}$ 的图像所围成的区域，求区域 $R$ 的面积。

In Example 2.1, we defined the interval of interest as part of the problem statement. Quite often, though, we want to define our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.

在示例 2.1 中，我们已将所关心的区间作为问题陈述的一部分给定。然而，很多时候我们更希望根据两函数图像的交点来界定所关心的区间。下面的例子说明了这一点。

Finding the Area of a Region between Two Curves 2 求两曲线之间区域的面积 2

If $R$ is the region bounded above by the graph of the function $f(x) = 9 - \left( {x\text{/}2} \right)^{2}$ and below by the graph of the function $g(x) = 6 - x,$ find the area of region $R.$

若 $R$ 是由函数 $f(x) = 9 - \left( {x\text{/}2} \right)^{2}$ 的图像在上方、函数 $g(x) = 6 - x$ 的图像在下方所围成的区域，求区域 $R$ 的面积。

Solution 解答

The region is depicted in the following figure.

该区域如下图所示。

We first need to compute where the graphs of the functions intersect. Setting $f(x) = g(x),$ we get

我们首先需要计算两函数图像的交点在哪里。令 $f(x) = g(x),$ 我们得到

$$\begin{array}{rll} {f(x)} & = & {g(x)} \\ & & \\ {9 - \left( \frac{x}{2} \right)^{2}} & = & {6 - x} \\ {9 - \frac{x^{2}}{4}} & = & {6 - x} \\ {36 - x^{2}} & = & {24 - 4x} \\ {x^{2} - 4x - 12} & = & 0 \\ {\left( {x - 6} \right)\left( {x + 2} \right)} & = & 0. \end{array}$$

The graphs of the functions intersect when $x = 6$ or $x = -2,$ so we want to integrate from $-2$ to $6.$ Since $f(x) \geq g(x)$ for $-2 \leq x \leq 6,$ we obtain

两函数图像在 $x = 6$ 或 $x = -2$ 处相交，因此我们要从 $-2$ 到 $6$ 进行积分。由于在 $-2 \leq x \leq 6$ 上 $f(x) \geq g(x)$，我们得到

$$\begin{array}{cl} A & {= \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx} \\ & {= {\int_{-2}^{6}{\left\lbrack {9 - \left( \frac{x}{2} \right)^{2} - \left( {6 - x} \right)} \right\rbrack dx}} = {\int_{-2}^{6}{\left\lbrack {3 - \frac{x^{2}}{4} + x} \right\rbrack dx}}} \\ & {= \left. \left\lbrack {3x - \frac{x^{3}}{12} + \frac{x^{2}}{2}} \right\rbrack\ \right|_{-2}^{6} = \frac{64}{3}.} \end{array}$$

The area of the region is $64\text{/}3$ units².

该区域的面积为 $64\text{/}3$ 单位²。

If *R* is the region bounded above by the graph of the function $f(x) = x$ and below by the graph of the function $g(x) = x^{4},$ find the area of region $R.$

若 *R* 是由函数 $f(x) = x$ 的图像在上方、函数 $g(x) = x^{4}$ 的图像在下方所围成的区域，求区域 $R$ 的面积。

Areas of Compound Regions 复合区域的面积

So far, we have required $f(x) \geq g(x)$ over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.

到目前为止，我们都要求在整个所关心的区间上 $f(x) \geq g(x)$，但如果我们要考察由彼此相交的函数图像所围成的区域，该怎么办？此时，我们通过使用绝对值函数来修正刚才发展起来的方法。

Finding the Area of a Region between Curves That Cross 求相交曲线之间区域的面积

Let $f(x)$ and $g(x)$ be continuous functions over an interval $\left\lbrack {a,b} \right\rbrack.$ Let $R$ denote the region between the graphs of $f(x)$ and $g(x),$ and be bounded on the left and right by the lines $x = a$ and $x = b,$ respectively. Then, the area of $R$ is given by

设 $f(x)$ 与 $g(x)$ 是区间 $\left\lbrack {a,b} \right\rbrack$ 上的连续函数。令 $R$ 表示 $f(x)$ 与 $g(x)$ 图像之间的区域，其左右两侧分别由直线 $x = a$ 与 $x = b$ 所界定。则 $R$ 的面积由下式给出

$$A = \int_{a}^{b}\left| {f(x) - g(x)} \right|dx.$$

In practice, applying this theorem requires us to break up the interval $\left\lbrack {a,b} \right\rbrack$ and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.

在实际应用中，应用这一定理需要我们将区间 $\left\lbrack {a,b} \right\rbrack$ 分段，并求若干个积分，具体取决于在该区间的哪一部分上哪个函数值更大。我们在下面的例子中研究这一过程。

Finding the Area of a Region Bounded by Functions That Cross 求由相交函数所围区域的面积

If *R* is the region between the graphs of the functions $f(x) = \text{sin}\ x$ and $g(x) = \text{cos}\ x$ over the interval $\left\lbrack {0,\pi} \right\rbrack,$ find the area of region $R.$

若 *R* 是区间 $\left\lbrack {0,\pi} \right\rbrack$ 上函数 $f(x) = \text{sin}\ x$ 与 $g(x) = \text{cos}\ x$ 图像之间的区域，求区域 $R$ 的面积。

Solution 解答

The region is depicted in the following figure.

该区域如下图所示。

The graphs of the functions intersect at $x = {\pi\text{/}4}.$ For $x \in \left\lbrack {0,{\pi\text{/}4}} \right\rbrack,$ $\text{cos}\ x \geq \text{sin}\ x,$ so

两函数图像在 $x = {\pi\text{/}4}$ 处相交。对于 $x \in \left\lbrack {0,{\pi\text{/}4}} \right\rbrack,$ 有 $\text{cos}\ x \geq \text{sin}\ x,$ 因此

$$\left| {f(x) - g(x)} \right| = \left| {\text{sin}\ x - \text{cos}\ x} \right| = \text{cos}\ x - \text{sin}\ x.$$

On the other hand, for $x \in \left\lbrack {{\pi\text{/}4},\pi} \right\rbrack,$ $\text{sin}\ x \geq \text{cos}\ x,$ so

另一方面，对于 $x \in \left\lbrack {{\pi\text{/}4},\pi} \right\rbrack,$ 有 $\text{sin}\ x \geq \text{cos}\ x,$ 因此

$$\left| {f(x) - g(x)} \right| = \left| {\text{sin}\ x - \text{cos}\ x} \right| = \text{sin}\ x - \text{cos}\ x.$$

Then

于是

$$\begin{array}{cl} A & {= \int_{a}^{b}\left| {f(x) - g(x)} \right|dx} \\ & {= \int_{0}^{\pi}\left| {\text{sin}\ x - \text{cos}\ x} \right|dx = \int_{0}^{\pi\text{/}4}\left( {\text{cos}\ x - \text{sin}\ x} \right)dx + \int_{\pi\text{/}4}^{\pi}\left( {\text{sin}\ x - \text{cos}\ x} \right)dx} \\ & {= \left. \left\lbrack {\text{sin}\ x + \text{cos}\ x} \right\rbrack\ \right|_{0}^{\pi\text{/}4} + \left. \left\lbrack {\text{−}\text{cos}\ x - \text{sin}\ x} \right\rbrack\ \right|_{\pi\text{/}4}^{\pi}} \\ & {= \left( {\sqrt{2} - 1} \right) + \left( {1 + \sqrt{2}} \right) = 2\sqrt{2}.} \end{array}$$

The area of the region is $2\sqrt{2}$ units².

该区域的面积为 $2\sqrt{2}$ 单位²。

If *R* is the region between the graphs of the functions $f(x) = \text{sin}\ x$ and $g(x) = \text{cos}\ x$ over the interval $\left\lbrack {{\pi\text{/}2},2\pi} \right\rbrack,$ find the area of region $R.$

若 *R* 是区间 $\left\lbrack {{\pi\text{/}2},2\pi} \right\rbrack$ 上函数 $f(x) = \text{sin}\ x$ 与 $g(x) = \text{cos}\ x$ 图像之间的区域，求区域 $R$ 的面积。

Finding the Area of a Complex Region 求复杂区域的面积

Consider the region depicted in Figure 2.7. Find the area of $R.$

考察图 2.7 所示的区域。求 $R$ 的面积。

Solution 解答

As with Example 2.3, we need to divide the interval into two pieces. The graphs of the functions intersect at $x = 1$ (set $f(x) = g(x)$ and solve for *x*), so we evaluate two separate integrals: one over the interval $\left\lbrack {0,1} \right\rbrack$ and one over the interval $\left\lbrack {1,2} \right\rbrack.$

与示例 2.3 类似，我们需要将区间分成两段。两函数图像在 $x = 1$ 处相交（令 $f(x) = g(x)$ 并解 *x*），因此我们计算两个独立的积分：一个在区间 $\left\lbrack {0,1} \right\rbrack$ 上，另一个在区间 $\left\lbrack {1,2} \right\rbrack$ 上。

Over the interval $\left\lbrack {0,1} \right\rbrack,$ the region is bounded above by $f(x) = x^{2}$ and below by the *x*-axis, so we have

在区间 $\left\lbrack {0,1} \right\rbrack$ 上，区域的上边界为 $f(x) = x^{2}$，下边界为 *x* 轴，因此我们有

$$A_{1} = \int_{0}^{1}x^{2}dx = \left. \frac{x^{3}}{3}\ \right|_{0}^{1} = \frac{1}{3}.$$

Over the interval $\left\lbrack {1,2} \right\rbrack,$ the region is bounded above by $g(x) = 2 - x$ and below by the $x\text{-axis,}$ so we have

在区间 $\left\lbrack {1,2} \right\rbrack$ 上，区域的上边界为 $g(x) = 2 - x$，下边界为 $x\text{-轴,}$ 因此我们有

$$A_{2} = \int_{1}^{2}\left( {2 - x} \right)dx = \left. \left\lbrack {2x - \frac{x^{2}}{2}} \right\rbrack\ \right|_{1}^{2} = \frac{1}{2}.$$

Adding these areas together, we obtain

将这些面积相加，我们得到

$$A = A_{1} + A_{2} = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}.$$

The area of the region is $5\text{/}6$ units².

该区域的面积为 $5\text{/}6$ 单位²。

Consider the region depicted in the following figure. Find the area of $R.$

考察下图所示的区域。求 $R$ 的面积。

Regions Defined with Respect to y 关于 y 定义的区域

In Example 2.4, we had to evaluate two separate integrals to calculate the area of the region. However, there is another approach that requires only one integral. What if we treat the curves as functions of $y,$ instead of as functions of $x?$ Review Figure 2.7. Note that the left graph, shown in red, is represented by the function $y = f(x) = x^{2}.$ We could just as easily solve this for $x$ and represent the curve by the function $x = v(y) = \sqrt{y}.$ (Note that $x = \text{−}\sqrt{y}$ is also a valid representation of the function $y = f(x) = x^{2}$ as a function of $y.$ However, based on the graph, it is clear we are interested in the positive square root.) Similarly, the right graph is represented by the function $y = g(x) = 2 - x,$ but could just as easily be represented by the function $x = u(y) = 2 - y.$ When the graphs are represented as functions of $y,$ we see the region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if we integrate with respect to $y,$ we need to evaluate one integral only. Let’s develop a formula for this type of integration.

在示例 2.4 中，我们不得不计算两个独立的积分来求出该区域的面积。然而，还有另一种只需要一个积分的方法。如果我们把曲线视为 $y$ 的函数，而非 $x$ 的函数，会怎样？回顾图 2.7。注意，左侧（红色所示）的图像由函数 $y = f(x) = x^{2}$ 表示。我们同样可以就此解出 $x$，并用函数 $x = v(y) = \sqrt{y}$ 来表示这条曲线。（注意，$x = \text{−}\sqrt{y}$ 也是函数 $y = f(x) = x^{2}$ 作为 $y$ 的函数的有效表示。不过，从图像上看，显然我们关心的是正的平方根。）类似地，右侧的图像由函数 $y = g(x) = 2 - x$ 表示，但同样可以容易地用函数 $x = u(y) = 2 - y$ 来表示。当图像被表示为 $y$ 的函数时，我们看到该区域左侧由其中一个函数的图像界定，右侧由另一个函数的图像界定。因此，如果我们对 $y$ 积分，就只需要计算一个积分。让我们为这种积分推导一个公式。

Let $u(y)$ and $v(y)$ be continuous functions over an interval $\left\lbrack {c,d} \right\rbrack$ such that $u(y) \geq v(y)$ for all $y \in \left\lbrack {c,d} \right\rbrack.$ We want to find the area between the graphs of the functions, as shown in the following figure.

设 $u(y)$ 与 $v(y)$ 是区间 $\left\lbrack {c,d} \right\rbrack$ 上的连续函数，且对所有 $y \in \left\lbrack {c,d} \right\rbrack$ 满足 $u(y) \geq v(y)$。我们想求这两个函数图像之间的面积，如下图所示。

This time, we are going to partition the interval on the $y\text{-axis}$ and use horizontal rectangles to approximate the area between the functions. So, for $i = 0,1,2\text{,…},n,$ let $Q = \left\{ y_{i} \right\}$ be a regular partition of $\left\lbrack {c,d} \right\rbrack.$ Then, for $i = 1,2\text{,…},n,$ choose a point $y_{i}^{*} \in \left\lbrack {y_{i - 1},y_{i}} \right\rbrack,$ then over each interval $\left\lbrack {y_{i - 1},y_{i}} \right\rbrack$ construct a rectangle that extends horizontally from $v\left( y_{i}^{*} \right)$ to $u\left( y_{i}^{*} \right).$ Figure 2.9(a) shows the rectangles when $y_{i}^{*}$ is selected to be the lower endpoint of the interval and $n = 10.$ Figure 2.9(b) shows a representative rectangle in detail.

这一次，我们将 $y\text{-轴}$ 上的区间进行分割，并用水平矩形来逼近两函数之间的面积。于是，对于 $i = 0,1,2\text{,…},n,$ 令 $Q = \left\{ y_{i} \right\}$ 为 $\left\lbrack {c,d} \right\rbrack$ 的一个正则分割。接着，对于 $i = 1,2\text{,…},n,$ 选取一点 $y_{i}^{*} \in \left\lbrack {y_{i - 1},y_{i}} \right\rbrack,$ 然后在每个区间 $\left\lbrack {y_{i - 1},y_{i}} \right\rbrack$ 上构造一个水平方向上从 $v\left( y_{i}^{*} \right)$ 延伸到 $u\left( y_{i}^{*} \right)$ 的矩形。图 2.9(a) 展示了当 $y_{i}^{*}$ 取为区间下端点且 $n = 10$ 时的矩形。图 2.9(b) 详细地展示了一个代表性的矩形。

The height of each individual rectangle is $\text{Δ}y$ and the width of each rectangle is $u\left( y_{i}^{*} \right) - v\left( y_{i}^{*} \right).$ Therefore, the area between the curves is approximately

每个矩形的高度为 $\text{Δ}y$，每个矩形的宽度为 $u\left( y_{i}^{*} \right) - v\left( y_{i}^{*} \right)$。因此，两曲线之间的面积近似为

$$A \approx \sum\limits_{i = 1}^{n}\left\lbrack {u\left( y_{i}^{*} \right) - v\left( y_{i}^{*} \right)} \right\rbrack\text{Δ}y.$$

This is a Riemann sum, so we take the limit as $n\rightarrow\infty,$ obtaining

这是一个黎曼和，因此取极限 $n\rightarrow\infty,$ 得到

$$A = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\left\lbrack {u\left( y_{i}^{*} \right) - v\left( y_{i}^{*} \right)} \right\rbrack\text{Δ}y = {\int_{c}^{d}{\left\lbrack {u(y) - v(y)} \right\rbrack dy}}.$$

These findings are summarized in the following theorem.

这些结论总结于下列定理中。

Finding the Area between Two Curves, Integrating along the y-axis 求两曲线之间的面积（沿 y 轴积分）

Let $u(y)$ and $v(y)$ be continuous functions such that $u(y) \geq v(y)$ for all $y \in \left\lbrack {c,d} \right\rbrack.$ Let $R$ denote the region bounded on the right by the graph of $u(y),$ on the left by the graph of $v(y),$ and above and below by the lines $y = d$ and $y = c,$ respectively. Then, the area of $R$ is given by

设 $u(y)$ 与 $v(y)$ 为连续函数，且对所有 $y \in \left\lbrack {c,d} \right\rbrack$ 满足 $u(y) \geq v(y)$。令 $R$ 表示右侧由 $u(y)$ 的图像、左侧由 $v(y)$ 的图像、上下两侧分别由直线 $y = d$ 与 $y = c$ 所围成的区域。则 $R$ 的面积由下式给出

$$A = \int_{c}^{d}\left\lbrack {u(y) - v(y)} \right\rbrack dy.$$ (2.2)

Integrating with Respect to y 关于 y 积分

Let’s revisit Example 2.4, only this time let’s integrate with respect to $y.$ Let $R$ be the region depicted in Figure 2.10. Find the area of $R$ by integrating with respect to $y.$

让我们重新考察示例 2.4，只是这次我们对 $y$ 积分。令 $R$ 为图 2.10 所示的区域。通过对 $y$ 积分求 $R$ 的面积。

Solution 解答

We must first express the graphs as functions of $y.$ As we saw at the beginning of this section, the curve on the left can be represented by the function $x = v(y) = \sqrt{y},$ and the curve on the right can be represented by the function $x = u(y) = 2 - y.$

我们必须首先将图像表示为 $y$ 的函数。正如本节开头所见，左侧的曲线可以用函数 $x = v(y) = \sqrt{y}$ 表示，右侧的曲线可以用函数 $x = u(y) = 2 - y$ 表示。

Now we have to determine the limits of integration. The region is bounded below by the *x*-axis, so the lower limit of integration is $y = 0.$ The upper limit of integration is determined by the point where the two graphs intersect, which is the point $\left( {1,1} \right),$ so the upper limit of integration is $y = 1.$ Thus, we have $\left\lbrack {c,d} \right\rbrack = \left\lbrack {0,1} \right\rbrack.$

现在我们需要确定积分的上下限。该区域下方由 *x* 轴界定，因此积分下限为 $y = 0$。积分上限由两图像的交点决定，该交点为 $\left( {1,1} \right),$ 因此积分上限为 $y = 1$。于是我们有 $\left\lbrack {c,d} \right\rbrack = \left\lbrack {0,1} \right\rbrack$。

Calculating the area of the region, we get

计算该区域的面积，我们得到

$$\begin{array}{cl} A & {= \int_{c}^{d}\left\lbrack {u(y) - v(y)} \right\rbrack dy} \\ & {= \int_{0}^{1}\left\lbrack {\left( {2 - y} \right) - \sqrt{y}} \right\rbrack dy = \left. \left\lbrack {2y - \frac{y^{2}}{2} - \frac{2}{3}y^{3\text{/}2}} \right\rbrack\ \right|_{0}^{1}} \\ & {= \frac{5}{6}.} \end{array}$$

The area of the region is $5\text{/}6$ units².

该区域的面积为 $5\text{/}6$ 单位²。

Let’s revisit the checkpoint associated with Example 2.4, only this time, let’s integrate with respect to $y.$ Let $R$ be the region depicted in the following figure. Find the area of $R$ by integrating with respect to $y.$

让我们重新考察与示例 2.4 相关的检查点，只是这次我们对 $y$ 积分。令 $R$ 为下图所示的区域。通过对 $y$ 积分求 $R$ 的面积。

Section 2.1 Exercises 2.1 节习题

For the following exercises, determine the area of the region between the two curves in the given figure by integrating over the $x\text{-axis}\text{.}$

对于下列习题，通过对 $x\text{-axis}\text{.}$ 积分，确定给定图形中两条曲线之间区域的面积。

$y = x^{2} - 3\ \text{and}\ y = 1$

2\.

$y = x^{2}\ \text{and}\ y = 3x + 4$

For the following exercises, split the region between the two curves into two smaller regions, then determine the area by integrating over the $x\text{-axis}.$ Note that you will have two integrals to solve.

对于下列习题，将两条曲线之间的区域分割成两个较小的区域，然后通过对 $x\text{-axis}.$ 积分确定面积。注意你将需要计算两个积分。

$y = x^{3}$ and $y = x^{2} + x$

4\.

$y = \text{cos}\ \theta$ and $y = 0.5,$ for $0 \leq \theta \leq \pi$

For the following exercises, determine the area of the region between the two curves by integrating over the $y\text{-axis}.$

对于下列习题，通过对 $y\text{-axis}.$ 积分确定两条曲线之间区域的面积。

$x = y^{2}\ \text{and}\ x = 9$

6\.

$y = x\ \text{and}\ x = y^{2}$

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the $x\text{-axis}.$

对于下列习题，画出方程的图像并阴影标出曲线之间区域的面积。通过对 $x\text{-axis}.$ 积分确定其面积。

$y = x^{2}\ \text{and}\ y = \text{−}x^{2} + 18x$

8\.

$y = \frac{1}{x},y = \frac{1}{x^{2}},\ \text{and}\ x = 3$

$y = \text{cos}\ x$ and $y = \text{cos}^{2}x$ on $x = \left\lbrack {\text{−}\pi,\pi} \right\rbrack$

10\.

$y = e^{x},y = e^{2x - 1},\ \text{and}\ x = 0$

11.

$y = e^{x},y = e^{\text{−}x},x = -1\ \text{and}\ x = 1$

12\.

$y = e,y = e^{x},\ \text{and}\ y = e^{\text{−}x}$

13.

$y = |x|\ \text{and}\ y = x^{2}$

For the following exercises, graph the equations and shade the area of the region between the curves. If necessary, break the region into sub-regions to determine its entire area.

对于下列习题，画出方程的图像并阴影标出曲线之间区域的面积。如有必要，将区域分割成若干子区域以确定其总面积。

14\.

$y = \text{sin}\left( {\pi x} \right),y = 2x,\ \text{and}\ x > 0$

15.

$y = 12 - x,y = \sqrt{x},\ \text{and}\ y = 1$

16\.

$y = \text{sin}\ x$ and $y = \text{cos}\ x$ over $x = \left\lbrack {\text{−}\pi,\pi} \right\rbrack$

17.

$y = x^{3}\ \text{and}\ y = x^{2} - 2x$ over $x = \left\lbrack {-1,1} \right\rbrack$

18\.

$y = x^{2} + 9\ \text{and}\ y = 10 + 2x$ over $x = \left\lbrack {-1,3} \right\rbrack$

19.

$y = x^{3} + 3x$ and $y = 4x$

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the $y\text{-axis}.$

对于下列习题，画出方程的图像并阴影标出曲线之间区域的面积。通过对 $y\text{-axis}.$ 积分确定其面积。

20\.

$x = y^{3}\ \text{and}\ x = 3y - 2$

21.

$x = 2y\ \text{and}\ x = y^{3} - y$

22\.

$x = -3 + y^{2}\ \text{and}\ x = y - y^{2}$

23.

$y^{2} = x\ \text{and}\ x = y + 2$

24\.

$x = |y|\ \text{and}\ 2x = \text{−}y^{2} + 2$

25.

$x = \text{sin}\ y,x = \text{cos}(2y),y = \pi\text{/}2,\text{and}\ y = \text{−}\pi\text{/}2$

For the following exercises, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the *x*-axis or *y*-axis, whichever seems more convenient.

对于下列习题，画出方程的图像并阴影标出曲线之间区域的面积。通过对 *x*-axis 或 *y*-axis 积分确定其面积，取看起来更方便者。

26\.

$x = y^{4}\text{and}\ x = y^{5}$

27.

$y = xe^{x},y = e^{x},x = 0,\ \text{and}\ x = 1$

28\.

$y = x^{6}\text{and}\ y = x^{4}$

29.

$x = y^{3} + 2y^{2} + 1\ \text{and}\ x = \text{−}y^{2} + 1$

30\.

$y = |x|\ \text{and}\ y = x^{2} - 1$

31.

$y = 4 - 3x\ \text{and}\ y = \frac{1}{x}$

32\.

$y = \text{sin}\ x,x = \text{−}\pi\text{/}6,x = \pi\text{/}6,\text{and}\ y = \text{cos}^{3}x$

33.

$y = x^{2} - 3x + 2\ \text{and}\ y = x^{3} - 2x^{2} - x + 2$

34\.

$y = 2\ \text{cos}^{3}\left( {3x} \right),y = -1,x = \frac{\pi}{4},\ \text{and}\ x = - \frac{\pi}{4}$

35.

$y + y^{3} = x\ \text{and}\ 2y = x$

36\.

$y = \sqrt{1 - x^{2}}\ \text{and}\ y = x^{2} - 1$

37.

$y = \text{cos}^{-1}x,y = \text{sin}^{-1}x,x = -1,\ \text{and}\ x = 1$

For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region.

对于下列习题，若可能，求出由给定方程所围成区域的精确面积。若无法解析地确定交点，可使用计算器将交点近似到三位小数，并确定该区域的近似面积。

38\.

\[T\] $x = e^{y}\ \text{and}\ y = x - 2$

39.

\[T\] $y = x^{2}\ \text{and}\ y = \sqrt{1 - x^{2}}$

40\.

\[T\] $y = 3x^{2} + 8x + 9\ \text{and}\ 3y = x + 24$

41.

\[T\] $x = \sqrt{4 - y^{2}}\ \text{and}\ y^{2} = 1 + x^{2}$

42\.

\[T\] $x^{2} = y^{3}\ \text{and}\ x = 3y$

43.

\[T\] $y = \text{sin}^{3}x + 2,y = \text{tan}\ x,x = -1.5,\ \text{and}\ x = 1.5$

44\.

\[T\] $y = \sqrt{1 - x^{2}}\ \text{and}\ y^{2} = x^{2}$

45.

\[T\] $y = \sqrt{1 - x^{2}}\ \text{and}\ y = x^{2} + 2x + 1$

46\.

\[T\] $x = 4 - y^{2}\ \text{and}\ x = 1 + 3y + y^{2}$

47.

\[T\] $y = \text{cos}\ x,y = e^{x},x = \text{−}\pi,\ \text{and}\ x = 0$

48\.

The largest triangle with a base on the $x\text{-axis}$ that fits inside the upper half of the unit circle $y^{2} + x^{2} = 1$ is given by $y = 1 + x$ and $y = 1 - x.$ See the following figure. What is the area inside the semicircle but outside the triangle?

底边位于 $x\text{-axis}$ 上、且内接于单位圆上半部分 $y^{2} + x^{2} = 1$ 的最大三角形由 $y = 1 + x$ 与 $y = 1 - x$ 给出。见下图。半圆内部但三角形外部的面积是多少？

49.

A factory selling cell phones has a marginal cost function $C(x) = 0.01x^{2} - 3x + 229,$ where $x$ represents the number of cell phones, $C$ is the marginal cost, in dollars, of selling $x$ phones, and a marginal revenue function given by $R(x) = 429 - 2x$, where $R$ is the revenue, in dollars, earned by selling $x$ cell phones. Find the area between the graphs of these curves and $x = 0.$ What does this area represent?

一家销售手机的工厂有边际成本函数 $C(x) = 0.01x^{2} - 3x + 229,$ 其中 $x$ 表示手机数量，$C$ 是销售 $x$ 部手机的边际成本（美元），还有边际收益函数 $R(x) = 429 - 2x$，其中 $R$ 是销售 $x$ 部手机所获得的收益（美元）。求这些曲线图像之间与 $x = 0$ 之间的面积。这个面积代表什么？

50\.

An amusement park has a marginal cost function $C(x) = 1000e^{\text{−}x} + 5,$ where $x$ represents the number of tickets sold, and a marginal revenue function given by $R(x) = 60 - 0.1x.$ Find the total profit generated when selling $550$ tickets. Use a calculator to determine intersection points, if necessary, to two decimal places.

一家游乐园有边际成本函数 $C(x) = 1000e^{\text{−}x} + 5,$ 其中 $x$ 表示售出的票数，以及边际收益函数 $R(x) = 60 - 0.1x.$ 求售出 $550$ 张票时产生的总利润。如有必要，用计算器确定交点，精确到两位小数。

51.

The tortoise versus the hare: The speed of the hare is given by the sinusoidal function $H(t) = 1 - \text{cos}\left( {\left( {\pi t} \right)\text{/}2} \right)$ whereas the speed of the tortoise is $T(t)~ = ~0.1032t$ where $t$ is time measured in hours and the speed is measured in miles per hour. Find the area between the curves from time $t = 0$ to the first time after one hour when the tortoise and hare are traveling at the same speed. What does it represent? Use a calculator to determine the intersection points, if necessary, accurate to three decimal places.

龟兔赛跑：兔子的速度由正弦型函数 $H(t) = 1 - \text{cos}\left( {\left( {\pi t} \right)\text{/}2} \right)$ 给出，而乌龟的速度为 $T(t)~ = ~0.1032t$，其中 $t$ 以小时计，速度以英里每小时计。求从 $t = 0$ 到一小时后乌龟与兔子速度首次相同的时刻之间曲线所围成的面积。它代表什么？如有必要，用计算器确定交点，精确到三位小数。

52\.

The tortoise versus the hare: The speed of the hare is given by the sinusoidal function $H(t) = \left( {1\text{/}2} \right) - \left( {1\text{/}2} \right)\text{cos}\left( {2\pi t} \right)$ whereas the speed of the tortoise is $T(t) = \sqrt{t},$ where $t$ is time measured in hours and speed is measured in kilometers per hour. If the race is over in $1$ hour, who won the race and by how much? Use a calculator to determine the intersection points, if necessary, accurate to three decimal places.

龟兔赛跑：兔子的速度由正弦型函数 $H(t) = \left( {1\text{/}2} \right) - \left( {1\text{/}2} \right)\text{cos}\left( {2\pi t} \right)$ 给出，而乌龟的速度为 $T(t) = \sqrt{t},$ 其中 $t$ 以小时计，速度以千米每小时计。如果比赛在 $1$ 小时内结束，谁赢了比赛，赢了多少？如有必要，用计算器确定交点，精确到三位小数。

For the following exercises, find the area between the curves by integrating with respect to $x$ and then with respect to $y.$ Is one method easier than the other? Do you obtain the same answer?

对于下列习题，通过对 $x$ 积分再对 $y$ 积分来求曲线之间的面积。其中一种方法是否比另一种更容易？你得到的结果是否相同？

53.

$y = x^{2} + 2x + 1\ \text{and}\ y = \text{−}x^{2} - 3x + 4$

54\.

$y = x^{4}\text{and}\ x = y^{5}$

55.

$x = y^{2} - 2\ \text{and}\ x = 2y$

For the following exercises, solve using calculus, then check your answer with geometry.

对于下列习题，用微积分求解，然后用几何方法验证你的答案。

56\.

Determine the equations for the sides of the square that touches the unit circle on all four sides, as seen in the following figure. Find the area between the perimeter of this square and the unit circle. Is there another way to solve this without using calculus?

确定四条边都与单位圆相切的正方形的各边方程，如下图所示。求此正方形周长与单位圆之间的面积。是否有不用微积分也能解此题的其他方法？

57.

Find the area between the perimeter of the unit circle and the triangle created from $y = 2x + 1,y = 1 - 2x$ and $y = - \frac{3}{5},$ as seen in the following figure. Is there a way to solve this without using calculus?

求单位圆周长与由 $y = 2x + 1,y = 1 - 2x$ 和 $y = - \frac{3}{5}$ 所构成三角形之间的面积，如下图所示。是否有不用微积分也能解此题的方法？

2.2 Determining Volumes by Slicing 2.2 用切片法确定体积

2.2.1 Determine the volume of a solid by integrating a cross-section (the slicing method).
2.2.2 Find the volume of a solid of revolution using the disk method.
2.2.3 Find the volume of a solid of revolution with a cavity using the washer method.

2.2.1 通过对横截面（切片法）积分来确定几何体的体积。
2.2.2 用圆盘法求旋转体的体积。
2.2.3 用垫圈法求带空腔的旋转体的体积。

In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite integrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—for finding these volumes, depending on the characteristics of the solid.

在上一节中，我们用定积分求两条曲线之间的面积。本节中，我们用定积分求三维几何体的体积。根据几何体的特征，我们考虑三种求体积的方法——切片法、圆盘法和垫圈法。

Volume and the Slicing Method 体积与切片法

Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: $V = lwh.$ The formulas for the volume of a sphere $\left( {V = \frac{4}{3}\pi r^{3}} \right),$ a cone $\left( {V = \frac{1}{3}\pi r^{2}h} \right),$ and a pyramid $\left( {V = \frac{1}{3}Ah} \right)$ have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.

正如面积是二维区域的数值度量，体积也是三维几何体的数值度量。我们大多数人都用过基本几何公式算过几何体的体积。例如，长方体的体积可以通过长、宽、高相乘得到：$V = lwh.$ 球体体积公式 $\left( {V = \frac{4}{3}\pi r^{3}} \right)$、圆锥体积公式 $\left( {V = \frac{1}{3}\pi r^{2}h} \right)$ 以及棱锥体积公式 $\left( {V = \frac{1}{3}Ah} \right)$ 我们也都学过。尽管其中一些公式是仅用几何方法推导出来的，但这些公式都可以用积分得到。

We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics the word *cylinder* has a more general meaning. To discuss cylinders in this more general context, we first need to define some vocabulary.

我们也可以计算圆柱体的体积。尽管我们大多数人想到圆柱体时会认为它有一个圆形底面，比如罐头或金属棒，但在数学中，*cylinder*（圆柱）一词含义更一般。要在这种更一般的语境下讨论圆柱体，我们首先需要定义一些术语。

We define the cross-section of a solid to be the intersection of a plane with the solid. A *cylinder* is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region, called the *axis* of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 2.11 is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the cross-section by the height of the cylinder: $V = A \cdot h.$ In the case of a right circular cylinder (soup can), this becomes $V = \pi r^{2}h.$

我们将几何体的横截面定义为一个平面与该几何体的交集。*cylinder*（圆柱）定义为：将一个平面区域沿一条垂直于该区域的直线（称为圆柱的*axis*（轴））平移所生成的任意几何体。因此，所有垂直于圆柱轴的横截面都完全相同。图 2.11 所示的几何体是一个以非圆为底的圆柱体的例子。要计算圆柱体的体积，我们只需将横截面积乘以圆柱体的高：$V = A \cdot h.$ 对于正圆柱（罐头），这就成为 $V = \pi r^{2}h.$

If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid *S* shown in Figure 2.12, extending along the $x\text{-axis}\text{.}$

如果一个几何体没有恒定的横截面（并且它也不是其他基本几何体之一），我们可能没有现成公式求其体积。此时，我们可以用定积分来计算该几何体的体积。方法是将几何体切成若干薄片，估计每一片的体积，再把这些估计的体积相加。这些薄片应彼此平行，当我们把所有薄片合在一起时，应当得到整个几何体。例如，考虑图 2.12 所示的沿 $x\text{-axis}\text{.}$ 延伸的几何体 *S*。

We want to divide $S$ into slices perpendicular to the $x\text{-axis}\text{.}$ As we see later in the chapter, there may be times when we want to slice the solid in some other direction—say, with slices perpendicular to the *y*-axis. The decision of which way to slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the $x\text{-axis}\text{.}$

我们想把 $S$ 切成垂直于 $x\text{-axis}\text{.}$ 的薄片。正如本章稍后所见，有时我们可能想沿其他方向切片——例如垂直于 *y*-axis 的薄片。选择沿哪个方向切片非常重要。如果选错，计算会变得相当繁琐。本章稍后我们将详细考察其中一些情形，并研究如何决定沿哪个方向切片。不过就本节而言，我们使用垂直于 $x\text{-axis}\text{.}$ 的薄片。

Because the cross-sectional area is not constant, we let $A(x)$ represent the area of the cross-section at point $x.$ Now let $P = \left\{ {x_{0},x_{1}\text{…},X_{n}} \right\}$ be a regular partition of $\left\lbrack {a,b} \right\rbrack,$ and for $i = 1,2\text{,…}n,$ let $S_{i}$ represent the slice of $S$ stretching from $x_{i - 1}\text{to}\ x_{i}.$ The following figure shows the sliced solid with $n = 3.$

由于横截面积不是常数，我们令 $A(x)$ 表示点 $x$ 处横截面的面积。令 $P = \left\{ {x_{0},x_{1}\text{…},X_{n}} \right\}$ 为 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割，并对 $i = 1,2\text{,…}n,$ 令 $S_{i}$ 表示 $S$ 中从 $x_{i - 1}\text{to}\ x_{i}$ 拉伸出的薄片。下图显示了 $n = 3$ 时的切片几何体。

Finally, for $i = 1,2\text{,…}n,$ let $x_{i}^{*}$ be an arbitrary point in $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ Then the volume of slice $S_{i}$ can be estimated by $V\left( S_{i} \right) \approx A\left( x_{i}^{*} \right)\text{Δ}x.$ Adding these approximations together, we see the volume of the entire solid $S$ can be approximated by

最后，对 $i = 1,2\text{,…}n,$ 令 $x_{i}^{*}$ 为 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 中的任意一点。那么薄片 $S_{i}$ 的体积可估计为 $V\left( S_{i} \right) \approx A\left( x_{i}^{*} \right)\text{Δ}x.$ 将这些近似值相加，我们看到整个几何体 $S$ 的体积可近似为

$$V(S) \approx {\sum\limits_{i = 1}^{n}{A\left( x_{i}^{*} \right)}}\text{Δ}x.$$

By now, we can recognize this as a Riemann sum, and our next step is to take the limit as $n\rightarrow\infty.$ Then we have

至此，我们可以认出这是一个黎曼和，下一步是取 $n\rightarrow\infty$ 时的极限。于是我们有

$$V(S) = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}{A\left( x_{i}^{*} \right)}}\text{Δ}x = {\int\limits_{a}^{b}{A(x)dx}}.$$

The technique we have just described is called the slicing method. To apply it, we use the following strategy.

我们刚才描述的方法称为切片法。要应用它，我们使用以下策略。

Finding Volumes by the Slicing Method 用切片法求体积

1. Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.

1. 考察几何体并确定其横截面的形状。若未提供图形，画一张图通常很有帮助。

2. Determine a formula for the area of the cross-section.

2. 确定横截面积的计算公式。

3. Integrate the area formula over the appropriate interval to get the volume.

3. 在相应区间上对面积公式积分，得到体积。

Recall that in this section, we assume the slices are perpendicular to the $x\text{-axis}\text{.}$ Therefore, the area formula is in terms of *x* and the limits of integration lie on the $x\text{-axis}\text{.}$ However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.

回顾一下，本节中我们假设薄片垂直于 $x\text{-axis}\text{.}$ 因此，面积公式用 *x* 表示，积分限位于 $x\text{-axis}\text{.}$ 上。然而，这里所示的解题策略无论我们如何选择切片方向都有效。

Deriving the Formula for the Volume of a Pyramid 推导棱锥体积公式

We know from geometry that the formula for the volume of a pyramid is $V = \frac{1}{3}Ah.$ If the pyramid has a square base, this becomes $V = \frac{1}{3}a^{2}h,$ where $a$ denotes the length of one side of the base. We are going to use the slicing method to derive this formula.

我们从几何学知道，棱锥体积公式为 $V = \frac{1}{3}Ah.$ 若棱锥底面为正方形，则变为 $V = \frac{1}{3}a^{2}h,$ 其中 $a$ 表示底面一边的长度。我们将用切片法来推导这个公式。

Solution 解答

We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in Figure 2.14, oriented along the $x\text{-axis}\text{.}$

我们要将切片法应用于底面为正方形的棱锥。为建立积分，考虑图 2.14 所示沿 $x\text{-axis}\text{.}$ 定向的棱锥。

We first want to determine the shape of a cross-section of the pyramid. We know the base is a square, so the cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure 2.14(b), and using a proportion, since these are similar triangles, we have

我们首先要确定棱锥横截面的形状。已知底面是正方形，因此横截面也是正方形（步骤 1）。现在我们要确定其中一个横截面正方形的面积公式。观察图 2.14(b)，并利用比例关系，由于这些是相似三角形，我们有

$$\frac{s}{a} = \frac{x}{h}\ \text{or}\ s = \frac{ax}{h}.$$

Therefore, the area of one of the cross-sectional squares is

因此，其中一个横截面正方形的面积为

$$A(x) = s^{2} = \left( \frac{ax}{h} \right)^{2}\left( {\text{step}\ 2} \right).$$

Then we find the volume of the pyramid by integrating from $0\ \text{to}\ h$ (step $3)\text{:}$

然后我们从 $0\ \text{to}\ h$ 积分求得棱锥的体积（步骤 $3)\text{:}$

$$\begin{array}{cl} V & {= {\int\limits_{0}^{h}{A(x)}}dx} \\ & {= {\int\limits_{0}^{h}\left( \frac{ax}{h} \right)^{2}}dx = \frac{a^{2}}{h^{2}}{\int\limits_{0}^{h}x^{2}}dx} \\ & {= \left. \left\lbrack {\frac{a^{2}}{h^{2}}\left( {\frac{1}{3}x^{3}} \right)} \right\rbrack\ \right|_{0}^{h} = \frac{1}{3}a^{2}h.} \end{array}$$

This is the formula we were looking for.

这正是我们要找的公式。

Use the slicing method to derive the formula $V = \frac{1}{3}\pi r^{2}h$ for the volume of a circular cone.

用切片法推导圆锥体积公式 $V = \frac{1}{3}\pi r^{2}h$。

Solids of Revolution 旋转体

If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.

如果平面中的一个区域绕该平面内的一条直线旋转，所得的立体称为旋转体，如下图所示。

Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of revolution.

旋转体在机械应用中很常见，例如车床加工出的机器零件。本节余下的部分我们就来考察这类立体。下一个例子用切片法计算一个旋转体的体积。

Use an online integral calculator to learn more.

使用在线积分计算器可进一步了解。

Using the Slicing Method to find the Volume of a Solid of Revolution 用切片法求旋转体的体积

Use the slicing method to find the volume of the solid of revolution bounded by the graphs of $f(x) = x^{2} - 4x + 5,x = 1,\ \text{and}\ x = 4,$ and rotated about the $x\text{-axis}\text{.}$

用切片法求由 $f(x) = x^{2} - 4x + 5$、$x = 1$ 与 $x = 4$ 的图像所围成、并绕 $x\text{-axis}$ 旋转所得的旋转体的体积。

Solution 解答

Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval $\left\lbrack {1,4} \right\rbrack$ as shown in the following figure.

运用解题策略，我们首先画出该二次函数在区间 $\left\lbrack {1,4} \right\rbrack$ 上的图像，如下图所示。

Next, revolve the region around the *x*-axis, as shown in the following figure.

接着，将该区域绕 *x* 轴旋转，如下图所示。

Since the solid was formed by revolving the region around the $x\text{-axis,}$ the cross-sections are circles (step 1). The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by $f(x).$ Use the formula for the area of the circle:

由于该立体是由区域绕 $x\text{-axis}$ 旋转而成，其横截面是圆（步骤 1）。因此横截面的面积为圆的面积，且圆的半径为 $f(x)$。使用圆的面积公式：

$$A(x) = \pi r^{2} = \pi\left\lbrack {f(x)} \right\rbrack^{2} = \pi\left( {x^{2} - 4x + 5} \right)^{2}\ \text{(step 2)}.$$

The volume, then, is (step 3)

于是体积为（步骤 3）

$$\begin{array}{cl} V & {= {\int\limits_{a}^{b}{A(x)}}dx} \\ & {= {\int_{1}^{4}{\pi\left( {x^{2} - 4x + 5} \right)^{2}}}dx = \pi{\int_{1}^{4}\left( {x^{4} - 8x^{3} + 26x^{2} - 40x + 25} \right)}dx} \\ & {= \left. {\pi\left( {\frac{x^{5}}{5} - 2x^{4} + \frac{26x^{3}}{3} - 20x^{2} + 25x} \right)} \right|_{1}^{4} = \frac{78}{5}\pi.} \end{array}$$

The volume is $78\pi\text{/}5.$

体积为 $78\pi\text{/}5$。

Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function $f(x) = 1\text{/}x$ and the $x\text{-axis}$ over the interval $\left\lbrack {1,2} \right\rbrack$ around the $x\text{-axis}\text{.}$ See the following figure.

用切片法求由函数 $f(x) = 1\text{/}x$ 的图像与 $x\text{-axis}$ 在区间 $\left\lbrack {1,2} \right\rbrack$ 上所围成区域绕 $x\text{-axis}$ 旋转所得的旋转体体积。见下图。

The Disk Method 圆盘法

When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function $f(x) = \left( {x - 1} \right)^{2} + 1$ and the $x\text{-axis}$ over the interval $\left\lbrack {-1,3} \right\rbrack$ around the $x\text{-axis}\text{.}$ The graph of the function and a representative disk are shown in Figure 2.18(a) and (b). The region of revolution and the resulting solid are shown in Figure 2.18(c) and (d).

当我们将切片法用于旋转体时，它常被称为圆盘法，因为对于旋转体，用来近似立体体积的切片是圆盘。为说明这一点，考虑由函数 $f(x) = \left( {x - 1} \right)^{2} + 1$ 的图像与 $x\text{-axis}$ 在区间 $\left\lbrack {-1,3} \right\rbrack$ 上所围成区域绕 $x\text{-axis}$ 旋转生成的旋转体。该函数的图像与一个有代表性的圆盘分别显示在图 2.18(a) 与 (b) 中；旋转区域与所得立体显示在图 2.18(c) 与 (d) 中。

We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that

在推导切片法时，我们已经使用了体积公式的黎曼和严格推导。我们知道

$$V = {\int_{a}^{b}{A(x)dx}}.$$

The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. This gives the following rule.

圆盘法与切片法唯一的区别是，我们事先就知道横截面积公式；它是圆的面积。由此得到如下法则。

Let $f(x)$ be continuous and nonnegative. Define $R$ as the region bounded above by the graph of $f(x),$ below by the $x\text{-axis,}$ on the left by the line $x = a,$ and on the right by the line $x = b.$ Then, the volume of the solid of revolution formed by revolving $R$ around the $x\text{-axis}$ is given by

设 $f(x)$ 连续且非负。定义 $R$ 为：上界为 $f(x)$ 图像、下界为 $x\text{-axis}$、左界为直线 $x = a$、右界为直线 $x = b$ 的区域。则 $R$ 绕 $x\text{-axis}$ 旋转所得旋转体的体积为

$$V = {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}.$$ (2.3)

The volume of the solid we have been studying (Figure 2.18) is given by

我们一直在研究的立体（图 2.18）的体积为

$$\begin{array}{cl} V & {= {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}} \\ & {= {\int_{-1}^{3}{\pi\left\lbrack {\left( {x - 1} \right)^{2} + 1} \right\rbrack^{2}dx}} = \pi{\int_{-1}^{3}{\left\lbrack {\left( {x - 1} \right)^{4} + 2\left( {x - 1} \right)^{2} + 1} \right\rbrack dx}}} \\ & {= \pi\left. \left\lbrack {\frac{1}{5}\left( {x - 1} \right)^{5} + \frac{2}{3}\left( {x - 1} \right)^{3} + x} \right\rbrack\ \right|_{-1}^{3} = \pi\left\lbrack {\left( {\frac{32}{5} + \frac{16}{3} + 3} \right) - \left( {- \frac{32}{5} - \frac{16}{3} - 1} \right)} \right\rbrack = \frac{412\pi}{15}\ \text{units}^{3}.} \end{array}$$

Let’s look at some examples.

我们来看几个例子。

Using the Disk Method to Find the Volume of a Solid of Revolution 1 用圆盘法求旋转体的体积 1

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $f(x) = \sqrt{x}$ and the $x\text{-axis}$ over the interval $\left\lbrack {1,4} \right\rbrack$ around the $x\text{-axis}\text{.}$

用圆盘法求由 $f(x) = \sqrt{x}$ 的图像与 $x\text{-axis}$ 在区间 $\left\lbrack {1,4} \right\rbrack$ 上所围成区域绕 $x\text{-axis}$ 旋转生成的旋转体体积。

Solution 解答

The graphs of the function and the solid of revolution are shown in the following figure.

该函数的图像与旋转体显示在下图中。

We have

我们有

$$\begin{array}{cl} V & {= {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}} \\ & {= {\int_{1}^{4}{\pi\left\lbrack \sqrt{x} \right\rbrack^{2}}}dx = \pi{\int_{1}^{4}{x\ dx}}} \\ & {= \left. {\frac{\pi}{2}x^{2}} \right|_{1}^{4} = \frac{15\pi}{2}.} \end{array}$$

The volume is $\left( {15\pi} \right)\text{/}2$ units³.

体积为 $\left( {15\pi} \right)\text{/}2$ 立方单位。

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $f(x) = \sqrt{4 - x}$ and the $x\text{-axis}$ over the interval $\left\lbrack {0,\ 4} \right\rbrack$ around the $x\text{-axis}\text{.}$

用圆盘法求由 $f(x) = \sqrt{4 - x}$ 的图像与 $x\text{-axis}$ 在区间 $\left\lbrack {0,\ 4} \right\rbrack$ 上所围成区域绕 $x\text{-axis}$ 旋转生成的旋转体体积。

So far, our examples have all concerned regions revolved around the $x\text{-axis,}$ but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has been generated by revolving a region around the $y\text{-axis}\text{.}$ The mechanics of the disk method are nearly the same as when the $x\text{-axis}$ is the axis of revolution, but we express the function in terms of $y$ and we integrate with respect to *y* as well. This is summarized in the following rule.

到目前为止，我们的例子都涉及绕 $x\text{-axis}$ 旋转的区域，但我们也可以让一个平面区域绕任意水平或竖直直线旋转来生成旋转体。在下一个例子中，我们考察一个由区域绕 $y\text{-axis}$ 旋转生成的旋转体。圆盘法的机理与以 $x\text{-axis}$ 为旋转轴时几乎相同，只是我们要把函数用 $y$ 表示，并对 *y* 积分。这一点总结在下面的法则中。

Let $g(y)$ be continuous and nonnegative. Define $Q$ as the region bounded on the right by the graph of $g(y),$ on the left by the $y\text{-axis,}$ below by the line $y = c,$ and above by the line $y = d.$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $y\text{-axis}$ is given by

设 $g(y)$ 连续且非负。定义 $Q$ 为：右界为 $g(y)$ 图像、左界为 $y\text{-axis}$、下界为直线 $y = c$、上界为直线 $y = d$ 的区域。则 $Q$ 绕 $y\text{-axis}$ 旋转所得旋转体的体积为

$$V = {\int_{c}^{d}{\pi\left\lbrack {g(y)} \right\rbrack^{2}dy}}.$$ (2.4)

The next example shows how this rule works in practice.

下一个例子展示该法则在实际中如何运用。

Using the Disk Method to Find the Volume of a Solid of Revolution 2 用圆盘法求旋转体的体积 2

Let $R$ be the region bounded by the graph of $g(y) = \sqrt{4 - y}$ and the $y\text{-axis}$ over the $y\text{-axis}$ interval $\left\lbrack {0,4} \right\rbrack.$ Use the disk method to find the volume of the solid of revolution generated by rotating $R$ around the $y\text{-axis}\text{.}$

设 $R$ 为 $g(y) = \sqrt{4 - y}$ 的图像与 $y\text{-axis}$ 在 $y\text{-axis}$ 上的区间 $\left\lbrack {0,4} \right\rbrack$ 所围成的区域。用圆盘法求 $R$ 绕 $y\text{-axis}$ 旋转生成的旋转体体积。

Solution 解答

Figure 2.20 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the $y\text{-axis,}$ the disks are horizontal, rather than vertical.

图 2.20 显示了该函数以及一个可用于估计体积的、有代表性的圆盘。注意，由于我们是绕 $y\text{-axis}$ 旋转该函数，圆盘是水平的而非竖直的。

The region to be revolved and the full solid of revolution are depicted in the following figure.

待旋转的区域以及完整的旋转体显示在下图中。

To find the volume, we integrate with respect to $y.$ We obtain

为求体积，我们对 $y$ 积分。得到

$$\begin{array}{cl} V & {= {\int_{c}^{d}\pi}\left\lbrack {g(y)} \right\rbrack^{2}dy} \\ & {= {\int_{0}^{4}\pi}\left\lbrack \sqrt{4 - y} \right\rbrack^{2}dy = \pi{\int_{0}^{4}\left( {4 - y} \right)}dy} \\ & {= \left. {\pi\left\lbrack {4y - \frac{y^{2}}{2}} \right\rbrack}\ \right|_{0}^{4} = 8\pi.} \end{array}$$

The volume is $8\pi$ units³.

体积为 $8\pi$ 立方单位。

Use the disk method to find the volume of the solid of revolution generated by rotating the region between the graph of $g(y) = y$ and the $y\text{-axis}$ over the interval $\left\lbrack {1,4} \right\rbrack$ around the $y\text{-axis}\text{.}$

用圆盘法求由 $g(y) = y$ 的图像与 $y\text{-axis}$ 在区间 $\left\lbrack {1,4} \right\rbrack$ 上所围成区域绕 $y\text{-axis}$ 旋转生成的旋转体体积。

The Washer Method 垫圈法

Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes, this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than the $x\text{-axis}$ or $y\text{-axis}$ is selected.

有些旋转体内部是中空的；它们并非一直实心地延伸到旋转轴。有时，这仅仅是旋转区域相对于旋转轴的形状所致。在另一些情况下，当旋转区域被定义为两个函数图形之间的区域时，就会出现空腔。第三种出现空腔的情况是：所选的旋转轴既不是 $x\text{-axis}$ 也不是 $y\text{-axis}$。

When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). For example, consider the region bounded above by the graph of the function $f(x) = \sqrt{x}$ and below by the graph of the function $g(x) = 1$ over the interval $\left\lbrack {1,4} \right\rbrack.$ When this region is revolved around the $x\text{-axis,}$ the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative washer are shown in Figure 2.22(a) and (b). The region of revolution and the resulting solid are shown in Figure 2.22(c) and (d).

当旋转体内部有空腔时，用来近似体积的切片就不是圆盘，而是垫圈（中心有孔的圆盘）。例如，考虑在区间 $\left\lbrack {1,4} \right\rbrack$ 上、上界为函数 $f(x) = \sqrt{x}$ 的图像、下界为函数 $g(x) = 1$ 的图像的区域。当该区域绕 $x\text{-axis}$ 旋转时，结果是一个中间有空腔的立体，其切片是垫圈。该函数的图像与一个代表性的垫圈显示在图 2.22(a) 与 (b) 中；旋转区域与所得立体显示在图 2.22(c) 与 (d) 中。

The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,

于是横截面积为外圆面积减去内圆面积。在此情形下，

$$A(x) = \pi\left( \sqrt{x} \right)^{2} - \pi(1)^{2} = \pi\left( {x - 1} \right).$$

Then the volume of the solid is

于是该立体的体积为

$$\begin{array}{cl} V & {= {\int_{a}^{b}{A(x)}}dx} \\ & {= {\int_{1}^{4}\pi}\left( {x - 1} \right)dx = \left. {\pi\left\lbrack {\frac{x^{2}}{2} - x} \right\rbrack}\ \right|_{1}^{4} = \frac{9}{2}\pi\ \text{units}^{3}.} \end{array}$$

Generalizing this process gives the washer method.

将这一过程推广，便得到垫圈法。

Suppose $f(x)$ and $g(x)$ are continuous, nonnegative functions such that $f(x) \geq g(x)$ over $\left\lbrack {a,b} \right\rbrack.$ Let $R$ denote the region bounded above by the graph of $f(x),$ below by the graph of $g(x),$ on the left by the line $x = a,$ and on the right by the line $x = b.$ Then, the volume of the solid of revolution formed by revolving $R$ around the $x\text{-axis}$ is given by

设 $f(x)$ 与 $g(x)$ 为连续非负函数，且在区间 $\left\lbrack {a,b} \right\rbrack$ 上满足 $f(x) \geq g(x)$。令 $R$ 表示：上界为 $f(x)$ 图像、下界为 $g(x)$ 图像、左界为直线 $x = a$、右界为直线 $x = b$ 的区域。则 $R$ 绕 $x\text{-axis}$ 旋转所得旋转体的体积为

$$V = {\int_{a}^{b}{\pi\left\lbrack {\left( {f(x)} \right)^{2} - \left( {g(x)} \right)^{2}} \right\rbrack}}dx.$$ (2.5)

Using the Washer Method 用垫圈法

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of $f(x) = x$ and below by the graph of $g(x) = 1\text{/}x$ over the interval $\left\lbrack {1,4} \right\rbrack$ around the $x\text{-axis}\text{.}$

求由 $f(x) = x$ 的图像在上、$g(x) = 1\text{/}x$ 的图像在下、在区间 $\left\lbrack {1,4} \right\rbrack$ 上所围成区域绕 $x\text{-axis}$ 旋转所得的旋转体体积。

Solution 解答

The graphs of the functions and the solid of revolution are shown in the following figure.

这些函数的图像与旋转体显示在下图中。

We have

我们有

$$\begin{array}{cl} V & {= {\int_{a}^{b}{\pi\left\lbrack {\left( {f(x)} \right)^{2} - \left( {g(x)} \right)^{2}} \right\rbrack}}dx} \\ & {= \pi{\int_{1}^{4}{\left\lbrack {x^{2} - \left( \frac{1}{x} \right)^{2}} \right\rbrack dx}}\ \ = \left. {\pi\left\lbrack {\frac{x^{3}}{3} + \frac{1}{x}} \right\rbrack}\ \right|_{1}^{4} = \frac{81\pi}{4}\ \text{units}^{3}.} \end{array}$$

Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of $f(x) = \sqrt{x}$ and $g(x) = 1\text{/}x$ over the interval $\left\lbrack {1,3} \right\rbrack$ around the $x\text{-axis}\text{.}$

求由 $f(x) = \sqrt{x}$ 与 $g(x) = 1\text{/}x$ 的图像在区间 $\left\lbrack {1,3} \right\rbrack$ 上所围成区域绕 $x\text{-axis}$ 旋转所得的旋转体体积。

As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region around the *y*-axis. In this case, the following rule applies.

与圆盘法类似，我们也可以把垫圈法用于由区域绕 *y* 轴旋转所得的旋转体。在这种情况下，适用如下法则。

Suppose $u(y)$ and $v(y)$ are continuous, nonnegative functions such that $v(y) \leq u(y)$ for $y \in \left\lbrack {c,d} \right\rbrack.$ Let $Q$ denote the region bounded on the right by the graph of $u(y),$ on the left by the graph of $v(y),$ below by the line $y = c,$ and above by the line $y = d.$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $y\text{-axis}$ is given by

设 $u(y)$ 与 $v(y)$ 为连续非负函数，且对 $y \in \left\lbrack {c,d} \right\rbrack$ 满足 $v(y) \leq u(y)$。令 $Q$ 表示：右界为 $u(y)$ 图像、左界为 $v(y)$ 图像、下界为直线 $y = c$、上界为直线 $y = d$ 的区域。则 $Q$ 绕 $y\text{-axis}$ 旋转所得旋转体的体积为

$$V = {\int_{c}^{d}{\pi\left\lbrack {\left( {u(y)} \right)^{2} - \left( {v(y)} \right)^{2}} \right\rbrack}}dy.$$

Rather than looking at an example of the washer method with the $y\text{-axis}$ as the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.

我们不去考察以 $y\text{-axis}$ 为旋转轴的垫圈法例子，而是考虑一个旋转轴不是这两条坐标轴之一的例子。同样的一般方法仍然适用，但你可能需要想象如何去描述该立体的横截面积。

The Washer Method with a Different Axis of Revolution 旋转轴不同的垫圈法

Find the volume of a solid of revolution formed by revolving the region bounded above by $f(x) = 4 - x$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {0,4} \right\rbrack$ around the line $y = -2.$

求由 $f(x) = 4 - x$ 在上、$x\text{-axis}$ 在下、在区间 $\left\lbrack {0,4} \right\rbrack$ 上所围成区域绕直线 $y = -2$ 旋转所得的旋转体体积。

Solution 解答

The graph of the region and the solid of revolution are shown in the following figure.

该区域与旋转体的图像显示在下图中。

We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by $f(x) + 2,$ which simplifies to

我们不能直接把体积公式套用到这个问题上，因为旋转轴不是坐标轴之一。不过我们仍然知道，横截面积是外圆面积减去内圆面积。观察该函数的图像，我们看到外圆的半径为 $f(x) + 2$，化简得到

$$f(x) + 2 = \left( {4 - x} \right) + 2 = 6 - x.$$

The radius of the inner circle is $g(x) = 2.$ Therefore, we have

内圆的半径为 $g(x) = 2$。于是我们有

$$\begin{array}{cl} V & {= {\int_{0}^{4}{\pi\left\lbrack {\left( {6 - x} \right)^{2} - (2)^{2}} \right\rbrack}}dx} \\ & {= \pi{\int_{0}^{4}{\left( {x^{2} - 12x + 32} \right)dx}}\ \ = \left. {\pi\left\lbrack {\frac{x^{3}}{3} - 6x^{2} + 32x} \right\rbrack}\ \right|_{0}^{4} = \frac{160\pi}{3}\ \text{units}^{3}.} \end{array}$$

Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of $f(x) = x + 2$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {0,3} \right\rbrack$ around the line $y = -1.$

求由 $f(x) = x + 2$ 的图像在上、$x\text{-axis}$ 在下、在区间 $\left\lbrack {0,3} \right\rbrack$ 上所围成区域绕直线 $y = -1$ 旋转所得的旋转体体积。

Section 2.2 Exercises 2.2 节习题

58\.

Derive the formula for the volume of a sphere using the slicing method.

用切片法推导球体体积的公式。

59\.

Use the slicing method to derive the formula for the volume of a cone.

用切片法推导圆锥体积的公式。

60\.

Use the slicing method to derive the formula for the volume of a tetrahedron with side length $a.$

用切片法推导棱长为 $a.$ 的四面体的体积公式。

61\.

Use the disk method to derive the formula for the volume of a trapezoidal cylinder.

用圆盘法推导梯形柱体的体积公式。

62\.

Explain when you would use the disk method versus the washer method. When are they interchangeable?

说明何时使用圆盘法、何时使用垫圈法。它们在什么情形下可以互换？

For the following exercises, draw a typical slice and find the volume using the slicing method for the given volume.

在以下习题中，画出一个典型切片，并对给定的立体用切片法求其体积。

63.

A pyramid with height 6 units and square base of side 2 units, as pictured here.

一个高为 6 个单位、底面为边长 2 个单位的正方形的棱锥，如图所示。

64\.

A pyramid with height 4 units and a rectangular base with length 2 units and width 3 units, as pictured here.

一个高为 4 个单位、底面为长 2 个单位、宽 3 个单位的矩形的棱锥，如图所示。

65.

A tetrahedron with a base side of 4 units, as seen here.

一个底边为 4 个单位的四面体，如图所示。

66\.

A pyramid with height 5 units, and an isosceles triangular base with lengths of 6 units and 8 units, as seen here.

一个高为 5 个单位、底面为边长分别是 6 个单位与 8 个单位的等腰三角形的棱锥，如图所示。

67.

A cone of radius $r$ and height $h$ has a smaller cone of radius $r\text{/}2$ and height $h\text{/}2$ removed from the top, as seen here. The resulting solid is called a *frustum*.

一个半径为 $r$、高为 $h$ 的圆锥，从其顶部挖去一个半径为 $r\text{/}2$、高为 $h\text{/}2$ 的小圆锥，如图所示。所得立体称为*圆台*（frustum）。

For the following exercises, draw an outline of the solid and find the volume using the slicing method.

在以下习题中，画出该立体的轮廓，并用切片法求其体积。

68\.

The base is a circle of radius $a.$ The slices perpendicular to the base are squares.

底面是半径为 $a.$ 的圆。垂直于底面的切片都是正方形。

69.

The base is a triangle with vertices $\left( {0,0} \right),\left( {1,0} \right),$ and $\left( {0,1} \right).$ Slices perpendicular to the *x*-axis are semicircles.

底面是以 $\left( {0,0} \right),\left( {1,0} \right),$ 与 $\left( {0,1} \right).$ 为顶点的三角形。垂直于 *x* 轴的切片都是半圆。

70\.

The base is the region under the parabola $y = 1 - x^{2}$ in the first quadrant. Slices perpendicular to the *xy*-plane and parallel to the y-axis are squares.

底面是第一象限内抛物线 $y = 1 - x^{2}$ 下方的区域。垂直于 *xy* 平面且平行于 y 轴的切片都是正方形。

71.

The base is the region under the parabola $y = 1 - x^{2}$ and above the $x\text{-axis}\text{.}$ Slices perpendicular to the $y\text{-axis}$ are squares.

底面是抛物线 $y = 1 - x^{2}$ 下方且位于 $x\text{-axis}\text{.}$ 上方的区域。垂直于 $y\text{-axis}$ 的切片都是正方形。

72\.

The base is the region enclosed by $y = x^{2}$ and $y = 9.$ Slices perpendicular to the *x*-axis are right isosceles triangles. The intersection of one of these slices and the base is the leg of the triangle.

底面是由 $y = x^{2}$ 与 $y = 9.$ 围成的区域。垂直于 *x* 轴的切片都是等腰直角三角形。其中一个切片与底面的交线是该三角形的直角边。

73.

The base is the area between $y = x$ and $y = x^{2}.$ Slices perpendicular to the *x*-axis are semicircles.

底面是 $y = x$ 与 $y = x^{2}.$ 之间的区域。垂直于 *x* 轴的切片都是半圆。

For the following exercises, draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the *x*-axis.

在以下习题中，画出由这些曲线围成的区域。然后，用圆盘法求该区域绕 *x* 轴旋转所得立体的体积。

74\.

$x + y = 8,x = 0,\ \text{and}\ y = 0$

75.

$y = 2x^{2},x = 0,x = 4,\ \text{and}\ y = 0$

76\.

$y = e^{x} + 1,x = 0,x = 1,\ \text{and}\ y = 0$

77.

$y = x^{4},x = 0,\ \text{and}\ y = 1\text{for}\ x \geq 0$

78\.

$y = \sqrt{x},x = 0,x = 4,\ \text{and}\ y = 0$

79.

$y = \text{sin}\ x,y = \text{cos}\ x,\ \text{and}\ x = 0$

80\.

$y = \frac{1}{x},x = 2,\ \text{and}\ y = 3$

81.

$x^{2} - y^{2} = 9\ \text{and}\ x + y = 9,y = 0\ \text{and}\ x = 0$

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the *y*-axis.

在以下习题中，画出由这些曲线围成的区域。然后，求该区域绕 *y* 轴旋转所得立体的体积。

82\.

$y = 4 - \frac{1}{2}x,x = 0,\ \text{and}\ y = 0$

83.

$y = 2x^{3},x = 0,x = 1,\ \text{and}\ y = 0$

84\.

$y = 3x^{2},x = 0,\ \text{and}\ y = 3$

85.

$y = \sqrt{4 - x^{2}},y = 0,\ \text{and}\ x = 0$

86\.

$y = \frac{1}{\sqrt{x + 1}},x = 0,\ x = 3,\ \text{and}\ y = 0$

87.

$x = \text{sec}(y)\ \text{and}\ y = \frac{\pi}{4},\ y = 0\ \text{and}\ x = 0$

88\.

$y = \frac{1}{x + 1},x = 0,\ ,\ x = 2,\ \text{and}\ y = 0$

89.

$y = 4 - x,y = x,\ \text{and}\ x = 0$

For the following exercises, draw the region bounded by the curves. Then, find the volume when the region is rotated around the *x*-axis.

在以下习题中，画出由这些曲线围成的区域。然后，求该区域绕 *x* 轴旋转所得立体的体积。

90\.

$y = x + 2,y = x + 6,x = 0,\ \text{and}\ x = 5$

91.

$y = x^{2}\ \text{and}\ y = x + 2$

92\.

$x^{2} = y^{3}\ \text{and}\ x^{3} = y^{2}$

93.

$y = 4 - x^{2}\ \text{and}\ y = 2 - x$

94\.

\[T\] $y = \text{cos}\ x,y = e^{\text{−}x},x = 0,\ \text{and}\ x = 1.2927$

95.

$y = \sqrt{x}\ \text{and}\ y = x^{2}$

96\.

$y = \text{sin}\ x\text{,}\ y = 5\ \text{sin}\ x,x = 0\ \text{and}\ x = \pi$

97.

$y = \sqrt{1 + x^{2}}\ \text{and}\ y = \sqrt{4 - x^{2}}$

For the following exercises, draw the region bounded by the curves. Then, use the washer method to find the volume when the region is revolved around the *y*-axis.

在以下习题中，画出由这些曲线围成的区域。然后，用垫圈法求该区域绕 *y* 轴旋转所得立体的体积。

98\.

$y = \sqrt{x},x = 4,\ \text{and}\ y = 0$

99.

$y = x + 2,y = 2x - 1,\ \text{and}\ x = 0$

100\.

$y = \sqrt[3]{x}\ \text{and}\ y = x^{3}$

101.

$x = e^{2y},x = y^{2},y = 0,\ \text{and}\ y = \text{ln}(2)$

102\.

$x = \sqrt{9 - y^{2}},x = e^{\text{−}y},y = 0,\ \text{and}\ y = 3$

103.

Yogurt containers can be shaped like frustums. Rotate the line $y = \frac{1}{m}x$ around the *y*-axis to find the volume between $y = a\ \text{and}\ y = b.$

酸奶杯的形状可以像圆台。把直线 $y = \frac{1}{m}x$ 绕 *y* 轴旋转，求 $y = a\ \text{and}\ y = b.$ 之间的体积。

104\.

Rotate the ellipse $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1$ around the *x*-axis to approximate the volume of a football, as seen here.

把椭圆 $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1$ 绕 *x* 轴旋转，以近似一个橄榄球的体积，如图所示。

105.

Rotate the ellipse $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1$ around the *y*-axis to approximate the volume of a football.

把椭圆 $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1$ 绕 *y* 轴旋转，以近似一个橄榄球的体积。

106\.

A better approximation of the volume of a football is given by the solid that comes from rotating $y = \text{sin}\ x$ around the *x*-axis from $x = 0$ to $x = \pi.$ What is the volume of this football approximation, as seen here?

橄榄球体积的一个更好的近似，由 $y = \text{sin}\ x$ 绕 *x* 轴从 $x = 0$ 旋转到 $x = \pi.$ 所得的立体给出。如图所示，这个橄榄球近似体的体积是多少？

107.

What is the volume of the Bundt cake that comes from rotating $y = \text{sin}\ x$ around the *y*-axis from $x = 0$ to $x = \pi?$

由 $y = \text{sin}\ x$ 绕 *y* 轴从 $x = 0$ 旋转到 $x = \pi?$ 所得的环形蛋糕（Bundt cake）的体积是多少？

For the following exercises, find the volume of the solid described.

在以下习题中，求所描述立体的体积。

108\.

The base is the region between $y = x$ and $y = x^{2}.$ Slices perpendicular to the *x*-axis are semicircles.

底面是 $y = x$ 与 $y = x^{2}.$ 之间的区域。垂直于 *x* 轴的切片都是半圆。

109.

The base is the region enclosed by the generic ellipse $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1.$ Slices perpendicular to the *x*-axis are semicircles.

底面是由一般椭圆 $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1.$ 围成的区域。垂直于 *x* 轴的切片都是半圆。

110\.

Bore a hole of radius $a$ down the axis of a right cone of height $b$ and radius $b$ through the base of the cone as seen here.

如图所示，钻一个半径为 $a$ 的孔，沿高为 $b$、半径为 $b$ 的正圆锥的轴向下，贯穿圆锥的底面。

111.

Find the volume common to two spheres of radius $r$ with centers that are $2h$ apart, as shown here.

求两个半径为 $r$、球心相距 $2h$ 的球体所共有部分的体积，如图所示。

112\.

Find the volume of a spherical cap of height $h$ and radius $r$ where $h < r,$ as seen here.

如图所示，求高为 $h$、半径为 $r$ 且满足 $h < r,$ 的球冠的体积。

113.

Find the volume of a sphere of radius $R$ with a cap of height $h$ removed from the top, as seen here.

如图所示，求一个半径为 $R$ 的球体在顶部挖去一个高为 $h$ 的球冠后所剩部分的体积。

2.3 Volumes of Revolution: Cylindrical Shells 2.3 旋转体的体积：圆柱壳法

2.3.1 Calculate the volume of a solid of revolution by using the method of cylindrical shells.
2.3.2 Compare the different methods for calculating a volume of revolution.

2.3.1 使用圆柱壳法计算旋转体的体积。
2.3.2 比较计算旋转体体积的各种不同方法。

In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells, we integrate along the coordinate axis *perpendicular* to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation.

本节中，我们考察圆柱壳法，这是求旋转体体积的最后一种方法。我们可以对与圆盘法或垫圈法适用的同类立体使用这一方法；不过，在圆盘法与垫圈法中，我们沿着与旋转轴平行的坐标轴积分。而在圆柱壳法中，我们沿着与旋转轴*垂直*的坐标轴积分。能够自行选择使用哪个积分变量，在处理较复杂的函数时可以是一个显著的优势。此外，立体的具体几何形状有时会使圆柱壳法比垫圈法更具吸引力。在本节最后一部分，我们回顾已学过的所有求体积的方法，并给出一些指导原则，帮助你在给定情形下判断该使用哪种方法。

The Method of Cylindrical Shells 圆柱壳法

Again, we are working with a solid of revolution. As before, we define a region $R,$ bounded above by the graph of a function $y = f(x),$ below by the $x\text{-axis,}$ and on the left and right by the lines $x = a$ and $x = b,$ respectively, as shown in Figure 2.25(a). We then revolve this region around the *y*-axis, as shown in Figure 2.25(b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of $x$ were revolved around the $x\text{-axis}$ or a line parallel to it.

我们再次处理一个旋转体。与之前一样，我们定义一个区域 $R,$，其上界为函数 $y = f(x)$ 的图形，下界为 $x\text{-axis,}$，左、右两侧分别为直线 $x = a$ 与 $x = b,$，如图 2.25(a) 所示。然后我们将该区域绕 *y* 轴旋转，如图 2.25(b) 所示。注意，这与我们之前所做的不同。此前，由关于 $x$ 的函数所定义的区域是绕 $x\text{-axis}$ 或与之平行的直线旋转的。

As we have done many times before, partition the interval $\left\lbrack {a,b} \right\rbrack$ using a regular partition, $P = \left\{ {x_{0},x_{1}\text{,…},x_{n}} \right\}$ and, for $i = 1,2\text{,…},n,$ choose a point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ Then, construct a rectangle over the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ of height $f(x_{i}^{*})$ and width $\text{Δ}x.$ A representative rectangle is shown in Figure 2.26(a). When that rectangle is revolved around the *y*-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.

与之前多次所做的一样，用正则分割 $P = \left\{ {x_{0},x_{1}\text{,…},x_{n}} \right\}$ 对区间 $\left\lbrack {a,b} \right\rbrack$ 进行分割，并对 $i = 1,2\text{,…},n,$ 在区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 中选一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ 然后，在区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 上作一个高为 $f(x_{i}^{*})$、宽为 $\text{Δ}x$ 的矩形。图 2.26(a) 显示了一个代表性矩形。当该矩形绕 *y* 轴旋转时，我们得到的不是一个圆盘或垫圈，而是一个圆柱壳，如下图所示。

To calculate the volume of this shell, consider Figure 2.27.

要计算这个壳的体积，请考虑图 2.27。

The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius $x_{i}$ and inner radius $x_{i - 1}.$ Thus, the cross-sectional area is $\pi x_{i}^{2} - \pi x_{i - 1}^{2}.$ The height of the cylinder is $f(x_{i}^{*}).$ Then the volume of the shell is

这个壳是一个圆柱体，因此它的体积等于横截面积乘以圆柱的高。横截面是圆环（环状区域——本质上是中心有孔的圆），外半径为 $x_{i}$，内半径为 $x_{i - 1}.$ 因此，横截面积为 $\pi x_{i}^{2} - \pi x_{i - 1}^{2}.$ 圆柱的高为 $f(x_{i}^{*}).$ 于是该壳的体积为

$$\begin{array}{cl} V_{\text{shell}} & {= f(x_{i}^{*})(\pi x_{i}^{2} - \pi x_{i - 1}^{2})} \\ & {= \pi f(x_{i}^{*})\left( {x_{i}^{2} - x_{i - 1}^{2}} \right)} \\ & {= \pi f(x_{i}^{*})\left( {x_{i} + x_{i - 1}} \right)\left( {x_{i} - x_{i - 1}} \right)} \\ & {= 2\pi f(x_{i}^{*})\left( \frac{x_{i} + x_{i - 1}}{2} \right)\left( {x_{i} - x_{i - 1}} \right).} \end{array}$$

Note that $x_{i} - x_{i - 1} = \text{Δ}x,$ so we have

注意 $x_{i} - x_{i - 1} = \text{Δ}x,$ 于是我们有

$$V_{\text{shell}} = 2\pi f(x_{i}^{*})\left( \frac{x_{i} + x_{i - 1}}{2} \right)\text{Δ}x.$$

Furthermore, $\frac{x_{i} + x_{i - 1}}{2}$ is both the midpoint of the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ and the average radius of the shell, and we can approximate this by $x_{i}^{*}.$ We then have

此外，$\frac{x_{i} + x_{i - 1}}{2}$ 既是区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 的中点，也是该壳的平均半径，而我们可以用 $x_{i}^{*}$ 来近似它。于是我们有

$$V_{\text{shell}} \approx 2\pi f(x_{i}^{*})x_{i}^{*}\text{Δ}x.$$

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate (Figure 2.28).

另一种理解方式是想象在壳上作一个竖直切口，然后将其展开形成一个平板（图 2.28）。

In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height $f(x_{i}^{*}),$ width $2\pi x_{i}^{*},$ and thickness $\text{Δ}x$ (Figure 2.28). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

实际上，壳的外半径大于内半径，因此平板的后缘会略长于前缘。然而，我们可以用一个高为 $f(x_{i}^{*})$、宽为 $2\pi x_{i}^{*}$、厚为 $\text{Δ}x$ 的平板来近似这个压平的壳（图 2.28）。于是，该壳的体积近似等于该平板的体积。将平板的高、宽、深相乘，我们得到

$$V_{\text{shell}} \approx f(x_{i}^{*})\left( {2\pi x_{i}^{*}} \right)\text{Δ}x,$$

which is the same formula we had before.

这与我们之前得到的公式相同。

To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

要计算整个立体的体积，我们将所有壳的体积相加，得到

$$V \approx \sum\limits_{i = 1}^{n}\left( {2\pi x_{i}^{*}f(x_{i}^{*})\text{Δ}x} \right).$$

Here we have another Riemann sum, this time for the function $2\pi xf(x).$ Taking the limit as $n\rightarrow\infty$ gives us

这里我们又得到一个黎曼和，这次是对函数 $2\pi xf(x)$ 而言的。取极限 $n\rightarrow\infty$ 便得到

$$V = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\left( {2\pi x_{i}^{*}f(x_{i}^{*})\text{Δ}x} \right) = {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx.$$

This leads to the following rule for the method of cylindrical shells.

由此得到圆柱壳法的如下法则。

Let $f(x)$ be continuous and nonnegative. Define $R$ as the region bounded above by the graph of $f(x),$ below by the $x\text{-axis},$ on the left by the line $x = a,$ and on the right by the line $x = b.$ Then the volume of the solid of revolution formed by revolving $R$ around the *y*-axis is given by

设 $f(x)$ 连续且非负。定义 $R$ 为这样一个区域：其上界为 $f(x)$ 的图形，下界为 $x\text{-axis,}$，左界为直线 $x = a$，右界为直线 $x = b.$ 那么，将 $R$ 绕 *y* 轴旋转所得的旋转体的体积为

$$V = {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx.$$ (2.6)

Now let’s consider an example.

现在我们考虑一个示例。

The Method of Cylindrical Shells 1 圆柱壳法 1

Define $R$ as the region bounded above by the graph of $f(x) = {1\text{/}x}$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {1,3} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

定义 $R$ 为区间 $\left\lbrack {1,3} \right\rbrack$ 上由函数 $f(x) = {1\text{/}x}$ 的图形为上界、以 $x\text{-axis}$ 为下界的区域。求将 $R$ 绕 $y\text{-axis}$ 旋转所得的旋转体的体积。

Solution 解答

First we must graph the region $R$ and the associated solid of revolution, as shown in the following figure.

首先，我们必须画出区域 $R$ 以及相应的旋转体，如下图所示。

Then the volume of the solid is given by

于是该立体的体积为

$$\begin{array}{cl} V & {= {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx} \\ & {= {\int_{1}^{3}{\left( {2\pi x\left( \frac{1}{x} \right)} \right)dx}}} \\ & {= {\int_{1}^{3}2}\pi\ dx = \left. {2\pi x} \right|_{1}^{3} = 4\pi\ \text{units}^{3}\text{.}} \end{array}$$

Define *R* as the region bounded above by the graph of $f(x) = x^{2}$ and below by the *x*-axis over the interval $\left\lbrack {1,2} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

定义 *R* 为区间 $\left\lbrack {1,2} \right\rbrack$ 上由函数 $f(x) = x^{2}$ 的图形为上界、以 *x* 轴为下界的区域。求将 $R$ 绕 $y\text{-axis}$ 旋转所得的旋转体的体积。

The Method of Cylindrical Shells 2 圆柱壳法 2

Define *R* as the region bounded above by the graph of $f(x) = 2x - x^{2}$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {0,2} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

定义 *R* 为区间 $\left\lbrack {0,2} \right\rbrack$ 上由函数 $f(x) = 2x - x^{2}$ 的图形为上界、以 $x\text{-axis}$ 为下界的区域。求将 $R$ 绕 $y\text{-axis}$ 旋转所得的旋转体的体积。

Solution 解答

First graph the region $R$ and the associated solid of revolution, as shown in the following figure.

首先画出区域 $R$ 及相应的旋转体，如下图所示。

Then the volume of the solid is given by

于是该立体的体积为

$$\begin{array}{cl} V & {= {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx} \\ & {= {\int_{0}^{2}\left( {2\pi x\left( {2x - x^{2}} \right)} \right)}dx = 2\pi{\int_{0}^{2}\left( {2x^{2} - x^{3}} \right)}dx} \\ & {= \left. {2\pi\left\lbrack {\frac{2x^{3}}{3} - \frac{x^{4}}{4}} \right\rbrack}\ \right|_{0}^{2} = \frac{8\pi}{3}\ \text{units}^{3}\text{.}} \end{array}$$

Define $R$ as the region bounded above by the graph of $f(x) = 3x - x^{2}$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {0,2} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

定义 $R$ 为区间 $\left\lbrack {0,2} \right\rbrack$ 上由函数 $f(x) = 3x - x^{2}$ 的图形为上界、以 $x\text{-axis}$ 为下界的区域。求将 $R$ 绕 $y\text{-axis}$ 旋转所得的旋转体的体积。

As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the $x\text{-axis},$ when we want to integrate with respect to $y.$ The analogous rule for this type of solid is given here.

与圆盘法和垫圈法一样，当我们想关于 $y$ 积分时，也可以将圆柱壳法用于绕 $x\text{-axis}$ 旋转的旋转体。这类立体的相应法则如下。

Let $g(y)$ be continuous and nonnegative. Define $Q$ as the region bounded on the right by the graph of $g(y),$ on the left by the $y\text{-axis},$ below by the line $y = c,$ and above by the line $y = d.$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $x\text{-axis}$ is given by

设 $g(y)$ 连续且非负。定义 $Q$ 为这样一个区域：其右界为 $g(y)$ 的图形，左界为 $y\text{-axis}$，下界为直线 $y = c$，上界为直线 $y = d.$ 那么，将 $Q$ 绕 $x\text{-axis}$ 旋转所得的旋转体的体积为

$$V = {\int_{c}^{d}\left( {2\pi yg(y)} \right)}dy.$$

The Method of Cylindrical Shells for a Solid Revolved around the x-axis 绕 x 轴旋转的立体的圆柱壳法

Define $Q$ as the region bounded on the right by the graph of $g(y) = 2\sqrt{y}$ and on the left by the $y\text{-axis}$ for $y \in \left\lbrack {0,4} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $Q$ around the *x*-axis.

定义 $Q$ 为 $y \in \left\lbrack {0,4} \right\rbrack$ 上右界为 $g(y) = 2\sqrt{y}$ 的图形、左界为 $y\text{-axis}$ 的区域。求将 $Q$ 绕 *x* 轴旋转所得的旋转体的体积。

Solution 解答

First, we need to graph the region $Q$ and the associated solid of revolution, as shown in the following figure.

首先，我们需要画出区域 $Q$ 及相应的旋转体，如下图所示。

Label the shaded region $Q.$ Then the volume of the solid is given by

标出阴影区域 $Q.$ 那么，该立体的体积为

$$\begin{array}{cl} V & {= {\int_{c}^{d}\left( {2\pi yg(y)} \right)}dy} \\ & {= {\int_{0}^{4}\left( {2\pi y\left( {2\sqrt{y}} \right)} \right)}dy = 4\pi{\int_{0}^{4}y^{3\text{/}2}}dy} \\ & {= {\left. {4\pi\left\lbrack \frac{2y^{5\text{/}2}}{5} \right.} \right\rbrack\left. \ \right|}_{0}^{4} = \frac{256\pi}{5}\ \text{units}^{3}\text{.}} \end{array}$$

Define $Q$ as the region bounded on the right by the graph of $g(y) = {3\text{/}y}$ and on the left by the $y\text{-axis}$ for $y \in \left\lbrack {1,3} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $Q$ around the *x*-axis.

定义 $Q$ 为 $y \in \left\lbrack {1,3} \right\rbrack$ 上右界为 $g(y) = {3\text{/}y}$ 的图形、左界为 $y\text{-axis}$ 的区域。求将 $Q$ 绕 *x* 轴旋转所得的旋转体的体积。

For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall that we found the volume of one of the shells to be given by

在下一个示例中，我们考虑一个旋转体，其函数图形绕两条坐标轴之外的某条直线旋转。为建立模型，我们需要回顾圆柱壳法的发展过程。回忆我们曾得到其中一个壳的体积为

This was based on a shell with an outer radius of $x_{i}$ and an inner radius of $x_{i - 1}.$ If, however, we rotate the region around a line other than the $y\text{-axis},$ we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line $x = \text{−}k,$ where $k$ is some positive constant. Then, the outer radius of the shell is $x_{i} + k$ and the inner radius of the shell is $x_{i - 1} + k.$ Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line $x = \text{−}k,$ the volume of a shell is given by

这是基于外半径为 $x_{i}$、内半径为 $x_{i - 1}$ 的壳。然而，如果我们把该区域绕 $y\text{-axis}$ 以外的直线旋转，就会得到不同的外半径与内半径。例如，假设我们把该区域绕直线 $x = \text{−}k$ 旋转，其中 $k$ 是某个正常数。那么，该壳的外半径为 $x_{i} + k$，内半径为 $x_{i - 1} + k.$ 将这些项代入体积表达式中，我们看到，当一个平面区域绕直线 $x = \text{−}k$ 旋转时，一个壳的体积为

$$\begin{array}{cl} V_{\text{shell}} & {= 2\pi f(x_{i}^{*})\left( \frac{\left( {x_{i} + k} \right) + \left( {x_{i - 1} + k} \right)}{2} \right)\left( {\left( {x_{i} + k} \right) - \left( {x_{i - 1} + k} \right)} \right)} \\ & {= 2\pi f(x_{i}^{*})\left( {\left( \frac{x_{i} + x_{i - 1}}{2} \right) + k} \right)\text{Δ}x.} \end{array}$$

As before, we notice that $\frac{x_{i} + x_{i - 1}}{2}$ is the midpoint of the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ and can be approximated by $x_{i}^{*}.$ Then, the approximate volume of the shell is

与之前一样，我们注意到 $\frac{x_{i} + x_{i - 1}}{2}$ 是区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 的中点，并可用 $x_{i}^{*}$ 来近似。于是，该壳的近似体积为

$$V_{\text{shell}} \approx 2\pi\left( {x_{i}^{*} + k} \right)f(x_{i}^{*})\text{Δ}x.$$

The remainder of the development proceeds as before, and we see that

余下的推导与之前相同，我们得到

$$V = {\int_{a}^{b}\left( {2\pi\left( {x + k} \right)f(x)} \right)}dx.$$

We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the $x\text{-term}$ in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.

我们也可以把该区域绕其他水平或竖直直线旋转，例如右半平面中的一条竖直直线。在每种情形下，体积公式都必须作相应调整。具体而言，积分中的 $x\text{-term}$ 必须换成一个表示壳半径的表达式。要理解其运作方式，请看下面的示例。

A Region of Revolution Revolved around a Line 绕一条直线旋转的旋转区域

Define $R$ as the region bounded above by the graph of $f(x) = x$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {1,2} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the line $x = -1.$

定义 $R$ 为区间 $\left\lbrack {1,2} \right\rbrack$ 上由函数 $f(x) = x$ 的图形为上界、以 $x\text{-axis}$ 为下界的区域。求将 $R$ 绕直线 $x = -1$ 旋转所得的旋转体的体积。

Solution 解答

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

首先，画出区域 $R$ 及相应的旋转体，如下图所示。

Note that the radius of a shell is given by $x + 1.$ Then the volume of the solid is given by

注意，一个壳的半径由 $x + 1$ 给出。于是该立体的体积为

$$\begin{array}{cl} V & {= {\int_{1}^{2}\left( {2\pi\left( {x + 1} \right)f(x)} \right)}dx} \\ & {= {\int_{1}^{2}\left( {2\pi\left( {x + 1} \right)x} \right)}dx = 2\pi{\int_{1}^{2}\left( {x^{2} + x} \right)}dx} \\ & {= \left. {2\pi\left\lbrack {\frac{x^{3}}{3} + \frac{x^{2}}{2}} \right\rbrack}\ \right|_{1}^{2} = \frac{23\pi}{3}\ \text{units}^{3}\text{.}} \end{array}$$

Define $R$ as the region bounded above by the graph of $f(x) = x^{2}$ and below by the $x\text{-axis}$ over the interval $\left\lbrack {0,1} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the line $x = -2.$

定义 $R$ 为区间 $\left\lbrack {0,1} \right\rbrack$ 上由函数 $f(x) = x^{2}$ 的图形为上界、以 $x\text{-axis}$ 为下界的区域。求将 $R$ 绕直线 $x = -2$ 旋转所得的旋转体的体积。

For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.

在本节的最后一个示例中，我们来看一个旋转体，其旋转区域由两个函数的图形所界定。

A Region of Revolution Bounded by the Graphs of Two Functions 由两个函数图形所界定的旋转区域

Define $R$ as the region bounded above by the graph of the function $f(x) = \sqrt{x}$ and below by the graph of the function $g(x) = {1\text{/}x}$ over the interval $\left\lbrack {1,4} \right\rbrack.$ Find the volume of the solid of revolution generated by revolving $R$ around the $y\text{-axis}.$

定义 $R$ 为区间 $\left\lbrack {1,4} \right\rbrack$ 上由函数 $f(x) = \sqrt{x}$ 的图形为上界、由函数 $g(x) = {1\text{/}x}$ 的图形为下界的区域。求将 $R$ 绕 $y\text{-axis}$ 旋转所得的旋转体的体积。

Solution 解答

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

首先，画出区域 $R$ 及相应的旋转体，如下图所示。

Note that the axis of revolution is the $y\text{-axis},$ so the radius of a shell is given simply by $x.$ We don’t need to make any adjustments to the *x*-term of our integrand. The height of a shell, though, is given by $f(x) - g(x),$ so in this case we need to adjust the $f(x)$ term of the integrand. Then the volume of the solid is given by

注意，旋转轴是 $y\text{-axis}$，因此一个壳的半径简单地由 $x$ 给出。我们无需对积分式的 *x* 项作任何调整。然而，一个壳的高由 $f(x) - g(x)$ 给出，因此在这种情况下，我们需要调整积分式中的 $f(x)$ 项。于是该立体的体积为

$$\begin{array}{cl} V & {= {\int_{1}^{4}\left( {2\pi x\left( {f(x) - g(x)} \right)} \right)}dx} \\ & {= {\int_{1}^{4}{\left( {2\pi x\left( {\sqrt{x} - \frac{1}{x}} \right)} \right)dx}} = 2\pi{\int_{1}^{4}\left( {x^{3\text{/}2} - 1} \right)}dx} \\ & {= \left. {2\pi\left\lbrack {\frac{2x^{5\text{/}2}}{5} - x} \right\rbrack}\ \right|_{1}^{4} = \frac{94\pi}{5}\ \text{units}^{3}.} \end{array}$$

Define $R$ as the region bounded above by the graph of $f(x) = x$ and below by the graph of $g(x) = x^{2}$ over the interval $\left\lbrack {0,1} \right\rbrack.$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

定义 $R$ 为区间 $\left\lbrack {0,1} \right\rbrack$ 上由函数 $f(x) = x$ 的图形为上界、由函数 $g(x) = x^{2}$ 的图形为下界的区域。求将 $R$ 绕 $y\text{-axis}$ 旋转所得的旋转体的体积。

Which Method Should We Use? 该用哪种方法？

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. Figure 2.34 describes the different approaches for solids of revolution around the $x\text{-axis}.$ It’s up to you to develop the analogous table for solids of revolution around the $y\text{-axis}.$

我们已经学习了几种求旋转体体积的方法，但怎样才能知道该用哪一种呢？这往往归结为选择最容易计算的积分。图 2.34 描述了绕 $x\text{-axis}$ 旋转的旋转体的不同处理方法。绕 $y\text{-axis}$ 旋转的旋转体的对应表格，就留给你来完成了。

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

我们来看几个额外的问题，并决定求解它们的最佳方法。

Selecting the Best Method 选择最佳方法

For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the $x\text{-axis},$ and set up the integral to find the volume (do not evaluate the integral).

对于下列各个问题，选择求旋转体体积的最佳方法——该旋转体由给定区域绕 $x\text{-axis}$ 旋转生成——并建立求体积的积分（不要计算该积分）。

1. The region bounded by the graphs of $y = x,$ $y = 2 - x,$ and the $x\text{-axis}.$

1. 由 $y = x$、$y = 2 - x$ 与 $x\text{-axis}$ 的图形所界的区域。

2. The region bounded by the graphs of $y = 4x - x^{2}$ and the $x\text{-axis}.$

2. 由 $y = 4x - x^{2}$ 与 $x\text{-axis}$ 的图形所界的区域。

Solution 解答

1. First, sketch the region and the solid of revolution as shown.

1. 首先，画出该区域与旋转体，如图所示。

Looking at the region, if we want to integrate with respect to $x,$ we would have to break the integral into two pieces, because we have different functions bounding the region over $\left\lbrack {0,1} \right\rbrack$ and $\left\lbrack {1,2} \right\rbrack.$ In this case, using the disk method, we would have

观察该区域，如果我们想关于 $x$ 积分，就不得不把积分拆成两段，因为在区间 $\left\lbrack {0,1} \right\rbrack$ 与 $\left\lbrack {1,2} \right\rbrack$ 上界定该区域的函数不同。在这种情况下，若使用圆盘法，我们会得到

$$V = {\int_{0}^{1}\left( {\pi x^{2}} \right)}dx + {\int_{1}^{2}\left( {\pi{(2 - x)}^{2}} \right)}dx.$$

If we used the shell method instead, we would use functions of $y$ to represent the curves, producing

如果改用壳法，我们会用关于 $y$ 的函数来表示曲线，从而得到

$$\begin{array}{cl} V & {= {\int_{0}^{1}\left( {2\pi y\left\lbrack {\left( {2 - y} \right) - y} \right\rbrack} \right)}dy} \\ & {= {\int_{0}^{1}\left( {2\pi y\left\lbrack {2 - 2y} \right\rbrack} \right)}dy.} \end{array}$$

Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.

这两个积分都不算特别麻烦，但由于壳法只需要一个积分，且被积函数所需的化简更少，在此情形下我们大概应当选择壳法。

2. First, sketch the region and the solid of revolution as shown.

2. 首先，画出该区域与旋转体，如图所示。

Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then

观察该区域，用一个水平矩形来界定它会很麻烦；该区域的左右两侧都由同一个函数界定。因此，我们可以排除壳法。该立体中间没有空腔，所以我们可以使用圆盘法。于是

$$V = {\int_{0}^{4}\pi}\left( {4x - x^{2}} \right)^{2}dx.$$

Select the best method to find the volume of a solid of revolution generated by revolving the given region around the $x\text{-axis},$ and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of $y = 2 - x^{2}$ and $y = x^{2}.$

选择求旋转体体积的最佳方法——该旋转体由给定区域绕 $x\text{-axis}$ 旋转生成——并建立求体积的积分（不要计算该积分）：由 $y = 2 - x^{2}$ 与 $y = x^{2}$ 的图形所界的区域。

Section 2.3 Exercises 2.3 节习题

For the following exercises, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand.

在以下习题中，求由两条曲线之间的区域绕给定轴旋转所产生的体积。同时使用圆柱壳法和垫圈法。利用技术工具绘制函数图像，并手绘一个典型切片。

114\.

\[T\] Bounded by the curves $y = 3x,x = 0,$ and $y = 3$ rotated around the $y\text{-axis}.$

\[T\] 由曲线 $y = 3x,x = 0,$ 与 $y = 3$ 所围成，绕 $y\text{-axis}.$ 旋转。

115.

\[T\] Bounded by the curves $y = 3x,y = 0,\ \text{and}\ x = 3$ rotated around the $y\text{-axis}.$

\[T\] 由曲线 $y = 3x,y = 0,\ \text{and}\ x = 3$ 所围成，绕 $y\text{-axis}.$ 旋转。

116\.

\[T\] Bounded by the curves $y = 3x,y = 0,\ \text{and}\ y = 3$ rotated around the $x\text{-axis}.$

\[T\] 由曲线 $y = 3x,y = 0,\ \text{and}\ y = 3$ 所围成，绕 $x\text{-axis}.$ 旋转。

117.

\[T\] Bounded by the curves $y = 3x,y = 0,\ \text{and}\ x = 3$ rotated around the $x\text{-axis}.$

\[T\] 由曲线 $y = 3x,y = 0,\ \text{and}\ x = 3$ 所围成，绕 $x\text{-axis}.$ 旋转。

118\.

\[T\] Bounded by the curves $y = 2x^{3},y = 0,\ \text{and}\ x = 2$ rotated around the $y\text{-axis}.$

\[T\] 由曲线 $y = 2x^{3},y = 0,\ \text{and}\ x = 2$ 所围成，绕 $y\text{-axis}.$ 旋转。

119.

\[T\] Bounded by the curves $y = 2x^{3},y = 0,\ \text{and}\ x = 2$ rotated around the $x\text{-axis}.$

\[T\] 由曲线 $y = 2x^{3},y = 0,\ \text{and}\ x = 2$ 所围成，绕 $x\text{-axis}.$ 旋转。

For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the $x\text{-axis}$ and are rotated around the $y\text{-axis}.$

在以下习题中，使用圆柱壳法求给定立体的体积。注意，旋转区域位于曲线与 $x\text{-axis}$ 之间，并绕 $y\text{-axis}.$ 旋转。

120\.

$y = 1 - x^{2},x = 0,\ \text{and}\ x = 1$

121.

$y = 5x^{3},x = 0,\ \text{and}\ x = 1$

122\.

$y = \frac{1}{x},x = 1,\ \text{and}\ x = 100$

123.

$y = \sqrt{1 - x^{2}},x = 0,\ \text{and}\ x = 1$

124\.

$y = \frac{1}{1 + x^{2}},x = 0,\ \text{and}\ x = 3$

125.

$y = \text{sin}x^{2},x = 0,\ \text{and}\ x = \sqrt{\pi}$

126\.

$y = \frac{1}{\sqrt{1 - x^{2}}},x = 0,\ \text{and}\ x = \frac{1}{2}$

127.

$y = \sqrt{x},x = 0,\ \text{and}\ x = 1$

128\.

$y = \left( {1 + x^{2}} \right)^{3},x = 0,\ \text{and}\ x = 1$

129.

$y = 5x^{3} - 2x^{4},x = 0,\ \text{and}\ x = 2$

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and $y = 0$ around the $x\text{-axis}.$

在以下习题中，使用圆柱壳法求由给定曲线与 $y = 0$ 之间的区域绕 $x\text{-axis}.$ 旋转所产生的体积。

130\.

$y = \sqrt{1 - x^{2}},x = 0,\ x = 1$ and the *x*-axis

$y = \sqrt{1 - x^{2}},x = 0,\ x = 1$ 以及 *x* 轴

131.

$y = x^{2},x = 0,\ x = 2$ and the *x*-axis

$y = x^{2},x = 0,\ x = 2$ 以及 *x* 轴

132\.

$y = \frac{x^{3}}{2},\ x = 0,\ x = 2,$ and the *x*-axis

$y = \frac{x^{3}}{2},\ x = 0,\ x = 2,$ 以及 *x* 轴

133.

$y = \frac{2}{x^{2}},\ x = 1,\ x = 2,$ and the *x*-axis

$y = \frac{2}{x^{2}},\ x = 1,\ x = 2,$ 以及 *x* 轴

134\.

$x = \frac{1}{1 + y^{2}},y = 4$

135.

$x = \frac{1 + y^{2}}{y},y = 1,\ y = 4,$ and the *y*-axis

$x = \frac{1 + y^{2}}{y},y = 1,\ y = 4,$ 以及 *y* 轴

136\.

$x = \sqrt{4 - y^{2}}\text{,}x = 0\text{,}y = 0$

137.

$x = y^{3} - 2y^{2},\ x = 0,\ x = 9$

138\.

$x = \sqrt{y} + 1,\ x = 1,\ x = 3,$ and the *x*-axis

$x = \sqrt{y} + 1,\ x = 1,\ x = 3,$ 以及 *x* 轴

139.

$x = \sqrt[3]{27y}\text{and}\ x = \frac{3y}{4}$

For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.

在以下习题中，求由曲线之间的区域绕给定轴旋转所产生的体积。

140\.

$y = 3 - x,y = 0,x = 0,\ \text{and}\ x = 2$ rotated around the $y\text{-axis}.$

$y = 3 - x,y = 0,x = 0,\ \text{and}\ x = 2$ 绕 $y\text{-axis}.$ 旋转。

141.

$y = x^{3},x = 0,\ \text{and}\ y = 8$ rotated around the $y\text{-axis}.$

$y = x^{3},x = 0,\ \text{and}\ y = 8$ 绕 $y\text{-axis}.$ 旋转。

142\.

$y = x^{2},y = x,$ rotated around the $y\text{-axis}.$

$y = x^{2},y = x,$ 绕 $y\text{-axis}.$ 旋转。

143.

$y = \sqrt{x},y = 0,\ \text{and}\ x = 1$ rotated around the line $x = 2.$

$y = \sqrt{x},y = 0,\ \text{and}\ x = 1$ 绕直线 $x = 2.$ 旋转。

144\.

$y = \frac{1}{4 - x},x = 1,\ x = 2\ \text{and}\ y = 0$ rotated around the line $x = 4.$

$y = \frac{1}{4 - x},x = 1,\ x = 2\ \text{and}\ y = 0$ 绕直线 $x = 4.$ 旋转。

145.

$y = \sqrt{x}\ \text{and}\ y = x^{2}$ rotated around the $y\text{-axis}.$

$y = \sqrt{x}\ \text{and}\ y = x^{2}$ 绕 $y\text{-axis}.$ 旋转。

146\.

$y = \sqrt{x}\ \text{and}\ y = x^{2}$ rotated around the line $x = 2.$

$y = \sqrt{x}\ \text{and}\ y = x^{2}$ 绕直线 $x = 2.$ 旋转。

147.

$x = y^{3},x = \frac{1}{y},x = 1,\ \text{and}\ x = 2$ rotated around the $x\text{-axis}.$

$x = y^{3},x = \frac{1}{y},x = 1,\ \text{and}\ x = 2$ 绕 $x\text{-axis}.$ 旋转。

148\.

$x = y^{2}\ \text{and}\ y = x$ rotated around the line $y = 2.$

$x = y^{2}\ \text{and}\ y = x$ 绕直线 $y = 2.$ 旋转。

149.

\[T\] Left of $x = \text{sin}\left( {\pi y} \right),$ right of $y = x,$ around the $y\text{-axis}.$

\[T\] 以 $x = \text{sin}\left( {\pi y} \right)$ 为左边界、以 $y = x$ 为右边界，绕 $y\text{-axis}.$ 旋转。

For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume.

在以下习题中，利用技术工具绘制该区域。确定你认为哪种方法最容易用于计算函数绕指定轴旋转所产生的体积。然后，使用你选择的方法求体积。

150\.

\[T\] $y = x^{2}$ and $y = 4x$ rotated around the $y\text{-axis}.$

\[T\] $y = x^{2}$ 与 $y = 4x$ 绕 $y\text{-axis}.$ 旋转。

151.

\[T\] $y = \text{cos}\left( {\pi x} \right),y = \text{sin}\left( {\pi x} \right),x = \frac{1}{4},\ \text{and}\ x = \frac{5}{4}$ rotated around the $y\text{-axis}.$ This exercise requires advanced technique. You may use technology to perform the integration.

\[T\] $y = \text{cos}\left( {\pi x} \right),y = \text{sin}\left( {\pi x} \right),x = \frac{1}{4},\ \text{and}\ x = \frac{5}{4}$ 绕 $y\text{-axis}.$ 旋转。本习题需要高阶技巧。你可以使用技术工具来执行积分。

152\.

\[T\] $y = x^{2} - 2x,x = 2,\ \text{and}\ x = 4$ rotated around the $y\text{-axis}.$

\[T\] $y = x^{2} - 2x,x = 2,\ \text{and}\ x = 4$ 绕 $y\text{-axis}.$ 旋转。

153.

\[T\] $y = x^{2} - 2x,x = 2,\ \text{and}\ x = 4$ rotated around the $x\text{-axis}.$

\[T\] $y = x^{2} - 2x,x = 2,\ \text{and}\ x = 4$ 绕 $x\text{-axis}.$ 旋转。

154\.

\[T\] $y = 3x^{3} - 2,y = x,\ \text{and}\ x = 2$ rotated around the $x\text{-axis}.$

\[T\] $y = 3x^{3} - 2,y = x,\ \text{and}\ x = 2$ 绕 $x\text{-axis}.$ 旋转。

155.

\[T\] $y = 3x^{3} - 2,y = x,\ \text{and}\ x = 2$ rotated around the $y\text{-axis}.$

\[T\] $y = 3x^{3} - 2,y = x,\ \text{and}\ x = 2$ 绕 $y\text{-axis}.$ 旋转。

156\.

\[T\] $x = \text{sin}\left( {\pi y^{2}} \right)$ and $x = \sqrt{2}y$ rotated around the $x\text{-axis}.$

\[T\] $x = \text{sin}\left( {\pi y^{2}} \right)$ 与 $x = \sqrt{2}y$ 绕 $x\text{-axis}.$ 旋转。

157.

\[T\] $x = y^{2},x = y^{2} - 2y + 1,\ \text{and}\ x = 2$ rotated around the $y\text{-axis}.$

\[T\] $x = y^{2},x = y^{2} - 2y + 1,\ \text{and}\ x = 2$ 绕 $y\text{-axis}.$ 旋转。

For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.

在以下习题中，使用圆柱壳法来近似求一些常见物体的体积，这些物体如附图所示。

158\.

Use the method of shells to find the volume of a sphere of radius $r.$

使用圆柱壳法求半径为 $r$ 的球体的体积。

159.

Use the method of shells to find the volume of a cone with radius $r$ and height $h.$

使用圆柱壳法求半径为 $r$、高为 $h$ 的圆锥体的体积。

160\.

Use the method of shells to find the volume of an ellipsoid $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1$ rotated around the $x\text{-axis}.$

使用圆柱壳法求椭球面 $\left( {x^{2}\text{/}a^{2}} \right) + \left( {y^{2}\text{/}b^{2}} \right) = 1$ 绕 $x\text{-axis}.$ 旋转所得立体的体积。

161.

Use the method of shells to find the volume of a cylinder with radius $r$ and height $h.$

使用圆柱壳法求半径为 $r$、高为 $h$ 的圆柱体的体积。

162\.

Use the method of shells to find the volume of the donut created when the circle $x^{2} + y^{2} = 4$ is rotated around the line $x = 4.$

使用圆柱壳法求圆 $x^{2} + y^{2} = 4$ 绕直线 $x = 4$ 旋转所成环面（甜甜圈形）的体积。

163.

Consider the region enclosed by the graphs of $y = f(x),y = 1 + f(x),x = 0,y = 0,$ and $x = a > 0.$ What is the volume of the solid generated when this region is rotated around the $y\text{-axis}?$ Assume that the function is defined over the interval $\lbrack 0,a\rbrack.$

考虑由 $y = f(x),y = 1 + f(x),x = 0,y = 0,$ 与 $x = a > 0$ 的图像所围成的区域。当该区域绕 $y\text{-axis}?$ 旋转时，所生成的立体体积为多少？假设该函数定义在区间 $\lbrack 0,a\rbrack$ 上。

164\.

Consider the function $y = f(x),$ which decreases from $f(0) = b$ to $f(1) = 0.$ Set up the integrals for determining the volume, using both the shell method and the disk method, of the solid generated when this region, with $x = 0$ and $y = 0,$ is rotated around the $y\text{-axis}.$ Prove that both methods approximate the same volume. Which method is easier to apply? (*Hint:* Since $f(x)$ is one-to-one, there exists an inverse $f^{-1}(y).)$

考虑函数 $y = f(x)$，它从 $f(0) = b$ 单调递减到 $f(1) = 0$。分别使用圆柱壳法和圆盘法，建立定积分来表示当该区域（边界为 $x = 0$ 与 $y = 0$）绕 $y\text{-axis}.$ 旋转时所生成立体的体积。证明两种方法逼近的是同一体积。哪种方法更容易应用？（*提示：*由于 $f(x)$ 是一一对应的，故存在反函数 $f^{-1}(y)$。）

2.4 Arc Length of a Curve and Surface Area 2.4 曲线的弧长与表面积

2.4.1 Determine the length of a curve, $y = f(x),$ between two points.
2.4.2 Determine the length of a curve, $x = g(y),$ between two points.
2.4.3 Find the surface area of a solid of revolution.

2.4.1 确定曲线 $y = f(x)$ 在两点之间的长度。
2.4.2 确定曲线 $x = g(y)$ 在两点之间的长度。
2.4.3 求旋转体的表面积。

In this section, we use definite integrals to find the arc length of a curve. We can think of arc length as the distance you would travel if you were walking along the path of the curve. Many real-world applications involve arc length. If a rocket is launched along a parabolic path, we might want to know how far the rocket travels. Or, if a curve on a map represents a road, we might want to know how far we have to drive to reach our destination.

在本节中，我们使用定积分来求曲线的弧长。我们可以把弧长想象成沿着曲线路径行走时所经过的距离。许多现实世界的应用都涉及弧长。如果一枚火箭沿抛物线轨道发射，我们可能想知道火箭飞行了多远。或者，如果地图上的某条曲线代表一条道路，我们可能想知道需要驾驶多远才能到达目的地。

We begin by calculating the arc length of curves defined as functions of $x,$ then we examine the same process for curves defined as functions of $y.$ (The process is identical, with the roles of $x$ and $y$ reversed.) The techniques we use to find arc length can be extended to find the surface area of a surface of revolution, and we close the section with an examination of this concept.

我们首先计算定义为 $x$ 的函数的曲线的弧长，然后研究定义为 $y$ 的函数的曲线的同一过程。（该过程完全相同，只是 $x$ 与 $y$ 的角色互换。）我们用来求弧长的技巧可以推广到求旋转面的表面积，我们在本节末尾将考察这一概念。

Arc Length of the Curve y = f(x) 曲线 y = f(x) 的弧长

In previous applications of integration, we required the function $f(x)$ to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for $f(x).$ Here, we require $f(x)$ to be differentiable, and furthermore we require its derivative, $f^{\prime}(x),$ to be continuous. Functions like this, which have continuous derivatives, are called *smooth*. (This property comes up again in later chapters.)

在前面积分的应用中，我们要求函数 $f(x)$ 可积，或至多连续。然而，对于计算弧长，我们对 $f(x)$ 有更严格的要求。这里，我们要求 $f(x)$ 可微，并且进一步要求其导数 $f^{\prime}(x)$ 连续。像这样具有连续导数的函数称为 *smooth*（光滑）。（这一性质在后面的章节中会再次出现。）

Let $f(x)$ be a smooth function defined over $\left\lbrack {a,b} \right\rbrack.$ We want to calculate the length of the curve from the point $\left( {a,f(a)} \right)$ to the point $\left( {b,f(b)} \right).$ We start by using line segments to approximate the length of the curve. For $i = 0,\ 1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of $\left\lbrack {a,b} \right\rbrack.$ Then, for $i = 1,2\text{,…},n,$ construct a line segment from the point $\left( {x_{i - 1},f(x_{i - 1})} \right)$ to the point $\left( {x_{i},f(x_{i})} \right).$ Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 2.37 depicts this construct for $n = 5.$

设 $f(x)$ 是定义在区间 $\left\lbrack {a,b} \right\rbrack$ 上的光滑函数。我们想计算从点 $\left( {a,f(a)} \right)$ 到点 $\left( {b,f(b)} \right)$ 的曲线长度。我们首先用线段来逼近曲线的长度。对 $i = 0,\ 1,2\text{,…},n,$ 设 $P = \left\{ x_{i} \right\}$ 是 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割。然后，对 $i = 1,2\text{,…},n,$ 从点 $\left( {x_{i - 1},f(x_{i - 1})} \right)$ 向点 $\left( {x_{i},f(x_{i})} \right)$ 构造一条线段。尽管使用水平或垂直线段看似合理，但我们希望线段尽可能逼近曲线。图 2.37 描绘了对 $n = 5$ 时的这一构造。

To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by $\text{Δ}x.$ The change in vertical distance varies from interval to interval, though, so we use $\text{Δ}y_{i} = f(x_{i}) - f(x_{i - 1})$ to represent the change in vertical distance over the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack,$ as shown in Figure 2.38. Note that some (or all) $\text{Δ}y_{i}$ may be negative.

为了帮助我们求每条线段的长度，我们考察每个区间上垂直距离的变化以及水平距离的变化。由于我们使用了正则分割，每个区间上水平距离的变化由 $\text{Δ}x$ 给出。不过，垂直距离的变化因区间而异，因此我们用 $\text{Δ}y_{i} = f(x_{i}) - f(x_{i - 1})$ 表示区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 上的垂直距离变化，如图 2.38 所示。注意某些（或全部）$\text{Δ}y_{i}$ 可能为负。

By the Pythagorean theorem, the length of the line segment is $\sqrt{\left( {\text{Δ}x} \right)^{2} + \left( {\text{Δ}y_{i}} \right)^{2}}.$ We can also write this as $\text{Δ}x\sqrt{1 + \left( {\left( {\text{Δ}y_{i}} \right)\text{/}\left( {\text{Δ}x} \right)} \right)^{2}}.$ Now, by the Mean Value Theorem, there is a point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ such that $f^{\prime}(x_{i}^{*}) = {\left( {\text{Δ}y_{i}} \right)\text{/}\left( {\text{Δ}x} \right)}.$ Then the length of the line segment is given by $\text{Δ}x\sqrt{1 + \left\lbrack {f^{\prime}(x_{i}^{*})} \right\rbrack^{2}}.$ Adding up the lengths of all the line segments, we get

由勾股定理，该线段的长度为 $\sqrt{\left( {\text{Δ}x} \right)^{2} + \left( {\text{Δ}y_{i}} \right)^{2}}.$ 我们也可以将其写成 $\text{Δ}x\sqrt{1 + \left( {\left( {\text{Δ}y_{i}} \right)\text{/}\left( {\text{Δ}x} \right)} \right)^{2}}.$ 现在，由中值定理，存在一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 使得 $f^{\prime}(x_{i}^{*}) = {\left( {\text{Δ}y_{i}} \right)\text{/}\left( {\text{Δ}x} \right)}.$ 于是该线段的长度由 $\text{Δ}x\sqrt{1 + \left\lbrack {f^{\prime}(x_{i}^{*})} \right\rbrack^{2}}.$ 给出。将所有线段的长度相加，我们得到

$$\text{Arc Length}\ \approx {\sum\limits_{i = 1}^{n}{\sqrt{1 + \left\lbrack {f^{\prime}(x_{i}^{*})} \right\rbrack^{2}}\ \text{Δ}x}}.$$

This is a Riemann sum. Taking the limit as $n\rightarrow\infty,$ we have

这是一个黎曼和。令 $n\rightarrow\infty,$ 取极限，我们得到

$$\text{Arc Length} = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}{\sqrt{1 + \left\lbrack {f^{\prime}(x_{i}^{*})} \right\rbrack^{2}}\ \text{Δ}x}} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx.$$

We summarize these findings in the following theorem.

我们将这些结论总结于以下定理。

Arc Length for y = f(x) 曲线 y = f(x) 的弧长

Let $f(x)$ be a smooth function over the interval $\left\lbrack {a,b} \right\rbrack.$ Then the arc length of the portion of the graph of $f(x)$ from the point $\left( {a,f(a)} \right)$ to the point $\left( {b,\ f(b)} \right)$ is given by

设 $f(x)$ 是区间 $\left\lbrack {a,b} \right\rbrack$ 上的光滑函数。则 $f(x)$ 的图像从点 $\left( {a,f(a)} \right)$ 到点 $\left( {b,\ f(b)} \right)$ 那一段的弧长由下式给出

$$\text{Arc Length} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx.$$ (2.7)

Note that we are integrating an expression involving $f^{\prime}(x),$ so we need to be sure $f^{\prime}(x)$ is integrable. This is why we require $f(x)$ to be smooth. The following example shows how to apply the theorem.

注意，我们积分的是一个含有 $f^{\prime}(x)$ 的表达式，因此必须确保 $f^{\prime}(x)$ 可积。这正是我们要求 $f(x)$ 光滑的原因。下面的示例展示了如何应用该定理。

Calculating the Arc Length of a Function of x 计算关于 x 的函数的弧长

Let $f(x) = 2x^{3\text{/}2}.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left\lbrack {0,1} \right\rbrack.$ Round the answer to three decimal places.

设 $f(x) = 2x^{3\text{/}2}.$ 计算 $f(x)$ 的图像在区间 $\left\lbrack {0,1} \right\rbrack$ 上的弧长。将答案四舍五入到三位小数。

Solution 解答

We have $f^{\prime}(x) = 3x^{1\text{/}2},$ so $\left\lbrack {f^{\prime}(x)} \right\rbrack^{2} = 9x.$ Then, the arc length is

我们有 $f^{\prime}(x) = 3x^{1\text{/}2},$ 因此 $\left\lbrack {f^{\prime}(x)} \right\rbrack^{2} = 9x.$ 于是弧长为

$$\begin{array}{cl} \text{Arc Length} & {= {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx} \\ & {= {\int_{0}^{1}\sqrt{1 + 9x}}\ dx.} \end{array}$$

Substitute $u = 1 + 9x.$ Then, $du = 9\ dx.$ When $x = 0,$ then $u = 1,$ and when $x = 1,$ then $u = 10.$ Thus,

令 $u = 1 + 9x.$ 则 $du = 9\ dx.$ 当 $x = 0$ 时，$u = 1$；当 $x = 1$ 时，$u = 10.$ 因此，

$$\begin{array}{cl} \text{Arc Length} & {= {\int_{0}^{1}\sqrt{1 + 9x}}\ dx} \\ & {= \frac{1}{9}\int_{0}^{1}\sqrt{1 + 9x}9dx = \frac{1}{9}{\int_{1}^{10}\sqrt{u}}\ du} \\ & {= \left. {\frac{1}{9} \cdot \frac{2}{3}u^{3\text{/}2}} \right|_{1}^{10} = \frac{2}{27}\left\lbrack {10\sqrt{10} - 1} \right\rbrack \approx 2.268\ \text{units}.} \end{array}$$

Let $f(x) = \left( {4\text{/}3} \right)x^{3\text{/}2}.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left\lbrack {0,1} \right\rbrack.$ Round the answer to three decimal places.

设 $f(x) = \left( {4\text{/}3} \right)x^{3\text{/}2}.$ 计算 $f(x)$ 的图像在区间 $\left\lbrack {0,1} \right\rbrack$ 上的弧长。将答案四舍五入到三位小数。

Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration. In some cases, we may have to use a computer or calculator to approximate the value of the integral.

尽管拥有一个计算弧长的公式很方便，但这个特定定理可能会产生难以积分的表达式。我们将在《积分技巧导论》中学习一些积分技巧。在某些情况下，我们可能必须借助计算机或计算器来近似积分的值。

Using a Computer or Calculator to Determine the Arc Length of a Function of x 使用计算机或计算器确定关于 x 的函数的弧长

Let $f(x) = x^{2}.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left\lbrack {1,3} \right\rbrack.$

设 $f(x) = x^{2}.$ 计算 $f(x)$ 的图像在区间 $\left\lbrack {1,3} \right\rbrack$ 上的弧长。

Solution 解答

We have $f^{\prime}(x) = 2x,$ so $\left\lbrack {f^{\prime}(x)} \right\rbrack^{2} = 4x^{2}.$ Then the arc length is given by

我们有 $f^{\prime}(x) = 2x,$ 因此 $\left\lbrack {f^{\prime}(x)} \right\rbrack^{2} = 4x^{2}.$ 于是弧长由下式给出

$$\text{Arc Length} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx = {\int_{1}^{3}\sqrt{1 + 4x^{2}}}\ dx.$$

Using a computer to approximate the value of this integral, we get

使用计算机来近似该积分的值，我们得到

$${\int_{1}^{3}\sqrt{1 + 4x^{2}}}\ dx \approx 8.26815.$$

Let $f(x) = \text{sin}\ x.$ Calculate the arc length of the graph of $f(x)$ over the interval $\left\lbrack {0,\pi} \right\rbrack.$ Use a computer or calculator to approximate the value of the integral.

设 $f(x) = \text{sin}\ x.$ 计算 $f(x)$ 的图像在区间 $\left\lbrack {0,\pi} \right\rbrack$ 上的弧长。使用计算机或计算器来近似该积分的值。

Arc Length of the Curve x = g(y) 曲线 x = g(y) 的弧长

We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of $y,$ we can repeat the same process, except we partition the $y\text{-axis}$ instead of the $x\text{-axis}.$ Figure 2.39 shows a representative line segment.

我们刚刚看到了如何用线段来逼近曲线的长度。如果我们想求一个关于 $y$ 的函数的图像弧长，可以重复同样的过程，只不过我们是对 $y\text{-axis}$ 而不是 $x\text{-axis}$ 进行分割。图 2.39 显示了一条具有代表性的线段。

Then the length of the line segment is $\sqrt{\left( {\text{Δ}y} \right)^{2} + \left( {\text{Δ}x_{i}} \right)^{2}},$ which can also be written as $\text{Δ}y\sqrt{1 + \left( {\left( {\text{Δ}x_{i}} \right)\text{/}\left( {\text{Δ}y} \right)} \right)^{2}}.$ If we now follow the same development we did earlier, we get a formula for arc length of a function $x = g(y).$

于是该线段的长度为 $\sqrt{\left( {\text{Δ}y} \right)^{2} + \left( {\text{Δ}x_{i}} \right)^{2}},$ 也可以写成 $\text{Δ}y\sqrt{1 + \left( {\left( {\text{Δ}x_{i}} \right)\text{/}\left( {\text{Δ}y} \right)} \right)^{2}}.$ 如果我们现在按照先前所做的同样推导，就可以得到关于函数 $x = g(y)$ 的弧长公式。

Arc Length for x = g(y) 曲线 x = g(y) 的弧长

Let $g(y)$ be a smooth function over a $y$ interval $\left\lbrack {c,d} \right\rbrack.$ Then, the arc length of the graph of $g(y)$ from the point $\left( g(d),~d \right)$ to the point $\left( g(c),~c \right)$ is given by

设 $g(y)$ 是 $y$ 区间 $\left\lbrack {c,d} \right\rbrack$ 上的光滑函数。则 $g(y)$ 的图像从点 $\left( g(d),~d \right)$ 到点 $\left( g(c),~c \right)$ 的弧长由下式给出

$$\text{Arc Length} = {\int_{c}^{d}\sqrt{1 + \left\lbrack {g^{\prime}(y)} \right\rbrack^{2}}}\ dy.$$ (2.8)

Calculating the Arc Length of a Function of y 计算关于 y 的函数的弧长

Let $g(y) = 3y^{3}.$ Calculate the arc length of the graph of $g(y)$ over the interval $\left\lbrack {1,2} \right\rbrack.$

设 $g(y) = 3y^{3}.$ 计算 $g(y)$ 的图像在区间 $\left\lbrack {1,2} \right\rbrack$ 上的弧长。

Solution 解答

We have $g^{\prime}(y) = 9y^{2},$ so $\left\lbrack {g^{\prime}(y)} \right\rbrack^{2} = 81y^{4}.$ Then the arc length is

我们有 $g^{\prime}(y) = 9y^{2},$ 因此 $\left\lbrack {g^{\prime}(y)} \right\rbrack^{2} = 81y^{4}.$ 于是弧长为

$$\text{Arc Length} = {\int_{c}^{d}\sqrt{1 + \left\lbrack {g^{\prime}(y)} \right\rbrack^{2}}}\ dy = {\int_{1}^{2}\sqrt{1 + 81y^{4}}}\ dy.$$

Using a computer to approximate the value of this integral, we obtain

使用计算机来近似该积分的值，我们得到

$${\int_{1}^{2}\sqrt{1 + 81y^{4}}}\ dy \approx 21.0277.$$

Let $g(y) = {1\text{/}y}.$ Calculate the arc length of the graph of $g(y)$ over the interval $\left\lbrack {1,4} \right\rbrack.$ Use a computer or calculator to approximate the value of the integral.

设 $g(y) = {1\text{/}y}.$ 计算 $g(y)$ 的图像在区间 $\left\lbrack {1,4} \right\rbrack$ 上的弧长。使用计算机或计算器来近似该积分的值。

Area of a Surface of Revolution 旋转面的面积

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let $f(x)$ be a nonnegative smooth function over the interval $\left\lbrack {a,b} \right\rbrack.$ We wish to find the surface area of the surface of revolution created by revolving the graph of $y = f(x)$ around the $x\text{-axis}$ as shown in the following figure.

我们用来求曲线弧长的概念，可以推广到求旋转面的面积。表面积是一个物体最外层的总面积。对于立方体或砖块这类物体，其表面积等于它所有面的面积之和。对于曲面，情况要稍微复杂一些。设 $f(x)$ 是定义在区间 $\left\lbrack {a,b} \right\rbrack$ 上的非负光滑函数。我们希望求出将图形 $y = f(x)$ 绕 $x\text{-axis}$ 旋转所生成的旋转面的面积，如下图所示。

As we have done many times before, we are going to partition the interval $\left\lbrack {a,b} \right\rbrack$ and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of $\left\lbrack {a,b} \right\rbrack.$ Then, for $i = 1,2\text{,…},n,$ construct a line segment from the point $\left( {x_{i - 1},f(x_{i - 1})} \right)$ to the point $\left( {x_{i},f(x_{i})} \right).$ Now, revolve these line segments around the $x\text{-axis}$ to generate an approximation of the surface of revolution as shown in the following figure.

如同我们之前多次所做的那样，我们将对区间 $\left\lbrack {a,b} \right\rbrack$ 作分割，并通过计算一些更简单形状的表面积来近似旋转面的面积。我们首先用线段来近似曲线，就如在前面本节中做过的那样。对于 $i = 0,1,2\text{,…},n,$ 设 $P = \left\{ x_{i} \right\}$ 为 $\left\lbrack {a,b} \right\rbrack$ 的一个等分分割。然后，对于 $i = 1,2\text{,…},n,$ 构造一条从点 $\left( {x_{i - 1},f(x_{i - 1})} \right)$ 到点 $\left( {x_{i},f(x_{i})} \right)$ 的线段。现在，将这些线段绕 $x\text{-axis}$ 旋转，生成旋转面的一个近似，如下图所示。

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

注意，当每条线段绕轴旋转时，它会生成一个环带。这些环带实际上是圆锥的若干部分（想象一个尖端被切掉的冰淇淋蛋筒）。像这样的圆锥的一部分称为圆台（frustum of a cone）。

To find the surface area of the band, we need to find the lateral surface area, $S,$ of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let $r_{1}$ and $r_{2}$ be the radii of the wide end and the narrow end of the frustum, respectively, and let $l$ be the slant height of the frustum as shown in the following figure.

为了求出这个环带的面积，我们需要求出圆台的侧面积 $S,$（即圆台斜面外表面的面积，不包括上、下底面的面积）。设 $r_{1}$ 与 $r_{2}$ 分别为圆台粗端和细端的半径，并设 $l$ 为圆台的斜高，如下图所示。

We know the lateral surface area of a cone is given by

我们知道圆锥的侧面积由下式给出

$$\text{Lateral Surface Area} = \pi rs,$$

where $r$ is the radius of the base of the cone and $s$ is the slant height (see the following figure).

其中 $r$ 是圆锥底面的半径，$s$ 是斜高（见下图）。

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the following figure).

由于圆台可视为圆锥的一部分，圆台的侧面积等于整个圆锥的侧面积减去被切掉的较小圆锥（尖顶部分）的侧面积（见下图）。

The cross-sections of the small cone and the large cone are similar triangles, so we see that

小圆锥与大圆锥的截面是相似三角形，因此我们有

$$\frac{r_{2}}{r_{1}} = \frac{s - l}{s}.$$

Solving for $s,$ we get

解出 $s,$ 我们得到

$$\begin{array}{rll} \frac{r_{2}}{r_{1}} & = & \frac{s - l}{s} \\ {r_{2}s} & = & {r_{1}\left( {s - l} \right)} \\ {r_{2}s} & = & {r_{1}s - r_{1}l} \\ {r_{1}l} & = & {r_{1}s - r_{2}s} \\ {r_{1}l} & = & {\left( {r_{1} - r_{2}} \right)s} \end{array}$$

Then the lateral surface area (SA) of the frustum is

于是圆台的侧面积（SA）为

$$\begin{array}{cl} S & {= \ \text{(Lateral SA of large cone)} - \text{(Lateral SA of small cone)}} \\ & {= \pi r_{1}s - \pi r_{2}\left( {s - l} \right)} \\ & {= \pi r_{1}s - \pi r_{2}s + \pi r_{2}l} \\ & {= \pi\left( {r_{1} - r_{2}} \right)s + \pi r_{2}l} \\ & {= \pi r_{1}l + \pi r_{2}l} \\ & {= \pi\left( {r_{1} + r_{2}} \right)l.} \end{array}$$

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the $x\text{-axis}\text{.}$ A representative band is shown in the following figure.

现在我们用这个公式来计算将各条线段绕 $x\text{-axis}\text{.}$ 旋转所形成的每个环带的面积。下图展示了一个具有代表性的环带。

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

注意，该圆台的斜高恰好就是用来生成它的那条直线段的长度。于是，应用面积公式，我们有

$$\begin{array}{cl} S & {= \pi\left( {r_{1} + r_{2}} \right)l} \\ & {= \pi\left( {f(x_{i - 1}) + f(x_{i})} \right)\sqrt{\text{Δ}x^{2} + \left( {\text{Δ}y_{i}} \right)^{2}}} \\ & {= \pi\left( {f(x_{i - 1}) + f(x_{i})} \right)\text{Δ}x\sqrt{1 + \left( \frac{\text{Δ}y_{i}}{\text{Δ}x} \right)^{2}}.} \end{array}$$

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ such that $f^{\prime}(x_{i}^{*}) = {\left( {\text{Δ}y_{i}} \right)\text{/}{\text{Δ}x}}.$ This gives us

现在，正如我们在推导弧长公式时所做的那样，我们应用均值定理选取 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$，使得 $f^{\prime}(x_{i}^{*}) = {\left( {\text{Δ}y_{i}} \right)\text{/}{\text{Δ}x}}.$ 这就给出

$$S = \pi\left( {f(x_{i - 1}) + f(x_{i})} \right)\text{Δ}x\sqrt{1 + \left( {f^{\prime}(x_{i}^{*})} \right)^{2}}.$$

Furthermore, since $f(x)$ is continuous, by the Intermediate Value Theorem, there is a point $x_{i}^{**} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ such that $f(x_{i}^{**}) = \left( {1\text{/}2} \right)\left\lbrack {f(x_{i - 1}) + f(x_{i})} \right\rbrack,$ so we get

此外，由于 $f(x)$ 连续，根据介值定理，存在一点 $x_{i}^{**} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 使得 $f(x_{i}^{**}) = \left( {1\text{/}2} \right)\left\lbrack {f(x_{i - 1}) + f(x_{i})} \right\rbrack,$ 于是我们得到

$$S = 2\pi f(x_{i}^{**})\text{Δ}x\sqrt{1 + \left( {f^{\prime}(x_{i}^{*})} \right)^{2}}.$$

Then the approximate surface area of the whole surface of revolution is given by

于是整个旋转面的近似表面积为

$$\text{Surface Area} \approx \sum\limits_{i = 1}^{n}2\pi f(x_{i}^{**})\text{Δ}x\sqrt{1 + \left( {f^{\prime}(x_{i}^{*})} \right)^{2}}.$$

This *almost* looks like a Riemann sum, except we have functions evaluated at two different points, $x_{i}^{*}$ and $x_{i}^{**},$ over the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ Although we do not examine the details here, it turns out that because $f(x)$ is smooth, if we let $n\rightarrow\infty,$ the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both $x_{i}^{*}$ and $x_{i}^{**}$ are in the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack,$ so it makes sense that as $n\rightarrow\infty,$ both $x_{i}^{*}$ and $x_{i}^{**}$ approach $x.$ Those of you who are interested in the details should consult an advanced calculus text.

这*几乎*像一个黎曼和，只是我们的函数在两个不同的点 $x_{i}^{*}$ 和 $x_{i}^{**}$ 处求值，区间为 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ 尽管我们在此不考察其细节，但结果是：由于 $f(x)$ 是光滑的，若令 $n\rightarrow\infty,$ 即使有两个不同求值的点，该极限的运作方式仍与黎曼和相同。这在直观上是合理的。$x_{i}^{*}$ 和 $x_{i}^{**}$ 都在区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 内，因此当 $n\rightarrow\infty$ 时，$x_{i}^{*}$ 和 $x_{i}^{**}$ 都趋近于 $x,$ 这是说得通的。对此细节感兴趣者，可查阅一本高等微积分教材。

Taking the limit as $n\rightarrow\infty,$ we get

取极限 $n\rightarrow\infty,$ 我们得到

$$\text{Surface Area} = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}2\pi f(x_{i}^{**})\text{Δ}x\sqrt{1 + \left( {f^{\prime}(x_{i}^{*})} \right)^{2}} = {\int_{a}^{b}{\left( {2\pi f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}} \right)dx}}.$$

As with arc length, we can conduct a similar development for functions of $y$ to get a formula for the surface area of surfaces of revolution about the $y\text{-axis}.$ These findings are summarized in the following theorem.

与弧长的情况一样，我们可以对关于 $y$ 的函数作类似的推导，从而得到绕 $y\text{-axis}$ 的旋转面面积公式。这些结果总结于下面的定理中。

Surface Area of a Surface of Revolution 旋转面的表面积

Let $f(x)$ be a nonnegative smooth function over the interval $\left\lbrack {a,b} \right\rbrack.$ Then, the surface area of the surface of revolution formed by revolving the graph of $f(x)$ around the *x*-axis is given by

设 $f(x)$ 是区间 $\left\lbrack {a,b} \right\rbrack$ 上的非负光滑函数。那么，将 $f(x)$ 的图形绕 *x* 轴旋转所形成的旋转面的表面积由下式给出

$$\text{Surface Area} = {\int_{a}^{b}{\left( {2\pi f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}} \right)dx}}.$$ (2.9)

Similarly, let $g(y)$ be a nonnegative smooth function over the interval $\left\lbrack {c,d} \right\rbrack.$ Then, the surface area of the surface of revolution formed by revolving the graph of $g(y)$ around the $y\text{-axis}$ is given by

类似地，设 $g(y)$ 是区间 $\left\lbrack {c,d} \right\rbrack$ 上的非负光滑函数。那么，将 $g(y)$ 的图形绕 $y\text{-axis}$ 旋转所形成的旋转面的表面积由下式给出

$$\text{Surface Area} = {\int_{c}^{d}{\left( {2\pi g(y)\sqrt{1 + \left( {g^{\prime}(y)} \right)^{2}}} \right)dy}}.$$

Calculating the Surface Area of a Surface of Revolution 1 计算旋转面的面积 1

Let $f(x) = \sqrt{x}$ over the interval $\left\lbrack {1,4} \right\rbrack.$ Find the surface area of the surface generated by revolving the graph of $f(x)$ around the $x\text{-axis}.$ Round the answer to three decimal places.

设 $f(x) = \sqrt{x}$，区间为 $\left\lbrack {1,4} \right\rbrack.$ 求将 $f(x)$ 的图形绕 $x\text{-axis}$ 旋转所生成曲面的面积。答案四舍五入到三位小数。

Solution 解答

The graph of $f(x)$ and the surface of rotation are shown in the following figure.

$f(x)$ 的图形与旋转面如下图所示。

We have $f(x) = \sqrt{x}.$ Then, $f^{\prime}(x) = {1\text{/}\left( {2\sqrt{x}} \right)}$ and $\left( {f^{\prime}(x)} \right)^{2} = {1\text{/}\left( {4x} \right)}.$ Then,

我们有 $f(x) = \sqrt{x}.$ 于是 $f^{\prime}(x) = {1\text{/}\left( {2\sqrt{x}} \right)}$，且 $\left( {f^{\prime}(x)} \right)^{2} = {1\text{/}\left( {4x} \right)}.$ 于是

$$\begin{array}{cl} \text{Surface Area} & {= {\int_{a}^{b}{\left( {2\pi f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}} \right)dx}}} \\ & {= {\int_{1}^{4}\left( {2\pi\sqrt{x}\sqrt{1 + \frac{1}{4x}}} \right)}dx} \\ & {= {\int_{1}^{4}{\left( {2\pi\sqrt{x + \frac{1}{4}}} \right)dx}}.} \end{array}$$

Let $u = x + {1\text{/}4}.$ Then, $du = dx.$ When $x = 1,$ $u = {5\text{/}4},$ and when $x = 4,$ $u = {17\text{/}4}.$ This gives us

令 $u = x + {1\text{/}4}.$ 则 $du = dx.$ 当 $x = 1$ 时，$u = {5\text{/}4},$ 当 $x = 4$ 时，$u = {17\text{/}4}.$ 这就给出

$$\begin{array}{cl} {\int_{1}^{4}{\left( {2\pi\sqrt{x + \frac{1}{4}}} \right)dx}} & {= {\int_{5\text{/}4}^{17\text{/}4}{2\pi\sqrt{u}}}\ du} \\ & {= 2\pi\left. \left\lbrack {\frac{2}{3}u^{3\text{/}2}} \right\rbrack\ \right|_{5\text{/}4}^{17\text{/}4} = \frac{\pi}{6}\left\lbrack {17\sqrt{17} - 5\sqrt{5}} \right\rbrack \approx 30.846.} \end{array}$$

Let $f(x) = \sqrt{1 - x}$ over the interval $\left\lbrack {0,{1\text{/}2}} \right\rbrack.$ Find the surface area of the surface generated by revolving the graph of $f(x)$ around the $x\text{-axis}.$ Round the answer to three decimal places.

设 $f(x) = \sqrt{1 - x}$，区间为 $\left\lbrack {0,{1\text{/}2}} \right\rbrack.$ 求将 $f(x)$ 的图形绕 $x\text{-axis}$ 旋转所生成曲面的面积。答案四舍五入到三位小数。

Calculating the Surface Area of a Surface of Revolution 2 计算旋转面的面积 2

Let $f(x) = y = \sqrt[3]{3x}.$ Consider the portion of the curve where $0 \leq y \leq 2.$ Find the surface area of the surface generated by revolving the graph of $f(x)$ around the $y\text{-axis}.$

设 $f(x) = y = \sqrt[3]{3x}.$ 考虑曲线上满足 $0 \leq y \leq 2$ 的部分。求将 $f(x)$ 的图形绕 $y\text{-axis}$ 旋转所生成曲面的面积。

Solution 解答

Notice that we are revolving the curve around the $y\text{-axis},$ and the interval is in terms of $y,$ so we want to rewrite the function as a function of *y*. We get $x = g(y) = \left( {1\text{/}3} \right)y^{3}.$ The graph of $g(y)$ and the surface of rotation are shown in the following figure.

注意，我们是将曲线绕 $y\text{-axis}$ 旋转，而区间是用 $y$ 表示的，因此我们要把函数改写为关于 *y* 的函数。我们得到 $x = g(y) = \left( {1\text{/}3} \right)y^{3}.$ $g(y)$ 的图形与旋转面如下图所示。

We have $g(y) = \left( {1\text{/}3} \right)y^{3},$ so $g^{\prime}(y) = y^{2}$ and $\left( {g^{\prime}(y)} \right)^{2} = y^{4}.$ Then

我们有 $g(y) = \left( {1\text{/}3} \right)y^{3},$ 故 $g^{\prime}(y) = y^{2}$，且 $\left( {g^{\prime}(y)} \right)^{2} = y^{4}.$ 于是

$$\begin{array}{cl} \text{Surface Area} & {= {\int_{c}^{d}{\left( {2\pi g(y)\sqrt{1 + \left( {g^{\prime}(y)} \right)^{2}}} \right)dy}}} \\ & {= {\int_{0}^{2}\left( {2\pi\left( {\frac{1}{3}y^{3}} \right)\sqrt{1 + y^{4}}} \right)}dy} \\ & {= \frac{2\pi}{3}{\int_{0}^{2}\left( {y^{3}\sqrt{1 + y^{4}}} \right)}dy.} \end{array}$$

Let $u = y^{4} + 1.$ Then $du = 4y^{3}dy.$ When $y = 0,$ $u = 1,$ and when $y = 2,$ $u = 17.$ Then

令 $u = y^{4} + 1.$ 则 $du = 4y^{3}dy.$ 当 $y = 0$ 时，$u = 1,$ 当 $y = 2$ 时，$u = 17.$ 于是

$$\begin{array}{cl} {\frac{2\pi}{3}{\int_{0}^{2}\left( {y^{3}\sqrt{1 + y^{4}}} \right)}dy} & {= \frac{2\pi}{3}{\int_{1}^{17}{\frac{1}{4}\sqrt{u}du}}} \\ & {= \frac{\pi}{6}\left. \left\lbrack {\frac{2}{3}u^{3\text{/}2}} \right\rbrack\ \right|_{1}^{17} = \frac{\pi}{9}\left\lbrack {(17)^{3\text{/}2} - 1} \right\rbrack \approx 24.118.} \end{array}$$

Let $g(y) = \sqrt{9 - y^{2}}$ over the interval $y \in \left\lbrack {0,2} \right\rbrack.$ Find the surface area of the surface generated by revolving the graph of $g(y)$ around the $y\text{-axis}.$

设 $g(y) = \sqrt{9 - y^{2}}$，区间为 $y \in \left\lbrack {0,2} \right\rbrack.$ 求将 $g(y)$ 的图形绕 $y\text{-axis}$ 旋转所生成曲面的面积。

Section 2.4 Exercises 2.4 节习题

For the following exercises, find the length of the functions over the given interval.

对于下列习题，求给定区间上函数的弧长。

165.

$y = 5x\ \text{from}\ x = 0\ \text{to}\ x = 2$

166\.

$y = - \frac{1}{2}x + 25\ \text{from}\ x = 1\ \text{to}\ x = 4$

167.

$x = 4y\ \text{from}\ y = -1\ \text{to}\ y = 1$

168\.

Pick an arbitrary linear function $x = g(y)$ over any interval of your choice $\left( {y_{1},y_{2}} \right).$ Determine the length of the function and then prove the length is correct by using geometry.

选取任意一个线性函数 $x = g(y)$，其定义在你任意选择的区间 $\left( {y_{1},y_{2}} \right)$ 上。确定该函数的弧长，然后用几何方法证明该弧长是正确的。

169.

Find the surface area of the volume generated when the curve $y = \sqrt{x}$ revolves around the $x\text{-axis}$ from $\left( {1,1} \right)$ to $(4,2),$ as seen here.

求曲线 $y = \sqrt{x}$ 绕 $x\text{-axis}$ 从 $\left( {1,1} \right)$ 到 $(4,2)$ 旋转所产生的立体的表面积，如图所示。

170\.

Find the surface area of the volume generated when the curve $y = x^{2}$ revolves around the $y\text{-axis}$ from $(1,\ 1)$ to $(3,9).$

求曲线 $y = x^{2}$ 绕 $y\text{-axis}$ 从 $(1,\ 1)$ 到 $(3,9)$ 旋转所产生的立体的表面积。

For the following exercises, find the lengths of the functions of $x$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

对于下列习题，求关于 $x$ 的函数在给定区间上的弧长。若不能精确求出积分值，可使用技术手段进行近似计算。

171.

$y = x^{3\text{/}2}$ from $\left( {0,0} \right)\ \text{to}\ \left( {1,1} \right)$

172\.

$y = x^{2\text{/}3}$ from $\left( {1,1} \right)\ \text{to}\ \left( {8,4} \right)$

173.

$y = \frac{1}{3}\left( {x^{2} + 2} \right)^{3\text{/}2}$ from $x = 0\ \text{to}\ x = 1$

174\.

$y = \frac{1}{3}\left( {x^{2} - 2} \right)^{3\text{/}2}$ from $x = 2$ to $x = 4$

175.

\[T\] $y = e^{x}$ on $x = 0$ to $x = 1$

\[T\] $y = e^{x}$ 在 $x = 0$ 到 $x = 1$

176\.

$y = \frac{x^{3}}{3} + \frac{1}{4x}$ from $x = 1\ \text{to}\ x = 3$

177.

$y = \frac{x^{4}}{4} + \frac{1}{8x^{2}}$ from $x = 1\ \text{to}\ x = 2$

178\.

$y = \frac{2x^{3\text{/}2}}{3} - \frac{x^{1\text{/}2}}{2}$ from $x = 1\ \text{to}\ x = 4$

179.

$y = \frac{1}{27}\left( {9x^{2} + 6} \right)^{3\text{/}2}$ from $x = 0\ \text{to}\ x = 2$

180\.

\[T\] $y = \text{sin}\ x$ on $x = 0\ \text{to}\ x = \pi$

\[T\] $y = \text{sin}\ x$ 在 $x = 0\ \text{to}\ x = \pi$

For the following exercises, find the lengths of the functions of $y$ over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

对于下列习题，求关于 $y$ 的函数在给定区间上的弧长。若不能精确求出积分值，可使用技术手段进行近似计算。

181.

$y = \frac{5 - 3x}{4}$ from $y = 0$ to $y = 4$

182\.

$x = \frac{1}{2}\left( {e^{y} + e^{\text{−}y}} \right)$ from $y = -1\ \text{to}\ y = 1$

183.

$x = 5y^{3\text{/}2}$ from $y = 0$ to $y = 1$

184\.

\[T\] $x = y^{2}$ from $y = 0$ to $y = 1$

185.

$x = \sqrt{y}$ from $y = 0\ \text{to}\ y = 1$

186\.

$x = \frac{2}{3}\left( {y^{2} + 1} \right)^{3\text{/}2}$ from $y = 1$ to $y = 3$

187.

\[T\] $x = \text{tan}\ y$ from $y = 0$ to $y = \frac{3}{4}$

188\.

\[T\] $x = \text{cos}^{2}y$ from $y = - \frac{\pi}{2}$ to $y = \frac{\pi}{2}$

189.

\[T\] $x = 4^{y}$ from $y = 0\ \text{to}\ y = 2$

190\.

\[T\] $x = \text{ln}(y)$ on $y = \frac{1}{e}$ to $y = e$

\[T\] $x = \text{ln}(y)$ 在 $y = \frac{1}{e}$ 到 $y = e$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $x\text{-axis}.$ If you cannot evaluate the integral exactly, use your calculator to approximate it.

对于下列习题，求下列曲线绕 $x\text{-axis}$ 旋转所产生的立体的表面积。若不能精确求出积分值，可用计算器近似计算。

191.

$y = \sqrt{x}$ from $x = 2$ to $x = 6$

192\.

$y = x^{3}$ from $x = 0$ to $x = 1$

193.

$y = 7x$ from $x = -1\ \text{to}\ x = 1$

194\.

\[T\] $y = \frac{1}{x^{2}}$ from $x = 1\ \text{to}\ x = 3$

195.

$y = \sqrt{4 - x^{2}}$ from $x = 0\ \text{to}\ x = 2$

196\.

$y = \sqrt{4 - x^{2}}$ from $x = -1\ \text{to}\ x = 1$

197.

$y = 5x$ from $x = 1\ \text{to}\ x = 5$

198\.

\[T\] $y = \text{tan}\ x$ from $x = - \frac{\pi}{4}\ \text{to}\ x = \frac{\pi}{4}$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $y\text{-axis}\text{.}$ If you cannot evaluate the integral exactly, use your calculator to approximate it.

对于下列习题，求下列曲线绕 $y\text{-axis}$ 旋转所产生的立体的表面积。若不能精确求出积分值，可用计算器近似计算。

199.

$y = x^{2}$ from $x = 0\ \text{to}\ x = 2$

200\.

$y = \frac{1}{2}x^{2} + \frac{1}{2}$ from $x = 0\ \text{to}\ x = 1$

201.

$y = x + 1$ from $x = 0\ \text{to}\ x = 3$

202\.

\[T\] $y = \frac{1}{x}$ from $x = \frac{1}{2}$ to $x = 1$

203.

$y = \sqrt[3]{x}$ from $x = 1\ \text{to}\ x = 27$

204\.

\[T\] $y = 3x^{4}$ from $x = 0$ to $x = 1$

205.

\[T\] $y = \frac{1}{\sqrt{x}}$ from $x = 1$ to $x = 3$

206\.

\[T\] $y = \text{cos}\ x$ from $x = 0$ to $x = \frac{\pi}{2}$

207.

The base of a lamp is constructed by revolving a quarter circle $y = \sqrt{2x - x^{2}}$ around the $y\text{-axis}$ from $x = 1$ to $x = 2,$ as seen here. Create an integral for the surface area of this curve and compute it.

一盏灯的底部由四分之一圆 $y = \sqrt{2x - x^{2}}$ 绕 $y\text{-axis}$ 从 $x = 1$ 到 $x = 2$ 旋转而成，如图所示。建立该曲线表面积的积分式并计算它。

208\.

A light bulb is a sphere with radius $1\text{/}2$ in. with the bottom sliced off to fit exactly onto a cylinder of radius $1\text{/}4$ in. and length $1\text{/}3$ in., as seen here. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is $1\text{/}4$ in. Find the surface area (not including the top or bottom of the cylinder).

一个灯泡是一个半径为 $1\text{/}2$ in. 的球体，其底部被切掉以恰好安装到一个半径为 $1\text{/}4$ in.、长度为 $1\text{/}3$ in. 的圆柱上，如图所示。球体在底部被切掉以恰好安装到圆柱上，因此切口的半径为 $1\text{/}4$ in.。求该灯泡的表面积（不包括圆柱的顶面和底面）。

209.

\[T\] A lampshade is constructed by rotating $y = {1\text{/}x}$ around the $x\text{-axis}$ from $y = 1$ to $y = 2,$ as seen here. Determine how much material you would need to construct this lampshade—that is, the surface area—accurate to four decimal places.

\[T\] 一个灯罩由曲线 $y = {1\text{/}x}$ 绕 $x\text{-axis}$ 从 $y = 1$ 到 $y = 2$ 旋转而成，如图所示。确定制作这个灯罩所需的材料量——即表面积——精确到小数点后四位。

210\.

\[T\] An anchor drags behind a boat according to the function $y = 24e^{{\text{−}x}\text{/}2} - 24,$ where $y$ represents the depth beneath the boat and $x$ is the horizontal distance of the anchor from the back of the boat. If the anchor is $23$ ft below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places.

\[T\] 一只锚按照函数 $y = 24e^{{\text{−}x}\text{/}2} - 24$ 拖在船后，其中 $y$ 表示锚在船下方的深度，$x$ 表示锚到船尾的水平距离。若锚在船下方 $23$ ft 处，你需要拉出多少绳子才能到达锚？将答案四舍五入到三位小数。

211.

\[T\] You are building a bridge that will span $10$ ft. You intend to add decorative rope in the shape of $y = 5\left| {\text{sin}\left( {\left( {x\pi} \right)\text{/}5} \right)} \right|,$ where $x$ is the distance in feet from one end of the bridge. Find out how much rope you need to buy, measured in a whole number of feet.

\[T\] 你正在建造一座跨度为 $10$ ft 的桥。你打算添加形状为 $y = 5\left| {\text{sin}\left( {\left( {x\pi} \right)\text{/}5} \right)} \right|$ 的装饰绳，其中 $x$ 是从桥一端起算的距离（单位：英尺）。求你需要购买的绳子长度，以整英尺计。

For the following exercises, find the exact arc length for the following problems over the given interval.

对于下列习题，求给定区间上下列问题的精确弧长。

212\.

$y = \text{ln}(\text{sin}\ x)$ from $x = {\pi\text{/}4}$ to $x = {\left( {3\pi} \right)\text{/}4}.$ (*Hint:* Recall trigonometric identities.)

$y = \text{ln}(\text{sin}\ x)$ from $x = {\pi\text{/}4}$ to $x = {\left( {3\pi} \right)\text{/}4}.$ （*提示：* 回忆三角恒等式。）

213.

\[T\] Draw graphs of $y = x^{2},$ $y = x^{6},$ and $y = x^{10}.$ For $y = x^{n},$ as $n$ increases, formulate a prediction on the arc length from $\left( {0,0} \right)$ to $\left( {1,1} \right).$ Now, compute the lengths of these three functions and determine whether your prediction is correct.

\[T\] 画出 $y = x^{2}$、$y = x^{6}$ 和 $y = x^{10}$ 的图像。对于 $y = x^{n}$，随着 $n$ 增大，预测从 $\left( {0,0} \right)$ 到 $\left( {1,1} \right)$ 的弧长。现在，计算这三个函数的弧长，并判断你的预测是否正确。

214\.

Compare the lengths of the parabola $x = y^{2}$ and the line $x = by$ from $\left( {0,0} \right)\ \text{to}\ \left( {b^{2},b} \right)$ as $b$ increases. What do you notice?

比较抛物线 $x = y^{2}$ 与直线 $x = by$ 从 $\left( {0,0} \right)\ \text{to}\ \left( {b^{2},b} \right)$ 随着 $b$ 增大时的弧长。你注意到了什么？

215.

Solve for the length of $x = y^{2}$ from $\left( {0,0} \right)\ \text{to}\ \left( {1,1} \right).$ Show that $x = \left( {1\text{/}2} \right)y^{2}$ from $\left( {0,0} \right)$ to $\left( {2,\ 2} \right)$ is twice as long. Graph both functions and explain why this is so.

求 $x = y^{2}$ 从 $\left( {0,0} \right)\ \text{to}\ \left( {1,1} \right)$ 的弧长。证明 $x = \left( {1\text{/}2} \right)y^{2}$ 从 $\left( {0,0} \right)$ 到 $\left( {2,\ 2} \right)$ 的弧长是它的两倍。画出两个函数的图像并解释原因。

216\.

\[T\] Which is longer between $\left( {1,\ 1} \right)$ and $\left( {2,{1\text{/}2}} \right)\text{:}$ the hyperbola $y = {1\text{/}x}$ or the graph of $x + 2y = 3?$

\[T\] 点 $\left( {1,\ 1} \right)$ 与 $\left( {2,{1\text{/}2}} \right)$ 之间，哪条更长：双曲线 $y = {1\text{/}x}$ 还是直线 $x + 2y = 3$ 的图像？

217.

Explain why the surface area is infinite when $y = {1\text{/}x}$ is rotated around the $x\text{-axis}$ for $1 \leq x < \infty,$ but the volume is finite.

解释为什么当 $y = {1\text{/}x}$ 绕 $x\text{-axis}$ 旋转时，对于 $1 \leq x < \infty$，其表面积是无穷大，但体积却是有限的。

2.5 Physical Applications 2.5 物理应用

2.5.1 Determine the mass of a one-dimensional object from its linear density function.
2.5.2 Determine the mass of a two-dimensional circular object from its radial density function.
2.5.3 Calculate the work done by a variable force acting along a line.
2.5.4 Calculate the work done in pumping a liquid from one height to another.
2.5.5 Find the hydrostatic force against a submerged vertical plate.

2.5.1 根据线密度函数确定一维物体的质量。
2.5.2 根据径向密度函数确定二维圆形物体的质量。
2.5.3 计算沿直线作用的可变力所做的功。
2.5.4 计算将液体从某一高度抽到另一高度所做的功。
2.5.5 求作用于浸没竖直平板的静水压力。

In this section, we examine some physical applications of integration. Let’s begin with a look at calculating mass from a density function. We then turn our attention to work, and close the section with a study of hydrostatic force.

本节中，我们研究积分的一些物理应用。我们先从由密度函数计算质量入手。然后我们将注意力转向功，并以研究静水压力结束本节。

Mass and Density 质量与密度

We can use integration to develop a formula for calculating mass based on a density function. First we consider a thin rod or wire. Orient the rod so it aligns with the $x\text{-axis,}$ with the left end of the rod at $x = a$ and the right end of the rod at $x = b$ (Figure 2.48). Note that although we depict the rod with some thickness in the figures, for mathematical purposes we assume the rod is thin enough to be treated as a one-dimensional object.

我们可以用积分来推导一个基于密度函数计算质量的公式。首先考虑一根细杆或细丝。将杆定向使其与 $x\text{-axis}$ 对齐，杆的左端在 $x = a$ 处，右端在 $x = b$ 处（图 2.48）。注意，尽管图中我们把杆画得有一定厚度，但从数学角度，我们假定杆足够细，可视为一维物体。

If the rod has constant density $\rho,$ given in terms of mass per unit length, then the mass of the rod is just the product of the density and the length of the rod: $\left( {b - a} \right)\rho.$ If the density of the rod is not constant, however, the problem becomes a little more challenging. When the density of the rod varies from point to point, we use a linear density function, $\rho(x),$ to denote the density of the rod at any point, $x.$ Let $\rho(x)$ be an integrable linear density function. Now, for $i = 0,1,2\text{,…},n$ let $P = \left\{ x_{i} \right\}$ be a regular partition of the interval $\left\lbrack {a,b} \right\rbrack,$ and for $i = 1,2\text{,…},n$ choose an arbitrary point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ Figure 2.49 shows a representative segment of the rod.

若杆具有恒定密度 $\rho$（以单位长度的质量给出），则杆的质量就是密度与杆长度的乘积：$\left( {b - a} \right)\rho.$ 然而，若杆的密度不是恒定的，问题就稍具挑战性。当杆的密度逐点变化时，我们用线密度函数 $\rho(x)$ 表示杆在任意点 $x$ 处的密度。设 $\rho(x)$ 是一个可积的线密度函数。现在，对 $i = 0,1,2\text{,…},n$，令 $P = \left\{ x_{i} \right\}$ 为区间 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割；对 $i = 1,2\text{,…},n$，任取一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ 图 2.49 显示了杆的一段代表性片段。

The mass $m_{i}$ of the segment of the rod from $x_{i - 1}$ to $x_{i}$ is approximated by

从 $x_{i - 1}$ 到 $x_{i}$ 的杆段的质量 $m_{i}$ 近似为

$$m_{i} \approx \rho(x_{i}^{*})\left( {x_{i} - x_{i - 1}} \right) = \rho(x_{i}^{*})\text{Δ}x.$$

Adding the masses of all the segments gives us an approximation for the mass of the entire rod:

将所有杆段的质量相加，得到整根杆质量的近似值：

$$m = {\sum\limits_{i = 1}^{n}m_{i}} \approx {\sum\limits_{i = 1}^{n}\rho}(x_{i}^{*})\text{Δ}x.$$

This is a Riemann sum. Taking the limit as $n\rightarrow\infty,$ we get an expression for the exact mass of the rod:

这是一个黎曼和。令 $n\rightarrow\infty$ 取极限，我们得到杆精确质量的表达式：

$$m = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\rho(x_{i}^{*})\text{Δ}x = {\int_{a}^{b}\rho}(x)dx.$$

We state this result in the following theorem.

我们将这一结果表述为下列定理。

Mass–Density Formula of a One-Dimensional Object 一维物体的质—密度公式

Given a thin rod oriented along the $x\text{-axis}$ over the interval $\left\lbrack {a,b} \right\rbrack,$ let $\rho(x)$ denote a linear density function giving the density of the rod at a point *x* in the interval. Then the mass of the rod is given by

给定一根沿 $x\text{-axis}$ 放置、定义在区间 $\left\lbrack {a,b} \right\rbrack$ 上的细杆，设 $\rho(x)$ 为线密度函数，表示杆在区间内点 *x* 处的密度。则该杆的质量由下式给出

$$m = {\int_{a}^{b}\rho}(x)dx.$$ (2.10)

We apply this theorem in the next example.

我们在下一个示例中应用此定理。

Calculating Mass from Linear Density 由线密度计算质量

Consider a thin rod oriented on the *x*-axis over the interval $\left\lbrack {{\pi\text{/}2},\pi} \right\rbrack.$ If the density of the rod is given by $\rho(x) = \text{sin}\ x,$ what is the mass of the rod?

考虑一根放置在 *x*-axis 上、定义在区间 $\left\lbrack {{\pi\text{/}2},\pi} \right\rbrack$ 上的细杆。若杆的密度由 $\rho(x) = \text{sin}\ x$ 给出，则该杆的质量是多少？

Solution 解答

Applying Equation 2.10 directly, we have

直接应用式 2.10，我们有

$$m = {\int_{a}^{b}\rho}(x)dx = \int_{\pi\text{/}2}^{\pi}\text{sin}\ x\ dx = \left. {\text{−}\text{cos}\ x} \right|_{\pi\text{/}2}^{\pi} = 1.$$

Consider a thin rod oriented on the *x*-axis over the interval $\left\lbrack {1,3} \right\rbrack.$ If the density of the rod is given by $\rho(x) = 2x^{2} + 3,$ what is the mass of the rod?

考虑一根放置在 *x*-axis 上、定义在区间 $\left\lbrack {1,3} \right\rbrack$ 上的细杆。若杆的密度由 $\rho(x) = 2x^{2} + 3$ 给出，则该杆的质量是多少？

We now extend this concept to find the mass of a two-dimensional disk of radius $r.$ As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object. We assume the density is given in terms of mass per unit area (called *area density*), and further assume the density varies only along the disk’s radius (called *radial density*). We orient the disk in the $xy\text{-plane,}$ with the center at the origin. Then, the density of the disk can be treated as a function of $x,$ denoted $\rho(x).$ We assume $\rho(x)$ is integrable. Because density is a function of $x,$ we partition the interval from $\left\lbrack {0,r} \right\rbrack$ along the $x\text{-axis}.$ For $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of the interval $\left\lbrack {0,r} \right\rbrack,$ and for $i = 1,2\text{,…},n,$ choose an arbitrary point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ Now, use the partition to break up the disk into thin (two-dimensional) washers. A disk and a representative washer are depicted in the following figure.

我们现在将这一概念推广，以求得半径为 $r$ 的二维圆盘的质量。与一维情形中研究的杆一样，这里我们假设圆盘足够薄，使得从数学角度可将其视为二维物体。我们假设密度以单位面积的质量给出（称为*面积密度*），并进一步假设密度仅沿圆盘的半径变化（称为*径向密度*）。我们将圆盘置于 $xy\text{-plane}$ 中，中心在原点。于是，圆盘的密度可视为 $x$ 的函数，记为 $\rho(x)$。我们假设 $\rho(x)$ 可积。因为密度是 $x$ 的函数，我们沿 $x\text{-axis}$ 对区间 $\left\lbrack {0,r} \right\rbrack$ 进行分割。对 $i = 0,1,2\text{,…},n$，令 $P = \left\{ x_{i} \right\}$ 为区间 $\left\lbrack {0,r} \right\rbrack$ 的一个正则分割；对 $i = 1,2\text{,…},n$，任取一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ 现在，利用该分割将圆盘划分为薄的（二维）垫圈。下图描绘了圆盘及一个有代表性的垫圈。

We now approximate the density and area of the washer to calculate an approximate mass, $m_{i}.$ Note that the area of the washer is given by

现在我们对垫圈的密度和面积进行近似，以计算近似质量 $m_{i}$。注意，垫圈的面积由下式给出

$$\begin{array}{cl} A_{i} & {= \pi{(x_{i})}^{2} - \pi{(x_{i - 1})}^{2}} \\ & {= \pi\left\lbrack {x_{i}^{2} - x_{i - 1}^{2}} \right\rbrack} \\ & {= \pi\left( {x_{i} + x_{i - 1}} \right)\left( {x_{i} - x_{i - 1}} \right)} \\ & {= \pi(x_{i} + x_{i - 1})\text{Δ}x.} \end{array}$$

You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use $x_{i}^{*} \approx (x_{i} + x_{i - 1})\text{/}2$ to approximate the average radius of the washer. We obtain

你或许还记得，我们在用壳法计算体积时曾遇到过类似的表达式。正如在那里所做的，我们用 $x_{i}^{*} \approx (x_{i} + x_{i - 1})\text{/}2$ 来近似垫圈的平均半径。我们得到

$$A_{i} = \pi(x_{i} + x_{i - 1})\text{Δ}x \approx 2\pi x_{i}^{*}\text{Δ}x.$$

Using $\rho(x_{i}^{*})$ to approximate the density of the washer, we approximate the mass of the washer by

用 $\rho(x_{i}^{*})$ 近似垫圈的密度，我们得到垫圈质量的近似值

$$m_{i} \approx 2\pi x_{i}^{*}\rho(x_{i}^{*})\text{Δ}x.$$

Adding up the masses of the washers, we see the mass $m$ of the entire disk is approximated by

将所有垫圈的质量相加，可知整张圆盘的质量 $m$ 近似为

$$m = {\sum\limits_{i = 1}^{n}m_{i}} \approx {\sum\limits_{i = 1}^{n}2}\pi x_{i}^{*}\rho(x_{i}^{*})\text{Δ}x.$$

We again recognize this as a Riemann sum, and take the limit as $n\rightarrow\infty.$ This gives us

我们再次将其视为一个黎曼和，并令 $n\rightarrow\infty$ 取极限。由此得到

$$m = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}2\pi x_{i}^{*}\rho(x_{i}^{*})\text{Δ}x = {\int_{0}^{r}2}\pi x\rho(x)dx.$$

We summarize these findings in the following theorem.

我们将这些结论总结为下列定理。

Mass–Density Formula of a Circular Object 圆形物体的质—密度公式

Let $\rho(x)$ be an integrable function representing the radial density of a disk of radius $r.$ Then the mass of the disk is given by

设 $\rho(x)$ 为可积函数，表示半径为 $r$ 的圆盘的径向密度。则该圆盘的质量由下式给出

$$m = {\int_{0}^{r}2}\pi x\rho(x)dx.$$ (2.11)

Calculating Mass from Radial Density 由径向密度计算质量

Let $\rho(x) = \sqrt{x}$ represent the radial density of a disk. Calculate the mass of a disk of radius 4.

设 $\rho(x) = \sqrt{x}$ 表示圆盘的径向密度。计算半径为 4 的圆盘的质量。

Solution 解答

Applying the formula, we find

应用公式，我们得到

$$\begin{array}{cl} m & {= {\int_{0}^{r}2}\pi x\rho(x)dx} \\ & {= {\int_{0}^{4}2}\pi x\sqrt{x}dx = 2\pi{\int_{0}^{4}x^{3\text{/}2}}dx} \\ & {= 2\pi\left. {\frac{2}{5}x^{5\text{/}2}} \right|_{0}^{4} = \frac{4\pi}{5}\lbrack 32\rbrack = \frac{128\pi}{5}.} \end{array}$$

Let $\rho(x) = 3x + 2$ represent the radial density of a disk. Calculate the mass of a disk of radius 2.

设 $\rho(x) = 3x + 2$ 表示圆盘的径向密度。计算半径为 2 的圆盘的质量。

Work Done by a Force 力做的功

We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.

我们现在来考虑功。在物理学中，功与力相关，力通常被直观地定义为对物体的推或拉。当力使物体移动时，我们说力对物体做了功。换言之，功可以理解为移动一个物体所需的能量大小。根据物理学，当有恒力时，功可以表示为力与距离的乘积。

In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate $1$ kilogram of mass at the rate of $1$ m/sec². Thus, the most common unit of work is the newton-meter. This same unit is also called the *joule*. Both are defined as kilograms times meters squared over seconds squared $\left( {{\text{kg} \cdot \text{m}^{2}}\text{/}\text{s}^{2}} \right).$

在英语单位制中，力的单位是磅，距离的单位是英尺，因此功用英尺-磅表示。在公制单位中，使用千克和米。1 牛顿是使 $1$ 千克质量以 $1$ m/sec² 的加速度加速所需的力。因此，最常用的功的单位是牛-米。这个单位也称为 *joule*（焦耳）。两者都定义为千克乘以米的平方再除以秒的平方 $\left( {{\text{kg} \cdot \text{m}^{2}}\text{/}\text{s}^{2}} \right).$

When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.

当力为恒力时，事情相当简单。然而，力很少是恒定的。例如，压缩（或伸长）弹簧所做的功，取决于弹簧已经被压缩（或拉伸）的程度。我们将在本节后面更详细地讨论弹簧。

Suppose we have a variable force $F(x)$ that moves an object in a positive direction along the *x*-axis from point $a$ to point $b.$ To calculate the work done, we partition the interval $\left\lbrack {a,b} \right\rbrack$ and estimate the work done over each subinterval. So, for $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of the interval $\left\lbrack {a,b} \right\rbrack,$ and for $i = 1,2\text{,…},n,$ choose an arbitrary point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ To calculate the work done to move an object from point $x_{i - 1}$ to point $x_{i},$ we assume the force is roughly constant over the interval, and use $F(x_{i}^{*})$ to approximate the force. The work done over the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack,$ then, is given by

假设我们有一个变力 $F(x)$，它沿 *x* 轴正向将一个物体从点 $a$ 移动到点 $b$。为计算所做的功，我们将区间 $\left\lbrack {a,b} \right\rbrack$ 分割，并估计每个子区间上做的功。于是，对于 $i = 0,1,2\text{,…},n,$ 令 $P = \left\{ x_{i} \right\}$ 为区间 $\left\lbrack {a,b} \right\rbrack$ 的正规分割，而对于 $i = 1,2\text{,…},n,$ 选取任意点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$。为计算将物体从点 $x_{i - 1}$ 移动到点 $x_{i}$ 所做的功，我们假设在该区间上力近似恒定，并用 $F(x_{i}^{*})$ 来近似力。那么，区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 上所做的功由下式给出

$$W_{i} \approx F(x_{i}^{*})\left( {x_{i} - x_{i - 1}} \right) = F(x_{i}^{*})\text{Δ}x.$$

Therefore, the work done over the interval $\left\lbrack {a,b} \right\rbrack$ is approximately

因此，区间 $\left\lbrack {a,b} \right\rbrack$ 上所做的功近似为

$$W = {\sum\limits_{i = 1}^{n}W_{i}} \approx {\sum\limits_{i = 1}^{n}F}(x_{i}^{*})\text{Δ}x.$$

Taking the limit of this expression as $n\rightarrow\infty$ gives us the exact value for work:

令该表达式取极限 $n\rightarrow\infty$，便得到功的精确值：

$$W = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}F}(x_{i}^{*})\text{Δ}x = {\int_{a}^{b}F}(x)dx.$$

Thus, we can define work as follows.

因此，我们可以如下定义功。

If a variable force $F(x)$ moves an object in a positive direction along the *x*-axis from point *a* to point *b*, then the work done on the object is

若变力 $F(x)$ 沿 *x* 轴正向将一个物体从点 *a* 移动到点 *b*，则对该物体所做的功为

$$W = {\int_{a}^{b}F}(x)dx.$$ (2.12)

Note that if *F* is constant, the integral evaluates to $F \cdot (b - a) = F \cdot d,$ which is the formula we stated at the beginning of this section.

注意，若 *F* 为常数，则该积分等于 $F \cdot (b - a) = F \cdot d,$ 这正是我们在本节开头给出的公式。

Now let’s look at the specific example of the work done to compress or elongate a spring. Consider a block attached to a horizontal spring. The block moves back and forth as the spring stretches and compresses. Although in the real world we would have to account for the force of friction between the block and the surface on which it is resting, we ignore friction here and assume the block is resting on a frictionless surface. When the spring is at its natural length (at rest), the system is said to be at equilibrium. In this state, the spring is neither elongated nor compressed, and in this equilibrium position the block does not move until some force is introduced. We orient the system such that $x = 0$ corresponds to the equilibrium position (see the following figure).

现在让我们看一个压缩或伸长弹簧做功的具体例子。考虑一个连接在水平弹簧上的物块。当弹簧拉伸和压缩时，物块来回运动。虽然在现实世界中我们必须考虑物块与其所搁置表面之间的摩擦力，但这里我们忽略摩擦，并假设物块位于无摩擦表面上。当弹簧处于其自然长度（静止）时，该系统称为处于平衡态。在这种状态下，弹簧既未被拉长也未被压缩，并且在该平衡位置，在引入某个力之前物块不会移动。我们对系统定向，使得 $x = 0$ 对应平衡位置（见下图）。

According to Hooke’s law, the force required to compress or stretch a spring from an equilibrium position is given by $F(x) = kx,$ for some constant $k.$ The value of $k$ depends on the physical characteristics of the spring. The constant $k$ is called the *spring constant* and is always positive. We can use this information to calculate the work done to compress or elongate a spring, as shown in the following example.

根据胡克定律，将弹簧从平衡位置压缩或拉伸所需的力由 $F(x) = kx,$ 给出，其中 $k$ 为某常数。$k$ 的值取决于弹簧的物理特性。常数 $k$ 称为 *spring constant*（弹簧常数），且恒为正。我们可以利用这一信息计算压缩或伸长弹簧所做的功，如下例所示。

The Work Required to Stretch or Compress a Spring 拉伸或压缩弹簧所需的功

Suppose it takes a force of $10$ N (in the negative direction) to compress a spring $0.2$ m from the equilibrium position. How much work is done to stretch the spring $0.5$ m from the equilibrium position?

假设用 $10$ N 的力（负方向）可将弹簧从平衡位置压缩 $0.2$ m。将弹簧从平衡位置拉伸 $0.5$ m 需要做多少功？

Solution

解答

First find the spring constant, $k.$ When $x = -0.2,$ we know $F(x) = -10,$ so

先求弹簧常数 $k$。当 $x = -0.2$ 时，我们知道 $F(x) = -10$，因此

$$\begin{array}{rll} {F(x)} & = & {kx} \\ {- 10} & = & {k(-0.2)} \\ k & = & 50 \end{array}$$

and $F(x) = 50x.$ Then, to calculate work, we integrate the force function, obtaining

于是 $F(x) = 50x$。接着，为计算功，我们对力函数积分，得到

$$W = {\int_{a}^{b}F}(x)dx = {\int_{0}^{0.5}50x\ dx} = \left. {25x^{2}} \right|_{0}^{0.5} = 6.25.$$

The work done to stretch the spring is $6.25$ J.

拉伸该弹簧所做的功为 $6.25$ J。

Suppose it takes a force of $8$ lb to stretch a spring $6$ in. from the equilibrium position. How much work is done to stretch the spring $1$ ft from the equilibrium position?

假设用 $8$ lb 的力可将弹簧从平衡位置拉伸 $6$ in.。将弹簧从平衡位置拉伸 $1$ ft 需要做多少功？

Work Done in Pumping 抽水做功

Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.

考虑将水（或其他液体）从水池中抽出所做的功。抽水问题比弹簧问题稍微复杂一些，因为许多计算依赖于水池的形状和大小。此外，我们关注的不是移动单个质量所做的功，而是移动一定体积的水所做的功，并且将水从池底移出比从池顶移出需要做更多的功。

We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of different shapes. Assume a cylindrical tank of radius $4$ m and height $10$ m is filled to a depth of 8 m. How much work does it take to pump all the water over the top edge of the tank?

我们在一个圆柱形水池的背景下考察这一过程，然后再看几个使用不同形状水池的例子。假设一个半径为 $4$ m、高为 $10$ m 的圆柱形水池被注入到 8 m 深。将所有水抽到水池顶边需要做多少功？

The first thing we need to do is define a frame of reference. We let $x$ represent the vertical distance below the top of the tank. That is, we orient the $x\text{-axis}$ vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).

我们需要做的第一件事是建立一个参考系。令 $x$ 表示距水池顶部的垂直距离。也就是说，我们将 $x\text{-axis}$ 竖直定向，原点位于水池顶部，向下方向为正（见下图）。

Using this coordinate system, the water extends from $x = 2$ to $x = 10.$ Therefore, we partition the interval $\left\lbrack {2,\ 10} \right\rbrack$ and look at the work required to lift each individual “layer” of water. So, for $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of the interval $\left\lbrack {2,\ 10} \right\rbrack,$ and for $i = 1,2\text{,…},n,$ choose an arbitrary point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ Figure 2.53 shows a representative layer.

使用这个坐标系，水从 $x = 2$ 延伸到 $x = 10$。因此，我们将区间 $\left\lbrack {2,\ 10} \right\rbrack$ 分割，并考察提升每一"层"水所需的功。于是，对于 $i = 0,1,2\text{,…},n,$ 令 $P = \left\{ x_{i} \right\}$ 为区间 $\left\lbrack {2,\ 10} \right\rbrack$ 的正规分割，而对于 $i = 1,2\text{,…},n,$ 选取任意点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$。图 2.53 显示了一层有代表性的水层。

In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, so it is equal to the weight of the water. Given that the weight-density of water is $9800$ N/m³, or $62.4$ lb/ft³, calculating the volume of each layer gives us the weight. In this case, we have

在抽水问题中，将水提升到水池顶部所需的力是克服重力所需的力，因此它等于水的重量。已知水的重度为 $9800$ N/m³，或 $62.4$ lb/ft³，计算出每一层的体积即可得到其重量。在本例中，我们有

$$V = \pi{(4)}^{2}\text{Δ}x = 16\pi\text{Δ}x.$$

Then, the force needed to lift each layer is

于是，提升每一层所需的力为

$$F = 9800 \cdot 16\pi\text{Δ}x = 156,800\pi\text{Δ}x.$$

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

注意，如果水池不是圆柱形的，这一步会稍微困难一些。我们将在下一个例子中考察一个非圆柱形水池。

We also need to know the distance the water must be lifted. Based on our choice of coordinate systems, we can use $x_{i}^{*}$ as an approximation of the distance the layer must be lifted. Then the work to lift the $i\text{th}$ layer of water $W_{i}$ is approximately

我们还需要知道水必须被提升的距离。根据我们所选取的坐标系，可以用 $x_{i}^{*}$ 来近似该层水必须被提升的距离。于是，提升第 $i\text{th}$ 层水 $W_{i}$ 所做的功近似为

$$W_{i} \approx 156,800\pi x_{i}^{*}\text{Δ}x.$$

Adding the work for each layer, we see the approximate work to empty the tank is given by

将每一层的功相加，我们得到抽空水池的近似功为

$$W = \sum\limits_{i = 1}^{n}W_{i} \approx \sum\limits_{i = 1}^{n}156,800\pi x_{i}^{*}\text{Δ}x.$$

This is a Riemann sum, so taking the limit as $n\rightarrow\infty,$ we get

这是一个黎曼和，因此取极限 $n\rightarrow\infty$，我们得到

$$\begin{array}{cl} W & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}156,800\pi x_{i}^{*}\text{Δ}x} \\ & {= 156,800\pi{\int_{2}^{10}x}dx} \\ & {= 156,800\pi\left. \left\lbrack \frac{x^{2}}{2} \right\rbrack\ \right|_{2}^{10} = 7,526,400\pi \approx 23,644,883.} \end{array}$$

The work required to empty the tank is approximately 23,650,000 J.

抽空该水池所需的功约为 23,650,000 J。

For pumping problems, the calculations vary depending on the shape of the tank or container. The following problem-solving strategy lays out a step-by-step process for solving pumping problems.

对于抽水问题，计算会因水池或容器的形状而异。下面的解题策略给出了求解抽水问题的一步一步的过程。

Solving Pumping Problems 抽水问题的求解

1. Sketch a picture of the tank and select an appropriate frame of reference.

1. 画出水池的示意图，并选择一个合适的参考系。

2. Calculate the volume of a representative layer of water.

2. 计算一层代表性水体的体积。

3. Multiply the volume by the weight-density of water to get the force.

3. 将体积乘以水的重度，得到力。

4. Calculate the distance the layer of water must be lifted.

4. 计算该层水必须被提升的距离。

5. Multiply the force and distance to get an estimate of the work needed to lift the layer of water.

5. 将力与距离相乘，得到提升该层水所需功的估计值。

6. Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum.

6. 将所有层所需提升的功相加。该表达式是抽出所需水量所需功的估计值，其形式为黎曼和。

7. Take the limit as $n\rightarrow\infty$ and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.

7. 取极限 $n\rightarrow\infty$ 并计算所得积分，得到抽出所需水量所需的精确功。

We now apply this problem-solving strategy in an example with a noncylindrical tank.

我们现在在一个非圆柱形水池的例子中应用这一解题策略。

A Pumping Problem with a Noncylindrical Tank 非圆柱形水池的抽水问题

Assume a tank in the shape of an inverted cone, with height $12$ ft and base radius $4$ ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is $4$ ft. How much work is required to pump out that amount of water?

假设一个倒置圆锥形的水池，高 $12$ ft，底半径 $4$ ft。开始时水池是满的，将水抽到水池上边缘，直到池中剩余水的高度为 $4$ ft。抽出这些水需要做多少功？

Solution

解答

The tank is depicted in Figure 2.54. As we did in the example with the cylindrical tank, we orient the $x\text{-axis}$ vertically, with the origin at the top of the tank and the downward direction being positive (step 1).

水池如图 2.54 所示。与圆柱形水池的例子一样，我们将 $x\text{-axis}$ 竖直定向，原点位于水池顶部，向下方向为正（步骤 1）。

The tank starts out full and ends with $4$ ft of water left, so, based on our chosen frame of reference, we need to partition the interval $\left\lbrack {0,8} \right\rbrack.$ Then, for $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of the interval $\left\lbrack {0,8} \right\rbrack,$ and for $i = 1,2\text{,…},n,$ choose an arbitrary point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).

水池开始时是满的，最后剩下 $4$ ft 的水，因此，根据我们所选取的参考系，我们需要将区间 $\left\lbrack {0,8} \right\rbrack$ 分割。于是，对于 $i = 0,1,2\text{,…},n,$ 令 $P = \left\{ x_{i} \right\}$ 为区间 $\left\lbrack {0,8} \right\rbrack$ 的正规分割，而对于 $i = 1,2\text{,…},n,$ 选取任意点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$。我们可以用一个圆盘来近似一层水的体积，然后利用相似三角形求出该圆盘的半径（见下图）。

From properties of similar triangles, we have

由相似三角形性质，我们有

$$\begin{array}{cll} \frac{r_{i}}{12 - x_{i}^{*}} & = & {\frac{4}{12} = \frac{1}{3}} \\ {3r_{i}} & = & {12 - x_{i}^{*}} \\ r_{i} & = & \frac{12 - x_{i}^{*}}{3} \\ & = & {4 - \frac{x_{i}^{*}}{3}.} \end{array}$$

Then the volume of the disk is

于是该圆盘的体积为

$$V_{i} = \pi\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 2).}$$

The weight-density of water is $62.4$ lb/ft³, so the force needed to lift each layer is approximately

水的重度为 $62.4$ lb/ft³，因此提升每一层所需的力近似为

$$F_{i} \approx 62.4\pi\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 3).}$$

Based on the diagram, the distance the water must be lifted is approximately $x_{i}^{*}$ feet (step 4), so the approximate work needed to lift the layer is

根据图示，水必须被提升的距离约为 $x_{i}^{*}$ 英尺（步骤 4），因此提升该层所需的近似功为

$$W_{i} \approx 62.4\pi x_{i}^{*}\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 5).}$$

Summing the work required to lift all the layers, we get an approximate value of the total work:

将所有层所需提升的功相加，我们得到总功的近似值：

$$W = \sum\limits_{i = 1}^{n}W_{i} \approx \sum\limits_{i = 1}^{n}62.4\pi x_{i}^{*}\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 6).}$$

Taking the limit as $n\rightarrow\infty,$ we obtain

取极限 $n\rightarrow\infty$，我们得到

$$\begin{array}{cl} W & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}62.4\pi x_{i}^{*}\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x} \\ & {= {\int_{0}^{8}{62.4\pi x\left( {4 - \frac{x}{3}} \right)^{2}dx}}} \\ & {= 62.4\pi{\int_{0}^{8}{x\left( {16 - \frac{8x}{3} + \frac{x^{2}}{9}} \right)dx}} = 62.4\pi{\int_{0}^{8}{\left( {16x - \frac{8x^{2}}{3} + \frac{x^{3}}{9}} \right)dx}}} \\ & {= 62.4\pi\left. \left\lbrack {8x^{2} - \frac{8x^{3}}{9} + \frac{x^{4}}{36}} \right\rbrack\ \right|_{0}^{8} = 10,649.6\pi \approx 33,456.7.} \end{array}$$

It takes approximately $33,450$ ft-lb of work to empty the tank to the desired level.

将水池抽到所需水位大约需要 $33,450$ ft-lb 的功。

A tank is in the shape of an inverted cone, with height $10$ ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?

一个倒置圆锥形的水池，高 $10$ ft，底半径 6 ft。开始时水池注入到 8 ft 深，将水抽到水池上边缘，直到池中剩余 3 ft 的水。抽出这些水需要做多少功？

Hydrostatic Force and Pressure 静水压力与压强

In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called *pascals*.

在最后一节中，我们考察作用在浸没于液体中的物体上的力与压强。在英语单位制中，力的单位为磅。在公制单位制中，力的单位为牛顿。压强是单位面积上的力，因此在英语单位制中，我们有磅每平方英尺（或者更常见的，磅每平方英寸，记作 psi）。在公制单位制中，我们有牛顿每平方米，也称为 *pascals*（帕斯卡）。

Let’s begin with the simple case of a plate of area $A$ submerged horizontally in water at a depth *s* (Figure 2.56). Then, the force exerted on the plate is simply the weight of the water above it, which is given by $F = \rho As,$ where $\rho$ is the weight density of water (weight per unit volume). To find the hydrostatic pressure—that is, the pressure exerted by water on a submerged object—we divide the force by the area. So the pressure is $p = {F\text{/}A} = \rho s.$

我们从一块面积为 $A$、水平浸没在深度 *s*（图 2.56）的水中的板开始。此时，作用在板上的力就是其上方水的重量，由 $F = \rho As,$ 给出，其中 $\rho$ 是水的重度（单位体积的重量）。为求静水压强——即水对浸没物体施加的压强——我们将力除以面积。于是压强为 $p = {F\text{/}A} = \rho s.$

By Pascal’s principle, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal’s principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula $F = \rho As$ directly, because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.

根据帕斯卡原理（Pascal’s principle），给定深度处的压强在各个方向上都相同，因此板是水平浸没还是竖直浸没并无区别。所以，只要我们知道深度，就知道压强。我们可以应用帕斯卡原理来求作用在诸如大坝等竖直朝向表面上的力。我们不能直接套用公式 $F = \rho As$，因为在一个竖直朝向的表面上，深度随点的位置而变化。因此，一如我们此前多次所做的那样，我们构造一个分割、一个黎曼和，并最终构造一个定积分来计算该力。

Suppose a thin plate is submerged in water. We choose our frame of reference such that the *x*-axis is oriented vertically, with the downward direction being positive, and point $x = 0$ corresponding to a logical reference point. Let $s(x)$ denote the depth at point *x*. Note we often let $x = 0$ correspond to the surface of the water. In this case, depth at any point is simply given by $s(x) = x.$ However, in some cases we may want to select a different reference point for $x = 0,$ so we proceed with the development in the more general case. Last, let $w(x)$ denote the width of the plate at the point $x.$

设一块薄板浸没在水中。我们选取坐标系，使 *x* 轴竖直方向，向下为正，并令点 $x = 0$ 对应某个合乎逻辑的参考点。记 $s(x)$ 为点 *x* 处的深度。注意，我们通常令 $x = 0$ 对应水面。此时，任意点处的深度简单地由 $s(x) = x$ 给出。然而，在某些情况下，我们可能想为 $x = 0$ 选取不同的参考点，因此我们按更一般的情况进行推导。最后，记 $w(x)$ 为点 $x$ 处板的宽度。

Assume the top edge of the plate is at point $x = a$ and the bottom edge of the plate is at point $x = b.$ Then, for $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of the interval $\left\lbrack {a,b} \right\rbrack,$ and for $i = 1,2\text{,…},n,$ choose an arbitrary point $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ The partition divides the plate into several thin, rectangular strips (see the following figure).

设板的顶边位于点 $x = a$，底边位于点 $x = b$。于是，对 $i = 0,1,2\text{,…},n,$ 令 $P = \left\{ x_{i} \right\}$ 为区间 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割，并对 $i = 1,2\text{,…},n,$ 在 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 中任取一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$。该分割把板分成若干细长的矩形条（见下图）。

Let’s now estimate the force on a representative strip. If the strip is thin enough, we can treat it as if it is at a constant depth, $s(x_{i}^{*}).$ We then have

现在我们来估计其中一条代表性窄条上的力。若该窄条足够薄，我们就可以把它当作处于恒定深度 $s(x_{i}^{*})$ 来处理。于是我们有

$$F_{i} = \rho As = \rho\left\lbrack {w(x_{i}^{*})\text{Δ}x} \right\rbrack s(x_{i}^{*}).$$

Adding the forces, we get an estimate for the force on the plate:

把这些力相加，我们得到板上所受力的一个估计：

$$F \approx {\sum\limits_{i = 1}^{n}F_{i}} = \sum\limits_{i = 1}^{n}\rho\left\lbrack {w(x_{i}^{*})\text{Δ}x} \right\rbrack s(x_{i}^{*}).$$

This is a Riemann sum, so taking the limit gives us the exact force. We obtain

这是一个黎曼和，因此取极限便得到精确的力。我们得到

$$F = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\rho\left\lbrack {w(x_{i}^{*})\text{Δ}x} \right\rbrack s(x_{i}^{*}) = {\int_{a}^{b}\rho}w(x)s(x)dx.$$ (2.13)

Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.

计算这个积分便得到作用在板上的力。我们把这一点总结在下面的解题策略中。

Finding Hydrostatic Force 求静水压力

1. Sketch a picture and select an appropriate frame of reference. (Note that if we select a frame of reference other than the one used earlier, we may have to adjust Equation 2.13 accordingly.)

1. 画出图形并选取合适的参考坐标系。（注意，如果我们选取的参考坐标系不同于前面所用的那个，则可能需要对方程 2.13 作相应调整。）

2. Determine the depth and width functions, $s(x)$ and $w(x).$

2. 确定深度函数与宽度函数，即 $s(x)$ 与 $w(x)$。

3. Determine the weight-density of whatever liquid with which you are working. The weight-density of water is $62.4$ lb/ft³, or 9800 N/m³.

3. 确定你所处理液体的重度。水的重度为 $62.4$ lb/ft³，或 9800 N/m³。

4. Use the equation to calculate the total force.

4. 使用该方程计算总力。

Finding Hydrostatic Force 求静水压力

A water trough 15 ft long has ends shaped like inverted isosceles triangles, with base 8 ft and height 3 ft. Find the force on one end of the trough if the trough is full of water.

一个水槽长 15 ft，其两端形状为倒置的等腰三角形，底边长 8 ft、高 3 ft。若水槽装满水，求作用在槽一端上的力。

Solution 解答

Figure 2.58 shows the trough and a more detailed view of one end.

图 2.58 显示了水槽以及其一端的更详细视图。

Select a frame of reference with the $x\text{-axis}$ oriented vertically and the downward direction being positive. Select the top of the trough as the point corresponding to $x = 0$ (step 1). The depth function, then, is $s(x) = x.$ Using similar triangles, we see that $w(x) = 8 - \left( {8\text{/}3} \right)x$ (step 2). Now, the weight density of water is $62.4$ lb/ft³ (step 3), so applying Equation 2.13, we obtain

选取一个参考坐标系，使 $x\text{-axis}$（x 轴）竖直方向、向下为正。选取水槽顶部为对应 $x = 0$ 的点（步骤 1）。于是深度函数为 $s(x) = x$。利用相似三角形，我们得到 $w(x) = 8 - \left( {8\text{/}3} \right)x$（步骤 2）。现在，水的重度为 $62.4$ lb/ft³（步骤 3），因此应用方程 2.13，我们得到

$$\begin{array}{cl} F & {= {\int_{a}^{b}\rho}w(x)s(x)dx} \\ & {= {\int_{0}^{3}{62.4\left( {8 - \frac{8}{3}x} \right)x\ dx}} = 62.4{\int_{0}^{3}{\left( {8x - \frac{8}{3}x^{2}} \right)dx}}} \\ & {= 62.4\left. \left\lbrack {4x^{2} - \frac{8}{9}x^{3}} \right\rbrack\ \right|_{0}^{3} = 748.8.} \end{array}$$

The water exerts a force of 748.8 lb on the end of the trough (step 4).

水对槽一端施加的力为 748.8 lb（步骤 4）。

A water trough 12 m long has ends shaped like inverted isosceles triangles, with base 6 m and height 4 m. Find the force on one end of the trough if the trough is full of water.

一个水槽长 12 m，其两端形状为倒置的等腰三角形，底边长 6 m、高 4 m。若水槽装满水，求作用在槽一端上的力。

Chapter Opener: Finding Hydrostatic Force 章节开篇：求静水压力

We now return our attention to the Hoover Dam, mentioned at the beginning of this chapter. The actual dam is arched, rather than flat, but we are going to make some simplifying assumptions to help us with the calculations. Assume the face of the Hoover Dam is shaped like an isosceles trapezoid with lower base $750$ ft, upper base $1250$ ft, and height $750$ ft (see the following figure).

现在我们把注意力转回本章开头提到的胡佛大坝（Hoover Dam）。实际的大坝是拱形的，而非平直的，但为了简化计算，我们将作一些简化假设。假设胡佛大坝的坝面形状为等腰梯形，下底 $750$ ft、上底 $1250$ ft、高 $750$ ft（见下图）。

When the reservoir is full, Lake Mead’s maximum depth is about 530 ft, and the surface of the lake is about 10 ft below the top of the dam (see the following figure).

当水库蓄满时，米德湖（Lake Mead）的最大深度约为 530 ft，湖面约在大坝顶部下方 10 ft 处（见下图）。

1. Find the force on the face of the dam when the reservoir is full.

1. 求当水库蓄满时作用在坝面上的力。

2. The southwest United States has been experiencing a drought, and the surface of Lake Mead is about 125 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

2. 美国西南部正经历干旱，米德湖的湖面比水库蓄满时低约 125 ft。在这种情形下，作用在坝面上的力是多少？

Solution 解答

1. We begin by establishing a frame of reference. As usual, we choose to orient the $x\text{-axis}$ vertically, with the downward direction being positive. This time, however, we are going to let $x = 0$ represent the top of the dam, rather than the surface of the water. When the reservoir is full, the surface of the water is $10$ ft below the top of the dam, so $s(x) = x - 10$ (see the following figure).

1. 我们先建立一个参考坐标系。与往常一样，我们选取 $x\text{-axis}$（x 轴）竖直方向、向下为正。不过这一次，我们令 $x = 0$ 代表坝顶，而不是水面。当水库蓄满时，水面在大坝顶部下方 $10$ ft 处，因此 $s(x) = x - 10$（见下图）。

To find the width function, we again turn to similar triangles as shown in the figure below.

为求宽度函数，我们再次借助下图所示相似三角形。

From the figure, we see that $w(x) = 750 + 2r.$ Using properties of similar triangles, we get $r = 250 - \left( {1\text{/}3} \right)x.$ Thus,

由图可见，$w(x) = 750 + 2r$。利用相似三角形的性质，我们得到 $r = 250 - \left( {1\text{/}3} \right)x$。于是，

$$w(x) = 1250 - \frac{2}{3}x\ \text{(step 2).}$$

Using a weight-density of $62.4$ lb/ft³ (step 3) and applying Equation 2.13, we get

使用重度 $62.4$ lb/ft³（步骤 3）并应用方程 2.13，我们得到

$$\begin{array}{cl} F & {= {\int_{a}^{b}\rho}w(x)s(x)dx} \\ & {= {\int_{10}^{540}{62.4\left( {1250 - \frac{2}{3}x} \right)\left( {x - 10} \right)dx}} = 62.4{\int_{10}^{540}{- \frac{2}{3}\left\lbrack {x^{2} - 1885x + 18750} \right\rbrack}}dx} \\ & {= -62.4\left( \frac{2}{3} \right)\left. \left\lbrack {\frac{x^{3}}{3} - \frac{1885x^{2}}{2} + 18750x} \right\rbrack\ \right|_{10}^{540} \approx 8,832,245,000\ \text{lb} = 4,416,122.5\ \text{t}\text{.}} \end{array}$$

Note the change from pounds to tons $(2000$ lb = $1$ ton) (step 4).

注意从磅到吨的换算 $(2000$ lb = $1$ ton)（步骤 4）。

2. Notice that the drought changes our depth function, $s(x),$ and our limits of integration. We have $s(x) = x - 135.$ The lower limit of integration is $135.$ The upper limit remains $540.$ Evaluating the integral, we get

2. 注意，干旱改变了我们的深度函数 $s(x)$ 以及积分限。我们有 $s(x) = x - 135$。积分下限为 $135$，上限仍为 $540$。计算该积分，我们得到

$$\begin{array}{cl} F & {= {\int_{a}^{b}\rho}w(x)s(x)dx} \\ & {= {\int_{135}^{540}{62.4\left( {1250 - \frac{2}{3}x} \right)\left( {x - 135} \right)dx}}} \\ & {= -62.4\left( \frac{2}{3} \right){\int_{135}^{540}{\left( {x - 1875} \right)\left( {x - 135} \right)dx}} = -62.4\left( \frac{2}{3} \right){\int_{135}^{540}{\left( {x^{2} - 2010x + 253125} \right)dx}}} \\ & {= -62.4\left( \frac{2}{3} \right)\left. \left\lbrack {\frac{x^{3}}{3} - 1005x^{2} + 253125x} \right\rbrack\ \right|_{135}^{540} \approx 5,015,230,000\ \text{lb} = \ 2,507,615\ \text{t}\text{.}} \end{array}$$

When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

当水库处于平均水平时，湖面比水库蓄满时低约 50 ft。在这种情形下，作用在坝面上的力是多少？

To learn more about Hoover Dam, see this article published by the History Channel.

想了解更多关于胡佛大坝的信息，请参阅历史频道（History Channel）发表的这篇文章。

Section 2.5 Exercises 2.5 节习题

For the following exercises, find the work done.

在以下习题中，求所做的功。

218\.

Find the work done when a constant force $F = 12$ lb moves a chair from $x = 0.9$ to $x = 1.1$ ft.

求当恒力 $F = 12$ lb 将一把椅子从 $x = 0.9$ 移动到 $x = 1.1$ ft 时所做的功。

219.

How much work is done when a person lifts a $50$ lb box of comics onto a truck that is $3$ ft off the ground?

当一个人把一箱重 $50$ lb 的漫画书搬到离地 $3$ ft 的卡车上时，做了多少功？

220\.

What is the work done lifting a $20$ kg child from the floor to a height of $2$ m? (Note that a mass of$1$ kg weighs $9.8$ N near the surface of the Earth.)

把一个 $20$ kg 的孩子从地面举到 $2$ m 的高度，做了多少功？（注意，在地球表面附近，质量为 $1$ kg 的重物重 $9.8$ N。）

221.

Find the work done when you push a box along the floor $2$ m, when you apply a constant force of $F = 100\ \text{N}.$

当你沿地板把一只箱子推过 $2$ m，且施加的恒力为 $F = 100\ \text{N}$ 时，求所做的功。

222\.

Compute the work done for a force $F = {12\text{/}x^{2}}$ N from $x = 1$ to $x = 2$ m.

计算力为 $F = {12\text{/}x^{2}}$ N、从 $x = 1$ 到 $x = 2$ m 时所做的功。

223.

What is the work done moving a particle from $x = 0$ to $x = 1$ m if the force acting on it is $F = 3x^{2}$ N?

若作用在粒子上的力为 $F = 3x^{2}$ N，将其从 $x = 0$ 移动到 $x = 1$ m 做了多少功？

For the following exercises, find the mass of the one-dimensional object.

在以下习题中，求该一维物体的质量。

224\.

A wire that is $2$ ft long (starting at $x = 0)$ and has a density function of $\rho(x) = x^{2} + 2x$ lb/ft

一根长 $2$ ft 的金属丝（从 $x = 0$ 开始），其密度函数为 $\rho(x) = x^{2} + 2x$ lb/ft

225.

A car antenna that is $3$ ft long (starting at $x = 0)$ and has a density function of $\rho(x) = 3x + 2$ lb/ft

一根长 $3$ ft 的汽车天线（从 $x = 0$ 开始），其密度函数为 $\rho(x) = 3x + 2$ lb/ft

226\.

A metal rod that is $8$ in. long (starting at $x = 0)$ and has a density function of $\rho(x) = e^{{({1\text{/}2})}x}$ lb/in.

一根长 $8$ in. 的金属杆（从 $x = 0$ 开始），其密度函数为 $\rho(x) = e^{{({1\text{/}2})}x}$ lb/in.

227.

A pencil that is $4$ in. long (starting at $x = 2)$ and has a density function of $\rho(x) = {5\text{/}x}$ oz/in.

一支长 $4$ in. 的铅笔（从 $x = 2$ 开始），其密度函数为 $\rho(x) = {5\text{/}x}$ oz/in.

228\.

A ruler that is $12$ in. long (starting at $x = 5)$ and has a density function of $\rho(x) = \text{ln}(x) + \left( {1\text{/}2} \right)x^{2}$ oz/in.

一把长 $12$ in. 的尺子（从 $x = 5$ 开始），其密度函数为 $\rho(x) = \text{ln}(x) + \left( {1\text{/}2} \right)x^{2}$ oz/in.

For the following exercises, find the mass of the two-dimensional object that is centered at the origin.

在以下习题中，求以原点为中心的二维物体的质量。

229.

An oversized hockey puck of radius $2$ in. with density function $\rho(x) = x^{3} - 2x + 5$

一个半径 $2$ in. 的超大冰球，其密度函数为 $\rho(x) = x^{3} - 2x + 5$

230\.

A frisbee of radius $6$ in. with density function $\rho(x) = e^{\text{−}x}$

一个半径 $6$ in. 的飞盘，其密度函数为 $\rho(x) = e^{\text{−}x}$

231.

A plate of radius $10$ in. with density function $\rho(x) = 1 + \text{cos}\left( {\pi x} \right)$

一个半径 $10$ in. 的圆盘，其密度函数为 $\rho(x) = 1 + \text{cos}\left( {\pi x} \right)$

232\.

A jar lid of radius $3$ in. with density function $\rho(x) = \text{ln}(x + 1)$

一个半径 $3$ in. 的罐盖，其密度函数为 $\rho(x) = \text{ln}(x + 1)$

233.

A disk of radius $5$ cm with density function $\rho(x) = \sqrt{3x}$

一个半径 $5$ cm 的圆盘，其密度函数为 $\rho(x) = \sqrt{3x}$

234\.

A $12$-in. spring is stretched to $15$ in. by a force of $75$ lb. What is the spring constant?

一根 $12$ in. 的弹簧在 $75$ lb 的力作用下被拉伸到 $15$ in.。弹簧常数是多少？

235.

A spring has a natural length of $10$ cm. It takes $2$ J to stretch the spring to $15$ cm. How much work would it take to stretch the spring from $15$ cm to $20$ cm?

一根弹簧的自然长度为 $10$ cm。将其拉伸到 $15$ cm 需要 $2$ J 的功。若要将弹簧从 $15$ cm 拉伸到 $20$ cm，需要做多少功？

236\.

A $1$-m spring requires $10$ J to stretch the spring to $1.1$ m. How much work would it take to stretch the spring from $1$ m to $1.2$ m?

一根 $1$ m 的弹簧被拉伸到 $1.1$ m 需要 $10$ J 的功。若要将弹簧从 $1$ m 拉伸到 $1.2$ m，需要做多少功？

237.

A spring requires $5$ J to stretch the spring from $8$ cm to $12$ cm, and an additional $4$ J to stretch the spring from $12$ cm to $14$ cm. What is the natural length of the spring?

将弹簧从 $8$ cm 拉伸到 $12$ cm 需要 $5$ J 的功，再将其从 $12$ cm 拉伸到 $14$ cm 还需额外 $4$ J 的功。弹簧的自然长度是多少？

238\.

A shock absorber is compressed 1 in. by a weight of 1 t. What is the spring constant?

一个减震器在 $1$ t 的重量下被压缩了 1 in.。弹簧常数是多少？

239.

A force of $F = 20x - x^{3}$ N stretches a nonlinear spring by $x$ meters. What work is required to stretch the spring from $x = 0$ to $x = 2$ m?

力 $F = 20x - x^{3}$ N 将一根非线性弹簧拉伸了 $x$ 米。将弹簧从 $x = 0$ 拉伸到 $x = 2$ m 需要做多少功？

240\.

Find the work done by winding up a hanging cable of length $100$ ft and weight-density $5$ lb/ft.

求卷起一根长 $100$ ft、重度为 $5$ lb/ft 的悬挂电缆所做的功。

241.

For the cable in the preceding exercise, how much work is done to lift the cable $50$ ft?

对于前一题中的电缆，将其提升 $50$ ft 需要做多少功？

242\.

For the cable in the preceding exercise, how much additional work is done by hanging a $200$ lb weight at the end of the cable?

对于前一题中的电缆，若在电缆末端悬挂一个 $200$ lb 的重物，会额外做多少功？

243.

\[T\] A pyramid of height $500$ ft has a square base $800$ ft by $800$ ft. Find the area $A$ at height $h.$ If the rock used to build the pyramid weighs approximately $w = 100\ \text{lb/ft}^{3},$ how much work did it take to lift all the rock?

\[T\] 一座高 $500$ ft 的金字塔，其正方形底面为 $800$ ft × $800$ ft。求高度 $h$ 处的截面积 $A$。若建造金字塔所用的岩石重量约为 $w = 100\ \text{lb/ft}^{3}$，那么将所有岩石提升上去需要做多少功？

244\.

\[T\] For the pyramid in the preceding exercise, assume there were $1000$ workers each working $10$ hours a day, $5$ days a week, $50$ weeks a year. If the workers, on average, lifted 10 100 lb rocks $2$ ft/hr, how long did it take to build the pyramid?

\[T\] 对于前一题中的金字塔，假设有 $1000$ 名工人，每人每天工作 $10$ 小时、每周工作 $5$ 天、每年工作 $50$ 周。若工人平均每小时将 $10$ 块 $100$ lb 的岩石提升 $2$ ft，那么建造这座金字塔花了多长时间？

245.

\[T\] The force of gravity on a mass $m$ is $F = \text{−}\left( {\left( {GMm} \right)\text{/}x^{2}} \right)$ newtons. For a rocket of mass $m = 1000\ \text{kg},$ compute the work to lift the rocket from $x = 6400$ to $x = 6500$ km. State your answers with three significant figures. (*Note*: $G = 6.67\ \times \ 10^{-11}\ \text{N m}^{2}\text{/}\text{kg}^{2}$ and $M = 6\ \times \ 10^{24}\ \text{kg}\text{.})$

\[T\] 质量为 $m$ 的物体所受的重力为 $F = \text{−}\left( {\left( {GMm} \right)\text{/}x^{2}} \right)$ 牛顿。对于质量 $m = 1000\ \text{kg}$ 的火箭，计算将其从 $x = 6400$ 提升到 $x = 6500$ km 所做的功。答案保留三位有效数字。（*注*：$G = 6.67\ \times \ 10^{-11}\ \text{N m}^{2}\text{/}\text{kg}^{2}$，$M = 6\ \times \ 10^{24}\ \text{kg}$。）

246\.

\[T\] For the rocket in the preceding exercise, find the work to lift the rocket from $x = 6400$ to $x = \infty.$

\[T\] 对于前一题中的火箭，求将其从 $x = 6400$ 提升到 $x = \infty$ 所做的功。

247.

\[T\] A rectangular dam is $40$ ft high and $60$ ft wide. Assume the weight density of water is 62.5 lbs/ft³. Compute the total force $F$ on the dam when

\[T\] 一座矩形大坝高 $40$ ft、宽 $60$ ft。假设水的重度为 62.5 lbs/ft³。计算大坝所受总力 $F$，当

1. the surface of the water is at the top of the dam and

1. 水面位于坝顶，且

2. the surface of the water is halfway down the dam.

2. 水面位于大坝一半高度处。

248\.

\[T\] Find the work required to pump all the water out of a cylinder that has a circular base of radius $5$ ft and height $200$ ft. Use the fact that the density of water is $62$ lb/ft³.

\[T\] 求将水从一个底面半径为 $5$ ft、高 $200$ ft 的圆柱体中全部抽出所需做的功。利用水的密度为 $62$ lb/ft³ 这一事实。

249.

\[T\] Find the work required to pump all the water out of the cylinder in the preceding exercise if the cylinder is only half full.

\[T\] 若前一题中的圆柱体只有半满，求将其中水全部抽出所需做的功。

250\.

\[T\] How much work is required to pump out a swimming pool if the area of the base is $800$ ft², the water is $4$ ft deep, and the top is $1$ ft above the water level? Assume that the density of water is $62$ lb/ft³.

\[T\] 若游泳池底面积为 $800$ ft²、水深 $4$ ft、池顶高出水面 $1$ ft，将水全部抽出需要做多少功？假设水的密度为 $62$ lb/ft³。

251.

A cylinder of depth $H$ and cross-sectional area $A$ stands full of water at density $\rho.$ Compute the work to pump all the water to the top.

一个深 $H$、横截面积为 $A$ 的圆柱体盛满密度为 $\rho$ 的水。计算将全部水抽到顶端所做的功。

252\.

For the cylinder in the preceding exercise, compute the work to pump all the water to the top if the cylinder is only half full.

对于前一题中的圆柱体，若只有半满，计算将全部水抽到顶端所做的功。

253.

A cone-shaped tank has a cross-sectional area that increases with its depth: $A = {\left( {\pi r^{2}h^{2}} \right)\text{/}{H^{3}.}}$ Show that the work to empty it is half the work for a cylinder with the same height and base.

一个锥形水箱的横截面积随深度增加而增大：$A = {\left( {\pi r^{2}h^{2}} \right)\text{/}{H^{3}.}}$ 证明将其排空所做的功，等于具有相同高度与底面的圆柱体的一半。

2.6 Moments and Centers of Mass 2.6 力矩与质心

2.6.1 Find the center of mass of objects distributed along a line.
2.6.2 Locate the center of mass of a thin plate.
2.6.3 Use symmetry to help locate the centroid of a thin plate.
2.6.4 Apply the theorem of Pappus for volume.

2.6.1 求沿一条直线分布的物体的质心。
2.6.2 确定薄板的质心。
2.6.3 利用对称性帮助确定薄板的形心。
2.6.4 应用帕普斯定理求体积。

In this section, we consider centers of mass (also called *centroids*, under certain conditions) and moments. The basic idea of the center of mass is the notion of a balancing point. Many of us have seen performers who spin plates on the ends of sticks. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate (without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhere other than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates; the spin helps keep the plates from falling even if the stick is not exactly in the right place.) Mathematically, that sweet spot is called the *center of mass of the plate*.

本节中，我们考虑质心（在某些条件下亦称形心）与力矩。质心的基本思想是一个平衡点的概念。我们许多人都见过在棍子顶端旋转盘子的表演者。表演者努力让其中若干个持续旋转，而不让任何一个掉落。如果我们观察单个盘子（不旋转它），盘子上有一个恰到好处的位置，使它能完美地平衡在棍子上。如果把棍子放在那个位置以外的任何地方，盘子便无法平衡而掉落到地上。（这正是表演者让盘子旋转的原因；旋转有助于即使棍子并不恰好位于正确位置也能防止盘子掉落。）在数学上，那个恰到好处的位置被称为*盘子的质心*。

In this section, we first examine these concepts in a one-dimensional context, then expand our development to consider centers of mass of two-dimensional regions and symmetry. Last, we use centroids to find the volume of certain solids by applying the theorem of Pappus.

本节中，我们首先在一维情形下考察这些概念，然后将推导拓展到考虑二维区域与对称性的质心。最后，我们应用帕普斯定理，利用形心来求某些立体的体积。

Center of Mass and Moments 质心与力矩

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in Figure 2.62(a). Now suppose we place objects having masses $m_{1}$ and $m_{2}$ at distances $d_{1}$ and $d_{2}$ from the fulcrum, respectively, as shown in Figure 2.62(b).

让我们从一维情形开始考察质心。考虑一根质量可忽略的长而细的金属丝或杆，搁置在支点上，如图 2.62(a) 所示。现在假设我们将质量为 $m_{1}$ 和 $m_{2}$ 的物体分别放置在距支点 $d_{1}$ 和 $d_{2}$ 处，如图 2.62(b) 所示。

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if $m_{1}d_{1} = m_{2}d_{2}.$

这类系统最贴近生活的例子是 playground 跷跷板（seesaw 或 teeter-totter），体重不同的孩子坐在距中心不同距离的位置上。在跷跷板上，若每个孩子各坐一端，较重的孩子会下沉，较轻的孩子被抬离地面。不过，若较重的孩子向中心滑动，跷跷板便会平衡。将这一概念应用于杆上的质量，我们注意到这些质量相互平衡当且仅当 $m_{1}d_{1} = m_{2}d_{2}.$

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, $m_{1}$ and $m_{2},$ located on a number line at points $x_{1}$ and $x_{2},$ respectively (Figure 2.63). The center of mass, $\overset{–}{x},$ is the point where the fulcrum should be placed to make the system balance.

在跷跷板例子中，我们通过相对于支点移动质量（孩子）来使系统平衡。然而，我们真正感兴趣的系统中，质量是固定不动的，我们转而通过移动支点来使系统平衡。假设有两个质点 $m_{1}$ 和 $m_{2},$ 分别位于数轴上的点 $x_{1}$ 和 $x_{2}$ 处（图 2.63）。质心 $\overset{–}{x}$ 就是为使系统平衡所应放置支点的位置。

Thus, we have

于是，我们有

$$\begin{matrix} {m_{1}\left| \overset{–}{x} - x_{1} \right|} & = & {m_{2}\left| x_{2} - \overset{–}{x} \right|} \\ {m_{1}\left( \overset{–}{x} - x_{1} \right)} & = & {m_{2}\left( x_{2} - \overset{–}{x} \right)} \\ {m_{1}\overset{–}{x} - m_{1}x_{1}} & = & {m_{2}x_{2} - m_{2}\overset{–}{x}} \\ {\overset{–}{x}\left( m_{1} + m_{2} \right)} & = & {m_{1}x_{1} + m_{2}x_{2}} \\ \overset{–}{x} & = & {\frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}.} \end{matrix}$$

The expression in the numerator, $m_{1}x_{1} + m_{2}x_{2},$ is called the *first moment of the system with respect to the origin.* If the context is clear, we often drop the word *first* and just refer to this expression as the moment of the system. The expression in the denominator, $m_{1} + m_{2},$ is the total mass of the system. Thus, the center of mass of the system is the point at which the total mass of the system could be concentrated without changing the moment.

分子里的表达式 $m_{1}x_{1} + m_{2}x_{2}$ 称为 *first moment of the system with respect to the origin.* 若语境清楚，我们常省去 *first* 一词，直接称此表达式为系统的矩。分母里的表达式 $m_{1} + m_{2}$ 是系统的总质量。因此，系统的质心就是系统的总质量可以集中于其上而不改变矩的那个点。

This idea is not limited just to two point masses. In general, if *n* masses, $m_{1},m_{2}\text{,…},m_{n},$ are placed on a number line at points $x_{1},x_{2}\text{,…},x_{n},$ respectively, then the center of mass of the system is given by

这一思想并不局限于两个质点。一般地，若 *n* 个质量 $m_{1},m_{2}\text{,…},m_{n}$ 分别被放置于数轴上的点 $x_{1},x_{2}\text{,…},x_{n}$ 处，则系统的质心由下式给出

$$\overset{–}{x} = \frac{{\sum\limits_{i = 1}^{n}m_{i}}x_{i}}{\sum\limits_{i = 1}^{n}m_{i}}.$$

Center of Mass of Objects on a Line 直线上物体的质心

Let $m_{1},m_{2}\text{,…},m_{n}$ be point masses placed on a number line at points $x_{1},x_{2}\text{,…},x_{n},$ respectively, and let $m = {\sum\limits_{i = 1}^{n}m_{i}}$ denote the total mass of the system. Then, the moment of the system with respect to the origin is given by

设 $m_{1},m_{2}\text{,…},m_{n}$ 为放置在数轴上点 $x_{1},x_{2}\text{,…},x_{n}$ 处的质点，并令 $m = {\sum\limits_{i = 1}^{n}m_{i}}$ 表示系统的总质量。则系统关于原点的矩由下式给出

$$M = \sum\limits_{i = 1}^{n}m_{i}x_{i}$$ (2.14)

and the center of mass of the system is given by

而系统的质心由下式给出

$$\overset{–}{x} = \frac{M}{m}.$$ (2.15)

We apply this theorem in the following example.

我们在下面的示例中应用这个定理。

Finding the Center of Mass of Objects along a Line 求直线上物体的质心

Suppose four point masses are placed on a number line as follows:

假设四个质点按如下方式放置在数轴上：

$$\begin{array}{lccl} {m_{1} = 30\ \text{kg,}\ \text{placed at}\ x_{1} = -2\ \text{m}} & & & {m_{2} = 5\ \text{kg,}\ \text{placed at}\ x_{2} = 3\ \text{m}} \\ {m_{3} = 10\ \text{kg,}\ \text{placed at}\ x_{3} = 6\ \text{m}} & & & {m_{4} = 15\ \text{kg,}\ \text{placed at}\ x_{4} = -3\ \text{m}.} \end{array}$$

Find the moment of the system with respect to the origin and find the center of mass of the system.

求系统关于原点的矩以及系统的质心。

Solution 解答

First, we need to calculate the moment of the system:

首先，我们需要计算系统的矩：

$$\begin{array}{cl} M & {= {\sum\limits_{i = 1}^{4}m_{i}}x_{i}} \\ & {= -60 + 15 + 60 - 45 = -30.} \end{array}$$

Now, to find the center of mass, we need the total mass of the system:

现在，为求质心，我们需要系统的总质量：

$$\begin{array}{cl} m & {= {\sum\limits_{i = 1}^{4}m_{i}}} \\ & {= 30 + 5 + 10 + 15 = 60\ \text{kg}\text{.}} \end{array}$$

Then we have

于是我们有

$$\overset{–}{x} = \frac{M}{m} = \frac{-30}{60} = - \frac{1}{2}.$$

The center of mass is located 1/2 m to the left of the origin.

质心位于原点左侧 1/2 m 处。

Suppose four point masses are placed on a number line as follows:

假设四个质点按如下方式放置在数轴上：

$$\begin{array}{lccl} {m_{1} = 12\ \text{kg,}\ \text{placed at}\ x_{1} = -4\ \text{m}} & & & {m_{2} = 12\ \text{kg,}\ \text{placed at}\ x_{2} = 4\ \text{m}} \\ {m_{3} = 30\ \text{kg,}\ \text{placed at}\ x_{3} = 2\ \text{m}} & & & {m_{4} = 6\ \text{kg,}\ \text{placed at}\ x_{4} = -6\ \text{m}.} \end{array}$$

Find the moment of the system with respect to the origin and find the center of mass of the system.

求系统关于原点的矩以及系统的质心。

We can generalize this concept to find the center of mass of a system of point masses in a plane. Let $m_{1}$ be a point mass located at point $\left( {x_{1},y_{1}} \right)$ in the plane. Then the moment $M_{x}$ of the mass with respect to the *x*-axis is given by $M_{x} = m_{1}y_{1}.$ Similarly, the moment $M_{y}$ with respect to the *y*-axis is given by $M_{y} = m_{1}x_{1}.$ Notice that the *x*-coordinate of the point is used to calculate the moment with respect to the *y*-axis, and vice versa. The reason is that the *x*-coordinate gives the distance from the point mass to the *y*-axis, and the *y*-coordinate gives the distance to the *x*-axis (see the following figure).

我们可以将这一概念推广，以求得平面上质点系的质心。设 $m_{1}$ 为位于平面内点 $\left( {x_{1},y_{1}} \right)$ 处的一个质点。则该质量关于 *x* 轴的矩 $M_{x}$ 由 $M_{x} = m_{1}y_{1}$ 给出。类似地，关于 *y* 轴的矩 $M_{y}$ 由 $M_{y} = m_{1}x_{1}$ 给出。注意，点的 *x* 坐标用于计算关于 *y* 轴的矩，反之亦然。原因在于 *x* 坐标给出该质点到 *y* 轴的距离，而 *y* 坐标给出到 *x* 轴的距离（见图）。

If we have several point masses in the *xy*-plane, we can use the moments with respect to the *x*- and *y*-axes to calculate the *x*- and *y*-coordinates of the center of mass of the system.

若在平面 *xy* 内有若干质点，我们可以利用关于 *x* 轴和 *y* 轴的矩来计算系统质心的 *x* 坐标和 *y* 坐标。

Center of Mass of Objects in a Plane 平面内物体的质心

Let $m_{1},m_{2}\text{,…},m_{n}$ be point masses located in the *xy*-plane at points $\left( {x_{1},y_{1}} \right),\left( {x_{2},y_{2}} \right)\text{,…},\left( {x_{n},y_{n}} \right),$ respectively, and let $m = {\sum\limits_{i = 1}^{n}m_{i}}$ denote the total mass of the system. Then the moments $M_{x}$ and $M_{y}$ of the system with respect to the *x*- and *y*-axes, respectively, are given by

设 $m_{1},m_{2}\text{,…},m_{n}$ 为位于平面 *xy* 内点 $\left( {x_{1},y_{1}} \right),\left( {x_{2},y_{2}} \right)\text{,…},\left( {x_{n},y_{n}} \right)$ 处的质点，并令 $m = {\sum\limits_{i = 1}^{n}m_{i}}$ 表示系统的总质量。则系统关于 *x* 轴和 *y* 轴的矩 $M_{x}$ 和 $M_{y}$ 分别由下式给出

$$M_{x} = \sum\limits_{i = 1}^{n}m_{i}y_{i}\quad\text{and}\quad M_{y} = \sum\limits_{i = 1}^{n}m_{i}x_{i}.$$ (2.16)

Also, the coordinates of the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ of the system are

此外，系统质心的坐标 $\left( {\overset{–}{x},\overset{–}{y}} \right)$ 为

$$\overset{–}{x} = \frac{M_{y}}{m}\quad\text{and}\quad\overset{–}{y} = \frac{M_{x}}{m}.$$ (2.17)

The next example demonstrates how to apply this theorem.

下一个示例展示如何应用这个定理。

Finding the Center of Mass of Objects in a Plane 求平面内物体的质心

Suppose three point masses are placed in the *xy*-plane as follows (assume coordinates are given in meters):

假设三个质点按如下方式放置在平面 *xy* 内（假设坐标以米为单位）：

$$\begin{array}{l} {m_{1} = 2\ \text{kg, placed at}\ {\left( {-1,3} \right),}} \\ {m_{2} = 6\ \text{kg, placed at}\ {\left( {1,1} \right),}} \\ {m_{3} = 4\ \text{kg, placed at}\ \left( {2,-2} \right).} \end{array}$$

Find the center of mass of the system.

求系统的质心。

Solution 解答

First we calculate the total mass of the system:

首先我们计算系统的总质量：

$$m = {\sum\limits_{i = 1}^{3}m_{i}} = 2 + 6 + 4 = 12\ \text{kg}\text{.}$$

Next we find the moments with respect to the *x*- and *y*-axes:

接下来我们求出关于 *x* 轴和 *y* 轴的矩：

$$\begin{array}{l} \\ \\ {M_{y} = {\sum\limits_{i = 1}^{3}m_{i}}x_{i} = -2 + 6 + 8 = 12,} \\ {M_{x} = {\sum\limits_{i = 1}^{3}m_{i}}y_{i} = 6 + 6 - 8 = 4.} \end{array}$$

Then we have

于是我们有

$$\overset{–}{x} = \frac{M_{y}}{m} = \frac{12}{12} = 1\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m} = \frac{4}{12} = \frac{1}{3}.$$

The center of mass of the system is $\left( {1,{1\text{/}3}} \right),$ in meters.

系统的质心为 $\left( {1,{1\text{/}3}} \right),$ 单位为米。

Suppose three point masses are placed on a number line as follows (assume coordinates are given in meters):

假设三个质点按如下方式放置在直线上（假设坐标以米为单位）：

$$\begin{array}{l} {m_{1} = 5\ \text{kg, placed at}\ {\left( {-2,-3} \right),}} \\ {m_{2} = 3\ \text{kg, placed at}\ {\left( {2,3} \right),}} \\ {m_{3} = 2\ \text{kg, placed at}\ \left( {-3,-2} \right).} \end{array}$$

Find the center of mass of the system.

求系统的质心。

Center of Mass of Thin Plates 薄板的质心

So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.

迄今为止，我们考察了一维直线和平面上的质点系系统。现在，我们不再把系统质量集中于离散的点上，而是要考察质量沿着一张薄材料片连续分布的系统。就我们的目的而言，我们假定该薄片的厚度足够小，可以当作二维对象来处理。这样的薄片称为薄片（lamina）。接下来我们发展一些技巧来求薄片的质心。在本节中，我们还假定薄片的密度是常数。

Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called its centroid. Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only on the shape of the corresponding region in the plane; it does not depend on the density. In this case, the center of mass of the lamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the *x*- and *y*-axes.

薄片常由一个平面中的二维区域来表示。这样一个区域的几何中心称为其形心（centroid）。由于我们已经假定薄片的密度是常数，薄片的质心只依赖于平面上相应区域的形状，而不依赖于密度。在这种情况下，薄片的质心对应于该平面中所界定区域的形心。与质点系系统一样，我们需要求出薄片的总质量，以及薄片关于 *x* 轴和 *y* 轴的矩。

We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the lamina balances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding, it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of the symmetry principle, and it is stated here without proof.

我们首先考虑一个矩形形状的薄片。回想一下，薄片的质心是薄片达到平衡的那个点。对矩形而言，这个点既是矩形的水平中心，也是垂直中心。基于这一认识，显然矩形薄片的质心就是其对角线相交的点，这是对称原理的一个结果，这里不加证明地给出。

The Symmetry Principle 对称原理

If a region *R* is symmetric about a line *l*, then the centroid of *R* lies on *l*.

若一个区域 *R* 关于一条直线 *l* 对称，则 *R* 的形心位于 *l* 上。

Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function $f(x),$ below by the *x*-axis, and on the left and right by the lines $x = a$ and $x = b,$ respectively, as shown in the following figure.

下面我们来看更一般的薄片。假设我们有一个薄片，其上界为连续函数 $f(x)$ 的图像，下界为 *x* 轴，左、右两侧分别为直线 $x = a$ 和 $x = b$，如下图所示。

As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the *x*- and *y*-axes. As we have done many times before, we approximate these quantities by partitioning the interval $\left\lbrack {a,b} \right\rbrack$ and constructing rectangles.

与质点系系统一样，要求薄片的质心，我们需要求出薄片的总质量，以及薄片关于 *x* 轴和 *y* 轴的矩。正如我们之前多次所做的，我们通过分割区间 $\left\lbrack {a,b} \right\rbrack$ 并构造矩形来近似这些量。

For $i = 0,1,2\text{,…},n,$ let $P = \left\{ x_{i} \right\}$ be a regular partition of $\left\lbrack {a,b} \right\rbrack.$ Recall that we can choose any point within the interval $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ as our $x_{i}^{*}.$ In this case, we want $x_{i}^{*}$ to be the *x*-coordinate of the centroid of our rectangles. Thus, for $i = 1,2\text{,…},n,$ we select $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ such that $x_{i}^{*}$ is the midpoint of the interval. That is, $x_{i}^{*} = {\left( {x_{i - 1} + x_{i}} \right)\text{/}2}.$ Now, for $i = 1,2\text{,…},n,$ construct a rectangle of height $f\left( x_{i}^{*} \right)$ on $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack.$ The center of mass of this rectangle is $\left( {x_{i}^{*},{\left( {f(x_{i}^{*})} \right)\text{/}2}} \right),$ as shown in the following figure.

对于 $i = 0,1,2\text{,…},n,$ 令 $P = \left\{ x_{i} \right\}$ 为 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割。回想一下，我们可以选取区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 内的任意点作为我们的 $x_{i}^{*}.$ 在此情形下，我们希望 $x_{i}^{*}$ 成为我们各个矩形形心的 *x* 坐标。于是，对 $i = 1,2\text{,…},n,$ 我们选取 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 使得 $x_{i}^{*}$ 为该区间的中点。即 $x_{i}^{*} = {\left( {x_{i - 1} + x_{i}} \right)\text{/}2}.$ 现在，对 $i = 1,2\text{,…},n,$ 在 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 上构造一个高度为 $f\left( x_{i}^{*} \right)$ 的矩形。该矩形的质心为 $\left( {x_{i}^{*},{\left( {f(x_{i}^{*})} \right)\text{/}2}} \right),$ 如下图所示。

Next, we need to find the total mass of the rectangle. Let $\rho$ represent the density of the lamina (note that $\rho$ is a constant). In this case, $\rho$ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area of the rectangle by $\rho.$ Then, the mass of the rectangle is given by $\rho f(x_{i}^{*})\text{Δ}x.$

接下来，我们需要求出矩形的总质量。设 $\rho$ 表示薄片的密度（注意 $\rho$ 是一个常数）。在此情形下，$\rho$ 以单位面积的质量来表示。因此，要求矩形的总质量，我们将矩形的面积乘以 $\rho.$ 于是，矩形的质量由 $\rho f(x_{i}^{*})\text{Δ}x$ 给出。

To get the approximate mass of the lamina, we add the masses of all the rectangles to get

为了得到薄片的近似质量，我们把所有矩形的质量相加，得到

$$m \approx {\sum\limits_{i = 1}^{n}\rho}f(x_{i}^{*})\text{Δ}x.$$

This is a Riemann sum. Taking the limit as $n\rightarrow\infty$ gives the exact mass of the lamina:

这是一个黎曼和。取 $n\rightarrow\infty$ 时的极限，便得到薄片的精确质量：

$$m = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}\rho}f(x_{i}^{*})\text{Δ}x = \rho\int_{a}^{b}f(x)dx.$$

Next, we calculate the moment of the lamina with respect to the *x*-axis. Returning to the representative rectangle, recall its center of mass is $\left( {x_{i}^{*},{\left( {f(x_{i}^{*})} \right)\text{/}2}} \right).$ Recall also that treating the rectangle as if it is a point mass located at the center of mass does not change the moment. Thus, the moment of the rectangle with respect to the *x*-axis is given by the mass of the rectangle, $\rho f(x_{i}^{*})\text{Δ}x,$ multiplied by the distance from the center of mass to the *x*-axis: ${\left( {f(x_{i}^{*})} \right)\text{/}2}.$ Therefore, the moment with respect to the *x*-axis of the rectangle is $\rho\left( {\left\lbrack {f(x_{i}^{*})} \right\rbrack^{2}\text{/}2} \right)\text{Δ}x.$ Adding the moments of the rectangles and taking the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the *x*-axis is

接下来，我们计算薄片关于 *x* 轴的矩。回到那个代表性矩形，回想其质心为 $\left( {x_{i}^{*},{\left( {f(x_{i}^{*})} \right)\text{/}2}} \right).$ 还要回想，把矩形当作位于质心处的一个质点来处理，并不会改变其矩。因此，该矩形关于 *x* 轴的矩由矩形的质量 $\rho f(x_{i}^{*})\text{Δ}x$ 乘以质心到 *x* 轴的距离 ${\left( {f(x_{i}^{*})} \right)\text{/}2}$ 给出。于是，该矩形关于 *x* 轴的矩为 $\rho\left( {\left\lbrack {f(x_{i}^{*})} \right\rbrack^{2}\text{/}2} \right)\text{Δ}x.$ 把各矩形的矩相加，并对所得黎曼和取极限，我们看到薄片关于 *x* 轴的矩为

$$M_{x} = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}\rho}\frac{\left\lbrack {f(x_{i}^{*})} \right\rbrack^{2}}{2}\text{Δ}x = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}.$$

We derive the moment with respect to the *y*-axis similarly, noting that the distance from the center of mass of the rectangle to the *y*-axis is $x_{i}^{*}.$ Then the moment of the lamina with respect to the *y*-axis is given by

类似地，我们导出关于 *y* 轴的矩，注意到矩形质心到 *y* 轴的距离为 $x_{i}^{*}.$ 于是薄片关于 *y* 轴的矩为

$$M_{y} = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\rho x_{i}^{*}f(x_{i}^{*})\text{Δ}x = \rho{\int_{a}^{b}{xf(x)dx}}.$$

We find the coordinates of the center of mass by dividing the moments by the total mass to give $\overset{–}{x} = {M_{y}\text{/}m}\ \text{and}\ \overset{–}{y} = {M_{x}\text{/}m}.$ If we look closely at the expressions for $M_{x},M_{y},\ \text{and}\ m,$ we notice that the constant $\rho$ cancels out when $\overset{–}{x}$ and $\overset{–}{y}$ are calculated.

我们把各矩除以总质量，得到质心的坐标：$\overset{–}{x} = {M_{y}\text{/}m}\ \text{and}\ \overset{–}{y} = {M_{x}\text{/}m}.$ 如果仔细审视 $M_{x},M_{y},\ \text{and}\ m$ 的表达式，我们会注意到，当计算 $\overset{–}{x}$ 和 $\overset{–}{y}$ 时，常数 $\rho$ 被约掉了。

We summarize these findings in the following theorem.

我们在下面的定理中总结这些结论。

Center of Mass of a Thin Plate in the xy-Plane xy 平面上薄板的质心

Let *R* denote a region bounded above by the graph of a continuous function $f(x),$ below by the *x*-axis, and on the left and right by the lines $x = a$ and $x = b,$ respectively. Let $\rho$ denote the density of the associated lamina. Then we can make the following statements:

设 *R* 表示一个区域，其上界为连续函数 $f(x)$ 的图像，下界为 *x* 轴，左、右两侧分别为直线 $x = a$ 和 $x = b$。设 $\rho$ 表示该相关薄片的密度。那么我们可以给出如下论断：

1. The mass of the lamina is

1. 薄板的质量为

$$m = \rho\int_{a}^{b}f(x)dx.$$ (2.18)

2. The moments $M_{x}$ and $M_{y}$ of the lamina with respect to the *x*- and *y*-axes, respectively, are

2. 薄片关于 *x* 轴和 *y* 轴的矩 $M_{x}$ 与 $M_{y}$ 分别为

$$M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}.$$ (2.19)

3. The coordinates of the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ are

3. 质心的坐标为 $\left( {\overset{–}{x},\overset{–}{y}} \right)$：

$$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}.$$ (2.20)

In the next example, we use this theorem to find the center of mass of a lamina.

在下一个例子中，我们用这个定理来求一个薄板的质心。

Finding the Center of Mass of a Lamina 求薄板的质心

Let *R* be the region bounded above by the graph of the function $f(x) = \sqrt{x}$ and below by the *x*-axis over the interval $\left\lbrack {0,4} \right\rbrack.$ Find the centroid of the region.

设 *R* 为在区间 $\left\lbrack {0,4} \right\rbrack$ 上、上界为函数 $f(x) = \sqrt{x}$ 的图像、下界为 *x* 轴所围成的区域。求该区域的形心。

Solution 解答

The region is depicted in the following figure.

该区域如下图所示。

Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant $\rho$ cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume $\rho = 1.$

由于我们只需该区域的形心，而不需要相关薄板的质量或矩，我们知道密度常数 $\rho$ 最终会在计算中被约掉。因此，为方便起见，我们假定 $\rho = 1.$

First, we need to calculate the total mass:

首先，我们需要计算总质量：

$$\begin{array}{cl} m & {= \rho\int_{a}^{b}f(x)dx = {\int_{0}^{4}\sqrt{x}}\ dx} \\ & {= \left. {\frac{2}{3}x^{3\text{/}2}} \right|_{0}^{4} = \frac{2}{3}\left\lbrack {8 - 0} \right\rbrack = \frac{16}{3}.} \end{array}$$

Next, we compute the moments:

接下来，我们计算各矩：

$$\begin{array}{cl} M_{x} & {= \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}} \\ & {= {\int_{0}^{4}{\frac{x}{2}dx}} = \left. {\frac{1}{4}x^{2}} \right|_{0}^{4} = 4} \end{array}$$

and

以及

$$\begin{array}{cl} M_{y} & {= \rho{\int_{a}^{b}{xf(x)dx}}} \\ & {= {\int_{0}^{4}{x\sqrt{x}dx}} = {\int_{0}^{4}x^{3\text{/}2}}dx} \\ & {= \left. {\frac{2}{5}x^{5\text{/}2}} \right|_{0}^{4} = \frac{2}{5}\left\lbrack {32 - 0} \right\rbrack = \frac{64}{5}.} \end{array}$$

Thus, we have

于是我们有

$$\overset{–}{x} = \frac{M_{y}}{m} = \frac{64\text{/}5}{16\text{/}3} = \frac{64}{5} \cdot \frac{3}{16} = \frac{12}{5}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m} = \frac{4}{16\text{/}3} = 4 \cdot \frac{3}{16} = \frac{3}{4}.$$

The centroid of the region is $\left( {{12\text{/}5},{3\text{/}4}} \right).$

该区域的形心为 $\left( {{12\text{/}5},{3\text{/}4}} \right).$

Let *R* be the region bounded above by the graph of the function $f(x) = x^{2}$ and below by the *x*-axis over the interval $\left\lbrack {0,2} \right\rbrack.$ Find the centroid of the region.

设 *R* 为在区间 $\left\lbrack {0,2} \right\rbrack$ 上、上界为函数 $f(x) = x^{2}$ 的图像、下界为 *x* 轴所围成的区域。求该区域的形心。

We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function $f(x),$ as before, but now, instead of having the lower bound for the region be the *x*-axis, suppose the region is bounded below by the graph of a second continuous function, $g(x),$ as shown in the following figure.

我们也可以调整这一方法来求更复杂区域的形心。假设我们的区域上界仍如前面一样是连续函数 $f(x)$ 的图像，但现在不再以 *x* 轴作为区域的下界，而是假设该区域下界为另一个连续函数 $g(x)$ 的图像，如下图所示。

Again, we partition the interval $\left\lbrack {a,b} \right\rbrack$ and construct rectangles. A representative rectangle is shown in the following figure.

同样地，我们分割区间 $\left\lbrack {a,b} \right\rbrack$ 并构造矩形。一个代表性矩形如下图所示。

Note that the centroid of this rectangle is $\left( {x_{i}^{*},{\left( {f(x_{i}^{*}) + g(x_{i}^{*})} \right)\text{/}2}} \right).$ We won’t go through all the details of the Riemann sum development, but let’s look at some of the key steps. In the development of the formulas for the mass of the lamina and the moment with respect to the *y*-axis, the height of each rectangle is given by $f(x_{i}^{*}) - g(x_{i}^{*}),$ which leads to the expression $f(x) - g(x)$ in the integrands.

注意该矩形的形心为 $\left( {x_{i}^{*},{\left( {f(x_{i}^{*}) + g(x_{i}^{*})} \right)\text{/}2}} \right).$ 我们不打算详尽推导黎曼和的全部细节，但来看其中一些关键步骤。在推导薄片质量公式以及关于 *y* 轴的矩公式时，每个矩形的高度由 $f(x_{i}^{*}) - g(x_{i}^{*})$ 给出，这就导致了被积函数中出现 $f(x) - g(x)$ 的表达式。

In the development of the formula for the moment with respect to the *x*-axis, the moment of each rectangle is found by multiplying the area of the rectangle, $\rho\left\lbrack {f(x_{i}^{*}) - g(x_{i}^{*})} \right\rbrack\text{Δ}x,$ by the distance of the centroid from the *x*-axis, ${\left( {f(x_{i}^{*}) + g(x_{i}^{*})} \right)\text{/}2},$ which gives $\rho\left( {1\text{/}2} \right)\left\{ {\left\lbrack {f(x_{i}^{*})} \right\rbrack^{2} - \left\lbrack {g(x_{i}^{*})} \right\rbrack^{2}} \right\}\text{Δ}x.$ Summarizing these findings, we arrive at the following theorem.

在推导关于 *x* 轴的矩公式时，每个矩形的矩由矩形的面积 $\rho\left\lbrack {f(x_{i}^{*}) - g(x_{i}^{*})} \right\rbrack\text{Δ}x$ 乘以其形心到 *x* 轴的距离 ${\left( {f(x_{i}^{*}) + g(x_{i}^{*})} \right)\text{/}2}$ 得到，即 $\rho\left( {1\text{/}2} \right)\left\{ {\left\lbrack {f(x_{i}^{*})} \right\rbrack^{2} - \left\lbrack {g(x_{i}^{*})} \right\rbrack^{2}} \right\}\text{Δ}x.$ 总结这些结论，我们得到如下定理。

Center of Mass of a Lamina Bounded by Two Functions 由两函数界定的薄板的质心

Let *R* denote a region bounded above by the graph of a continuous function $f(x),$ below by the graph of the continuous function $g(x),$ and on the left and right by the lines $x = a$ and $x = b,$ respectively. Let $\rho$ denote the density of the associated lamina. Then we can make the following statements:

设 *R* 表示一个区域，其上界为连续函数 $f(x)$ 的图像，下界为连续函数 $g(x)$ 的图像，左、右两侧分别为直线 $x = a$ 和 $x = b$。设 $\rho$ 表示该相关薄片的密度。那么我们可以给出如下论断：

1. The mass of the lamina is

1. 薄板的质量为

$$m = \rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx.$$ (2.21)

2. The moments $M_{x}$ and $M_{y}$ of the lamina with respect to the *x*- and *y*-axes, respectively, are

2. 薄片关于 *x* 轴和 *y* 轴的矩 $M_{x}$ 与 $M_{y}$ 分别为

$$M_{x} = \rho{\int_{a}^{b}{\frac{1}{2}\left( {\left\lbrack {f(x)} \right\rbrack^{2} - \left\lbrack {g(x)} \right\rbrack^{2}} \right)}}dx\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}.$$ (2.22)

3. The coordinates of the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ are

3. 质心的坐标为 $\left( {\overset{–}{x},\overset{–}{y}} \right)$：

$$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}.$$ (2.23)

We illustrate this theorem in the following example.

我们在下面的例子中说明这个定理。

Finding the Centroid of a Region Bounded by Two Functions 求由两函数界定的区域的形心

Let *R* be the region bounded above by the graph of the function $f(x) = 1 - x^{2}$ and below by the graph of the function $g(x) = x - 1.$ Find the centroid of the region.

设 *R* 为上界为函数 $f(x) = 1 - x^{2}$ 的图像、下界为函数 $g(x) = x - 1$ 的图像所围成的区域。求该区域的形心。

Solution 解答

The region is depicted in the following figure.

该区域如下图所示。

The graphs of the functions intersect at $\left( {-2,-3} \right)$ and $(1,0),$ so we integrate from −2 to 1. Once again, for the sake of convenience, assume $\rho = 1.$

这两个函数的图像相交于点 $\left( {-2,-3} \right)$ 和 $(1,0),$ 因此我们从 −2 积分到 1。再次为方便起见，假定 $\rho = 1.$

First, we need to calculate the total mass:

首先，我们需要计算总质量：

$$\begin{array}{cl} m & {= \rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx} \\ & {= {\int_{-2}^{1}\left\lbrack {1 - x^{2} - (x - 1)} \right\rbrack}dx = \int_{-2}^{1}(2 - x^{2} - x)dx} \\ & {= \left. \left\lbrack {2x - \frac{1}{3}x^{3} - \frac{1}{2}x^{2}} \right\rbrack\ \right|_{-2}^{1} = \left\lbrack {2 - \frac{1}{3} - \frac{1}{2}} \right\rbrack - \left\lbrack {-4 + \frac{8}{3} - 2} \right\rbrack = \frac{9}{2}.} \end{array}$$

Next, we compute the moments:

接下来，我们计算各矩：

$$\begin{array}{cl} M_{x} & {= \rho{\int_{a}^{b}{\frac{1}{2}\left( {\left\lbrack {f(x)} \right\rbrack^{2} - \left\lbrack {g(x)} \right\rbrack^{2}} \right)}}dx} \\ & {= \frac{1}{2}{\int_{-2}^{1}{\left( {\left( {1 - x^{2}} \right)^{2} - \left( {x - 1} \right)^{2}} \right)dx}} = \frac{1}{2}\int_{-2}^{1}\left( {x^{4} - 3x^{2} + 2x} \right)dx} \\ & {= \frac{1}{2}\left. \left\lbrack {\frac{x^{5}}{5} - x^{3} + x^{2}} \right\rbrack\ \right|_{-2}^{1} = - \frac{27}{10}} \end{array}$$

and

以及

$$\begin{matrix} M_{y} & {= \rho\int_{a}^{b}x\left\lbrack {f(x) - g(x)} \right\rbrack dx} \\ & {= \int_{-2}^{1}x\left\lbrack {\left( 1 - x^{2} \right) - (x - 1)} \right\rbrack dx = \int_{-2}^{1}x\left\lbrack 2 - x^{2} - x \right\rbrack dx} \\ {= \int_{-2}^{1}\left( 2x - x^{3} - x^{2} \right)dx} & \\ & {= \left. \left\lbrack x^{2} - \frac{x^{4}}{4} - \frac{x^{3}}{3} \right\rbrack\ \right|_{-2}^{1} = - \frac{9}{4}.} \end{matrix}$$

Therefore, we have

因此，我们有

$$\overset{–}{x} = \frac{M_{y}}{m} = - \frac{9}{4} \cdot \frac{2}{9} = - \frac{1}{2}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m} = - \frac{27}{10} \cdot \frac{2}{9} = - \frac{3}{5}.$$

The centroid of the region is $\left( {\text{−}\left( {1\text{/}2} \right),\text{−}\left( {3\text{/}5} \right)} \right).$

该区域的形心为 $\left( {\text{−}\left( {1\text{/}2} \right),\text{−}\left( {3\text{/}5} \right)} \right).$

Let *R* be the region bounded above by the graph of the function $f(x) = 6 - x^{2}$ and below by the graph of the function $g(x) = 3 - 2x.$ Find the centroid of the region.

设 *R* 为上界为函数 $f(x) = 6 - x^{2}$ 的图像、下界为函数 $g(x) = 3 - 2x$ 的图像所围成的区域。求该区域的形心。

The Symmetry Principle 对称原理

We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.

我们此前在考察矩形形心时已经陈述过对称原理。当求对称区域的形心时，对称原理可以帮上大忙。请看下面的例子。

Finding the Centroid of a Symmetric Region 求对称区域的形心

Let *R* be the region bounded above by the graph of the function $f(x) = 4 - x^{2}$ and below by the *x*-axis. Find the centroid of the region.

设 *R* 为上方由函数 $f(x) = 4 - x^{2}$ 的图像界定、下方由 *x* 轴界定的区域。求该区域的形心。

Solution 解答

The region is depicted in the following figure.

该区域如下图所示。

The region is symmetric with respect to the *y*-axis. Therefore, the *x*-coordinate of the centroid is zero. We need only calculate $\overset{–}{y}.$ Once again, for the sake of convenience, assume $\rho = 1.$

该区域关于 *y* 轴对称。因此形心的 *x* 坐标为零。我们只需计算 $\overset{–}{y}$。再一次，为方便起见，设 $\rho = 1$。

First, we calculate the total mass:

首先，我们计算总质量：

$$\begin{array}{cl} m & {= \rho\int_{a}^{b}f(x)dx} \\ & {= {\int_{-2}^{2}\left( {4 - x^{2}} \right)}dx} \\ & {= \left. \left\lbrack {4x - \frac{x^{3}}{3}} \right\rbrack\ \right|_{-2}^{2} = \frac{32}{3}.} \end{array}$$

Next, we calculate the moments. We only need $M_{x}\text{:}$

接下来，我们计算力矩。我们只需 $M_{x}\text{:}$

$$\begin{array}{cl} M_{x} & {= \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}} \\ & {= \frac{1}{2}\int_{-2}^{2}\left\lbrack {4 - x^{2}} \right\rbrack^{2}dx = \frac{1}{2}\int_{-2}^{2}\left( {16 - 8x^{2} + x^{4}} \right)dx} \\ & {= \frac{1}{2}\left. \left\lbrack {\frac{x^{5}}{5} - \frac{8x^{3}}{3} + 16x} \right\rbrack\ \right|_{-2}^{2} = \frac{256}{15}.} \end{array}$$

Then we have

于是我们有

$$\overset{–}{y} = \frac{M_{x}}{m} = \frac{256}{15} \cdot \frac{3}{32} = \frac{8}{5}.$$

The centroid of the region is $\left( {0,{8\text{/}5}} \right).$

该区域的形心为 $\left( {0,{8\text{/}5}} \right).$

Let *R* be the region bounded above by the graph of the function $f(x) = 1 - x^{2}$ and below by *x*-axis. Find the centroid of the region.

设 *R* 为上方由函数 $f(x) = 1 - x^{2}$ 的图像界定、下方由 *x* 轴界定的区域。求该区域的形心。

The Grand Canyon Skywalk 科罗拉多大峡谷空中步道

The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shaped observation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).

科罗拉多大峡谷空中步道于 2007 年 3 月 28 日向公众开放。这一工程奇迹是一座马蹄形的观景平台，悬于科罗拉多大峡谷西缘、科罗拉多河上方 4000 ft 处。其晶莹透明的玻璃地板可让人饱览下方峡谷的壮丽景色（见下图）。

The Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with no visible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineered to be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts that extend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquake within 50 mi, and is capable of supporting more than 70,000,000 lb.

空中步道采用悬臂设计，意味着观景平台伸出峡谷边缘之外，其下方没有任何可见的支撑结构。尽管没有可见的支柱或撑杆，悬臂结构在工程上被设计为非常稳定，空中步道也不例外。观景平台牢牢固定在向下伸入基岩 46 ft 的支撑柱上。该结构可抵御 50 mi 范围内的 100 英里/小时大风与 8.0 级地震，并能承载超过 70,000,000 lb。

One factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculate the center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto the observation platform.

影响空中步道稳定性的一个因素是该结构的重心。我们将计算空中步道的重心，并考察当游客走上观景平台时重心如何变化。

The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors’ center, 48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.

观景平台呈 U 形。U 的两臂宽 10 ft，起始于陆地之上、游客中心下方、距峡谷边缘 48 ft 处。平台向峡谷边缘外延伸 70 ft。

To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the *xy*-plane to represent the platform. We begin by dividing the region into three subregions so we can consider each subregion separately. The first region, denoted $R_{1},$ consists of the curved part of the U. We model $R_{1}$ as a semicircular annulus, with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).

为了计算该结构的质心，我们将其视为一块薄片，并用 *xy* 平面上的一个二维区域来表示该平台。我们首先将该区域分成三个子区域，以便分别考虑每个子区域。第一个区域记为 $R_{1}$，由 U 的弯曲部分构成。我们将 $R_{1}$ 建模为一个半圆环，内半径 25 ft、外半径 35 ft，以原点为圆心（见下图）。

The legs of the platform, extending 35 ft between $R_{1}$ and the canyon wall, comprise the second sub-region, $R_{2}.$ Last, the ends of the legs, which extend 48 ft under the visitor center, comprise the third sub-region, $R_{3}.$ Assume the density of the lamina is constant and assume the total weight of the platform is 1,200,000 lb (not including the weight of the visitor center; we will consider that later). Use $g = 32\ \text{ft/sec}^{2}.$

平台的两臂在 $R_{1}$ 与峡谷壁之间延伸 35 ft，构成第二个子区域 $R_{2}$。最后，两臂伸入游客中心下方 48 ft 的末端，构成第三个子区域 $R_{3}$。假设薄片的密度为常数，并设平台总重量为 1,200,000 lb（不含游客中心的重量；该重量我们稍后考虑）。取 $g = 32\ \text{ft/sec}^{2}$。

1. Compute the area of each of the three sub-regions. Note that the areas of regions $R_{2}$ and $R_{3}$ should include the areas of the legs only, not the open space between them. Round answers to the nearest square foot.

1. 计算三个子区域各自的面积。注意，区域 $R_{2}$ 与 $R_{3}$ 的面积应只包含两臂的面积，而不包含它们之间的空隙。将答案四舍五入到最接近的平方英尺。

2. Determine the mass associated with each of the three sub-regions.

2. 确定三个子区域各自相应的质量。

3. Calculate the center of mass of each of the three sub-regions.

3. 计算三个子区域各自的质心。

4. Now, treat each of the three sub-regions as a point mass located at the center of mass of the corresponding sub-region. Using this representation, calculate the center of mass of the entire platform.

4. 现在，将三个子区域各自视为位于其对应子区域质心处的一个质点。利用这种表示，计算整个平台的质心。

5. Assume the visitor center weighs 2,200,000 lb, with a center of mass corresponding to the center of mass of $R_{3}.$ Treating the visitor center as a point mass, recalculate the center of mass of the system. How does the center of mass change?

5. 假设游客中心重 2,200,000 lb，且其质心与 $R_{3}$ 的质心一致。将游客中心视为一个质点，重新计算系统的质心。质心发生了怎样的变化？

6. Although the Skywalk was built to limit the number of people on the observation platform to 120, the platform is capable of supporting up to 800 people weighing 200 lb each. If all 800 people were allowed on the platform, and all of them went to the farthest end of the platform, how would the center of gravity of the system be affected? (Include the visitor center in the calculations and represent the people by a point mass located at the farthest edge of the platform, 70 ft from the canyon wall.)

6. 尽管空中步道建造时将观景平台上的游客人数限制为 120 人，但该平台最多可承载 800 人，每人重 200 lb。如果允许全部 800 人来到平台，并且他们都走到平台最远端，系统的重心会受到怎样的影响？（计算时包含游客中心，并将人群表示为位于平台最远端、距峡谷壁 70 ft 处的一个质点。）

Theorem of Pappus 帕普斯定理

This section ends with a discussion of the theorem of Pappus for volume, which allows us to find the volume of particular kinds of solids by using the centroid. (There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.)

本节以关于体积的帕普斯定理的讨论作结，它使我们能够利用形心求出某些特定类型立体的体积。（也存在关于表面积的帕普斯定理，但它远不如体积定理有用。）

Theorem of Pappus for Volume 帕普斯体积定理

Let *R* be a region in the plane and let *l* be a line in the plane that does not intersect *R*. Then the volume of the solid of revolution formed by revolving *R* around *l* is equal to the area of *R* multiplied by the distance *d* traveled by the centroid of *R.*

设 *R* 为平面内一个区域，*l* 为平面内一条不与 *R* 相交的直线。则将 *R* 绕 *l* 旋转所形成的旋转体的体积，等于 *R* 的面积乘以 *R* 的形心所经过的距离 *d*。

Proof 证明

We can prove the case when the region is bounded above by the graph of a function $f(x)$ and below by the graph of a function $g(x)$ over an interval $\left\lbrack {a,b} \right\rbrack,$ and for which the axis of revolution is the *y*-axis. In this case, the area of the region is $A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx.$ Since the axis of rotation is the *y*-axis, the distance traveled by the centroid of the region depends only on the *x*-coordinate of the centroid, $\overset{–}{x},$ which is

我们可以证明如下情形：区域在上方由函数 $f(x)$ 的图像界定，在下方由函数 $g(x)$ 的图像界定，定义在区间 $\left\lbrack {a,b} \right\rbrack$ 上，且其旋转轴为 *y* 轴。此时，该区域的面积为 $A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx$。由于旋转轴是 *y* 轴，区域形心经过的距离仅取决于形心的 *x* 坐标 $\overset{–}{x}$，即

$$\overset{–}{x} = \frac{M_{y}}{m},$$

where

其中

$$m = \rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}.$$

Then,

于是

$$d = 2\pi\frac{\rho{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}}{\rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx}$$

and thus

因此

$$d \cdot A = 2\pi{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}.$$

However, using the method of cylindrical shells, we have

然而，利用圆柱壳法，我们有

$$V = 2\pi{\int_{a}^{b}x}\left\lbrack {f(x) - g(x)} \right\rbrack dx.$$

So,

所以

$$V = d \cdot A$$

and the proof is complete.

证明完毕。

□

Using the Theorem of Pappus for Volume 帕普斯体积定理的应用

Let *R* be a circle of radius 2 centered at $\left( {4,0} \right).$ Use the theorem of Pappus for volume to find the volume of the torus generated by revolving *R* around the *y*-axis.

设 *R* 为半径 2、圆心位于 $\left( {4,0} \right)$ 的圆。利用帕普斯体积定理求将 *R* 绕 *y* 轴旋转所生成的环面体积。

Solution 解答

The region and torus are depicted in the following figure.

区域与环面如下图所示。

The region *R* is a circle of radius 2, so the area of *R* is $A = 4\pi$ units². By the symmetry principle, the centroid of *R* is the center of the circle. The centroid travels around the *y*-axis in a circular path of radius 4, so the centroid travels $d = 8\pi$ units. Then, the volume of the torus is $A \cdot d = 32\pi^{2}$ units³.

区域 *R* 是一个半径为 2 的圆，所以 *R* 的面积为 $A = 4\pi$ 单位²。由对称原理，*R* 的形心就是该圆的圆心。形心绕 *y* 轴沿半径为 4 的圆周运动，因此形心经过的距离 $d = 8\pi$ 单位。于是，环面的体积为 $A \cdot d = 32\pi^{2}$ 单位³。

Let *R* be a circle of radius 1 centered at $\left( {3,0} \right).$ Use the theorem of Pappus for volume to find the volume of the torus generated by revolving *R* around the *y*-axis.

设 *R* 为半径 1、圆心位于 $\left( {3,0} \right)$ 的圆。利用帕普斯体积定理求将 *R* 绕 *y* 轴旋转所生成的环面体积。

Section 2.6 Exercises 2.6 节习题

For the following exercises, calculate the center of mass for the collection of masses given.

对于下列习题，计算所给质量集合的质心。

254\.

$m_{1} = 2$ at $x_{1} = 1$ and $m_{2} = 4$ at $x_{2} = 2$

$m_{1} = 2$ 位于 $x_{1} = 1$，$m_{2} = 4$ 位于 $x_{2} = 2$

255.

$m_{1} = 1$ at $x_{1} = -1$ and $m_{2} = 3$ at $x_{2} = 2$

$m_{1} = 1$ 位于 $x_{1} = -1$，$m_{2} = 3$ 位于 $x_{2} = 2$

256\.

$m = 3$ at $x = 0,1,2,6$

$m = 3$ 位于 $x = 0,1,2,6$

257.

Unit masses at $(x,y) = (1,0),(0,1),(1,1)$

单位质量位于 $(x,y) = (1,0),(0,1),(1,1)$

258\.

$m_{1} = 1$ at $(1,0)$ and $m_{2} = 4$ at $(0,1)$

$m_{1} = 1$ 位于 $(1,0)$，$m_{2} = 4$ 位于 $(0,1)$

259.

$m_{1} = 1$ at $(1,0)$ and $m_{2} = 3$ at $(2,2)$

$m_{1} = 1$ 位于 $(1,0)$，$m_{2} = 3$ 位于 $(2,2)$

For the following exercises, compute the center of mass $\overset{–}{x}.$

对于下列习题，计算质心 $\overset{–}{x}$。

260\.

$\rho = 1$ for $x \in (-1,3)$

$\rho = 1$，其中 $x \in (-1,3)$

261.

$\rho = x^{2}$ for $x \in (0,L)$

$\rho = x^{2}$，其中 $x \in (0,L)$

262\.

$\rho = 1$ for $x \in (0,1)$ and $\rho = 2$ for $x \in (1,2)$

$\rho = 1$，其中 $x \in (0,1)$；$\rho = 2$，其中 $x \in (1,2)$

263.

$\rho = \text{sin}\ x$ for $x \in (0,\pi)$

$\rho = \text{sin}\ x$，其中 $x \in (0,\pi)$

264\.

$\rho = \text{cos}\ x$ for $x \in \left( {0,\frac{\pi}{2}} \right)$

$\rho = \text{cos}\ x$，其中 $x \in \left( {0,\frac{\pi}{2}} \right)$

265.

$\rho = e^{x}$ for $x \in \left( {0,2} \right)$

$\rho = e^{x}$，其中 $x \in \left( {0,2} \right)$

266\.

$\rho = x^{3} + xe^{\text{−}x}$ for $x \in (0,1)$

$\rho = x^{3} + xe^{\text{−}x}$，其中 $x \in (0,1)$

267.

$\rho = x\ \text{sin}\ x$ for $x \in (0,\pi)$

$\rho = x\ \text{sin}\ x$，其中 $x \in (0,\pi)$

268\.

$\rho = \sqrt{x}$ for $x \in \left( {1,4} \right)$

$\rho = \sqrt{x}$，其中 $x \in \left( {1,4} \right)$

269.

$\rho = \text{ln}\ x$ for $x \in \left( {1,e} \right)$

$\rho = \text{ln}\ x$，其中 $x \in \left( {1,e} \right)$

For the following exercises, compute the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right).$ Use symmetry to help locate the center of mass whenever possible.

对于下列习题，计算质心 $\left( {\overset{–}{x},\overset{–}{y}} \right)$。尽可能利用对称性来确定质心的位置。

270\.

$\rho = 7$ in the square $0 \leq x \leq 1,$ $0 \leq y \leq 1$

$\rho = 7$，在正方形 $0 \leq x \leq 1,$ $0 \leq y \leq 1$ 内

271.

$\rho = 3$ in the triangle with vertices $(0,0),$ $(a,0),$ and $(0,b)$

$\rho = 3$，在顶点为 $(0,0),$ $(a,0),$ 与 $(0,b)$ 的三角形内

272\.

$\rho = 2$ for the region bounded by $y = \text{cos}(x),$ $y = \text{−}\text{cos}(x),$ $x = - \frac{\pi}{2},$ and $x = \frac{\pi}{2}$

$\rho = 2$，对于由 $y = \text{cos}(x),$ $y = \text{−}\text{cos}(x),$ $x = - \frac{\pi}{2},$ 与 $x = \frac{\pi}{2}$ 所围成的区域

For the following exercises, use a calculator to draw the region, then compute the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right).$ Use symmetry to help locate the center of mass whenever possible.

对于下列习题，使用计算器画出该区域，然后计算质心 $\left( {\overset{–}{x},\overset{–}{y}} \right)$。尽可能利用对称性来确定质心的位置。

273.

\[T\] The region bounded by $y = \text{cos}(2x),$ $x = - \frac{\pi}{4},$ and $x = \frac{\pi}{4}$

\[T\] 由 $y = \text{cos}(2x),$ $x = - \frac{\pi}{4},$ 与 $x = \frac{\pi}{4}$ 所围成的区域

274\.

\[T\] The region between $y = 2x^{2},$ $y = 0,$ $x = 0,$ and $x = 1$

\[T\] 介于 $y = 2x^{2},$ $y = 0,$ $x = 0,$ 与 $x = 1$ 之间的区域

275.

\[T\] The region between $y = \frac{5}{4}x^{2}$ and $y = 5$

\[T\] 介于 $y = \frac{5}{4}x^{2}$ 与 $y = 5$ 之间的区域

276\.

\[T\] Region between $y = \sqrt{x},$ $y = \text{ln}(x),$ $x = 1,$ and $x = 4$

\[T\] 介于 $y = \sqrt{x},$ $y = \text{ln}(x),$ $x = 1,$ 与 $x = 4$ 之间的区域

277.

\[T\] The region bounded by $y = 0,$ $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

\[T\] 由 $y = 0,$ $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ 所围成的区域

278\.

\[T\] The region bounded by $y = 0,$ $x = 0,$ and $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

\[T\] 由 $y = 0,$ $x = 0,$ 与 $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ 所围成的区域

279.

\[T\] The region bounded by $y = x^{2}$ and $y = x^{4}$ in the first quadrant

\[T\] 在第一象限内由 $y = x^{2}$ 与 $y = x^{4}$ 所围成的区域

For the following exercises, use the theorem of Pappus to determine the volume of the shape.

对于下列习题，利用帕普斯定理确定该形状的体积。

280\.

Rotating $y = mx$ around the $x$-axis between $x = 0$ and $x = 1$

将 $y = mx$ 绕 $x$ 轴在 $x = 0$ 与 $x = 1$ 之间旋转

281.

Rotating $y = mx$ around the $y$-axis between $x = 0$ and $x = 1$

将 $y = mx$ 绕 $y$ 轴在 $x = 0$ 与 $x = 1$ 之间旋转

282\.

A general cone created by rotating a triangle with vertices $(0,0),$ $(a,0),$ and $(0,b)$ around the $y$-axis. Does your answer agree with the volume of a cone?

将顶点为 $(0,0),$ $(a,0),$ 与 $(0,b)$ 的三角形绕 $y$ 轴旋转所得的通用圆锥。你的答案是否与圆锥体积一致？

283.

A general cylinder created by rotating a rectangle with vertices $(0,0),$ $(a,0),(0,b),$ and $(a,b)$ around the $y$-axis. Does your answer agree with the volume of a cylinder?

将顶点为 $(0,0),$ $(a,0),(0,b),$ 与 $(a,b)$ 的矩形绕 $y$ 轴旋转所得的通用圆柱。你的答案是否与圆柱体积一致？

284\.

A sphere created by rotating a semicircle with radius $a$ around the $y$-axis. Does your answer agree with the volume of a sphere?

将半径为 $a$ 的半圆绕 $y$ 轴旋转所得的球体。你的答案是否与球体积一致？

For the following exercises, use a calculator to draw the region enclosed by the curve. Find the area $M$ and the centroid $\left( {\overset{–}{x},\overset{–}{y}} \right)$ for the given shapes. Use symmetry to help locate the center of mass whenever possible.

对于下列习题，使用计算器画出曲线所围成的区域。求给定形状的相应面积 $M$ 与形心 $\left( {\overset{–}{x},\overset{–}{y}} \right)$。尽可能利用对称性来确定质心的位置。

285.

\[T\] Quarter-circle: $y = \sqrt{1 - x^{2}},$ $y = 0,$ and $x = 0$

\[T\] 四分之一圆：$y = \sqrt{1 - x^{2}},$ $y = 0,$ 与 $x = 0$

286\.

\[T\] Triangle: $y = x,$ $y = 2 - x,$ and $y = 0$

\[T\] 三角形：$y = x,$ $y = 2 - x,$ 与 $y = 0$

287.

\[T\] Lens: $y = x^{2}$ and $y = x$

\[T\] 透镜形：$y = x^{2}$ 与 $y = x$

288\.

\[T\] Ring: $y^{2} + x^{2} = 1$ and $y^{2} + x^{2} = 4$

\[T\] 圆环：$y^{2} + x^{2} = 1$ 与 $y^{2} + x^{2} = 4$

289.

\[T\] Half-ring: $y^{2} + x^{2} = 1,$ $y^{2} + x^{2} = 4,$ and $y = 0$

\[T\] 半环：$y^{2} + x^{2} = 1,$ $y^{2} + x^{2} = 4,$ 与 $y = 0$

290\.

Find the generalized center of mass in the sliver between $y = x^{a}$ and $y = x^{b}$ with $a > b.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the *y*-axis.

求介于 $y = x^{a}$ 与 $y = x^{b}$（$a > b$）之间的狭长区域的广义质心。然后，利用帕普斯定理求绕 *y* 轴旋转所生成的立体体积。

291.

Find the generalized center of mass between $y = a^{2} - x^{2},$ $x = 0,$ and $y = 0.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the *y*-axis.

求介于 $y = a^{2} - x^{2},$ $x = 0,$ 与 $y = 0$ 之间的广义质心。然后，利用帕普斯定理求绕 *y* 轴旋转所生成的立体体积。

292\.

Find the generalized center of mass between $y = b\ \text{sin}(ax),$ $x = 0,$ and $x = \frac{\pi}{a}.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the *y*-axis.

求介于 $y = b\ \text{sin}(ax),$ $x = 0,$ 与 $x = \frac{\pi}{a}$ 之间的广义质心。然后，利用帕普斯定理求绕 *y* 轴旋转所生成的立体体积。

293.

Use the theorem of Pappus to find the volume of a torus (pictured here). Assume that a disk of radius $a$ is positioned with the left end of the circle at $x = b,$ $b > 0,$ and is rotated around the *y*-axis.

利用帕普斯定理求环面（图见此处）的体积。假设半径为 $a$ 的圆盘，其圆的左端位于 $x = b,$ $b > 0,$ 并绕 *y* 轴旋转。

294\.

Find the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ for a thin wire along the semicircle $y = \sqrt{1 - x^{2}}$ with unit mass. (*Hint:* Use the theorem of Pappus.)

求沿半圆 $y = \sqrt{1 - x^{2}}$、具有单位质量的细丝的质心 $\left( {\overset{–}{x},\overset{–}{y}} \right)$。（*提示：* 使用帕普斯定理。）

2.7 Integrals, Exponential Functions, and Logarithms 2.7 积分、指数函数与对数函数

2.7.1 Write the definition of the natural logarithm as an integral.
2.7.2 Recognize the derivative of the natural logarithm.
2.7.3 Integrate functions involving the natural logarithmic function.
2.7.4 Define the number $e$ through an integral.
2.7.5 Recognize the derivative and integral of the exponential function.
2.7.6 Prove properties of logarithms and exponential functions using integrals.
2.7.7 Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.

2.7.1 将自然对数定义为一个积分。
2.7.2 识别自然对数的导数。
2.7.3 积分包含自然对数函数的函数。
2.7.4 通过一个积分定义数 $e$。
2.7.5 识别指数函数的导数与积分。
2.7.6 利用积分证明对数函数与指数函数的性质。
2.7.7 用自然对数与指数来表示一般的对数函数与指数函数。

We already examined exponential functions and logarithms in earlier chapters. However, we glossed over some key details in the previous discussions. For example, we did not study how to treat exponential functions with exponents that are irrational. The definition of the number *e* is another area where the previous development was somewhat incomplete. We now have the tools to deal with these concepts in a more mathematically rigorous way, and we do so in this section.

我们在前面的章节中已经考察过指数函数与对数函数。然而，在之前的讨论中我们略过了一些关键细节。例如，我们并未研究如何处理指数为无理数的指数函数。数 *e* 的定义也是先前讨论中不够完整的一个方面。现在我们拥有了以更严谨的数学方式处理这些概念的工具，本节就来这样做。

For purposes of this section, assume we have not yet defined the natural logarithm, the number *e*, or any of the integration and differentiation formulas associated with these functions. By the end of the section, we will have studied these concepts in a mathematically rigorous way (and we will see they are consistent with the concepts we learned earlier).

就本节而言，假设我们尚未定义自然对数、数 *e*，以及任何与这些函数相关的积分与微分公式。到本节结束时，我们将以严谨的数学方式研究这些概念（并且会看到它们与我们之前学到的概念是一致的）。

We begin the section by defining the natural logarithm in terms of an integral. This definition forms the foundation for the section. From this definition, we derive differentiation formulas, define the number $e,$ and expand these concepts to logarithms and exponential functions of any base.

我们通过对数形式用积分来定义自然对数，以此开始本节。这个定义构成本节的基础。由这个定义出发，我们推导出微分公式，定义数 $e$，并将这些概念推广到任意底的对数函数与指数函数。

The Natural Logarithm as an Integral 用积分定义的自然对数

Recall the power rule for integrals:

回顾积分的幂法则：

$${\int{x^{n}dx = \frac{x^{n + 1}}{n + 1}}} + C,\ n \neq \text{−}1.$$

Clearly, this does not work when $n = -1,$ as it would force us to divide by zero. So, what do we do with ${\int{\frac{1}{x}dx}}?$ Recall from the Fundamental Theorem of Calculus that $\int_{1}^{x}{\frac{1}{t}dt}$ is an antiderivative of $1\text{/}x.$ Therefore, we can make the following definition.

显然，当 $n = -1$ 时这不成立，因为它会迫使我们除以零。那么，对于 ${\int{\frac{1}{x}dx}}$ 我们该怎么办？由微积分基本定理可知，$\int_{1}^{x}{\frac{1}{t}dt}$ 是 $1\text{/}x$ 的一个原函数。因此，我们可以作出如下定义。

For $x > 0,$ define the natural logarithm function by

对于 $x > 0$，用积分定义自然对数函数如下

$$\text{ln}\ x = {\int_{1}^{x}{\frac{1}{t}dt}}.$$ (2.24)

For $x > 1,$ this is just the area under the curve $y = 1\text{/}t$ from $1$ to $x.$ For $x < 1,$ we have ${\int_{1}^{x}{\frac{1}{t}dt}} = \text{−}{\int_{x}^{1}{\frac{1}{t}dt}},$ so in this case it is the negative of the area under the curve from $x\ \text{to}\ 1$ (see the following figure).

对于 $x > 1$，这就是曲线 $y = 1\text{/}t$ 从 $1$ 到 $x$ 下方的面积。对于 $x < 1$，我们有 ${\int_{1}^{x}{\frac{1}{t}dt}} = \text{−}{\int_{x}^{1}{\frac{1}{t}dt}}$，因此在这种情况下，它是曲线从 $x\ \text{to}\ 1$ 下方面积的相反数（见下图）。

Notice that $\text{ln}\ 1 = 0.$ Furthermore, the function $y = 1\text{/}t > 0$ for $x > 0.$ Therefore, by the properties of integrals, it is clear that $\text{ln}\ x$ is increasing for $x > 0.$

注意 $\text{ln}\ 1 = 0$。此外，函数 $y = 1\text{/}t > 0$ 对于 $x > 0$ 成立。因此，由积分的性质可知，$\text{ln}\ x$ 在 $x > 0$ 时是递增的。

Properties of the Natural Logarithm 自然对数的性质

Because of the way we defined the natural logarithm, the following differentiation formula falls out immediately as a result of the Fundamental Theorem of Calculus.

由于我们定义自然对数的方式，下述求导公式作为微积分基本定理的直接结果立刻成立。

Derivative of the Natural Logarithm 自然对数求导

For $x > 0,$ the derivative of the natural logarithm is given by

当 $x > 0,$ 时，自然对数的导数由下式给出

$$\frac{d}{dx}\text{ln}\ x = \frac{1}{x}.$$

Corollary to the Derivative of the Natural Logarithm 自然对数求导的推论

The function $\text{ln}\ x$ is differentiable; therefore, it is continuous.

函数 $\text{ln}\ x$ 可微；因此它是连续的。

A graph of $\text{ln}\ x$ is shown in Figure 2.76. Notice that it is continuous throughout its domain of $\left( {0,\infty} \right).$

$\text{ln}\ x$ 的图像如图 2.76 所示。注意它在定义域 $\left( {0,\infty} \right)$ 上处处连续。

Calculating Derivatives of Natural Logarithms 自然对数导数的计算

Calculate the following derivatives:

计算下列导数：

1. $\frac{d}{dx}\text{ln}\left( {5x^{3} - 2} \right)$

2. $\frac{d}{dx}\left( {\text{ln}\left( {3x} \right)} \right)^{2}$

Solution 解答

We need to apply the chain rule in both cases.

这两种情形我们都需要使用链式法则。

1. $\frac{d}{dx}\text{ln}\left( {5x^{3} - 2} \right) = \frac{15x^{2}}{5x^{3} - 2}$

2. $\frac{d}{dx}\left( {\text{ln}\left( {3x} \right)} \right)^{2} = \frac{2\left( {\text{ln}\left( {3x} \right)} \right) \cdot 3}{3x} = \frac{2\left( {\text{ln}\left( {3x} \right)} \right)}{x}$

Calculate the following derivatives:

计算下列导数：

1. $\frac{d}{dx}\text{ln}\left( {2x^{2} + x} \right)$

2. $\frac{d}{dx}\left( {\text{ln}\left( x^{3} \right)} \right)^{2}$

Note that if we use the absolute value function and create a new function $\text{ln}\ |x|,$ we can extend the domain of the natural logarithm to include $x < 0.$ Then $\left( {d\text{/}\left( {dx} \right)} \right)\text{ln}\ |x| = 1\text{/}x.$ This gives rise to the familiar integration formula.

注意，若使用绝对值函数并构造新函数 $\text{ln}\ |x|,$ 我们便可将自然对数的定义域扩展到包含 $x < 0.$ 此时 $\left( {d\text{/}\left( {dx} \right)} \right)\text{ln}\ |x| = 1\text{/}x.$ 这就导出了我们熟悉的积分公式。

Integral of (1/u) du （1/u）du 的积分

The natural logarithm is the antiderivative of the function $f(u) = 1\text{/}u\text{:}$

自然对数是函数 $f(u) = 1\text{/}u\text{:}$ 的原函数：

$${\int\frac{1}{u}}du = \text{ln}\ |u| + C.$$

Calculating Integrals Involving Natural Logarithms 含自然对数的积分计算

Calculate the integral ${\int\frac{x}{x^{2} + 4}}dx.$

计算下列积分 ${\int\frac{x}{x^{2} + 4}}dx.$

Solution 解答

Using $u$-substitution, let $u = x^{2} + 4.$ Then $du = 2x\ dx$ and we have

使用换元法，令 $u = x^{2} + 4.$ 于是 $du = 2x\ dx$，我们得到

$${\int\frac{x}{x^{2} + 4}}dx = \frac{1}{2}{\int{\frac{1}{u}du = \frac{1}{2}\text{ln}\ |u| + C =}}\frac{1}{2}\text{ln}\ \left| {x^{2} + 4} \right| + C = \frac{1}{2}\text{ln}\left( {x^{2} + 4} \right) + C.$$

Calculate the integral ${\int\frac{x^{2}}{x^{3} + 6}}dx.$

计算下列积分 ${\int\frac{x^{2}}{x^{3} + 6}}dx.$

Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.

虽然我们把这个函数称为“对数”，但我们尚未真正证明对数性质对它都成立。我们在此证明。

Properties of the Natural Logarithm 自然对数的性质

If $a,b > 0$ and $r$ is a rational number, then

若 $a,b > 0$ 且 $r$ 为有理数，则

1. $\text{ln}\ 1 = 0$

2. $\text{ln}\left( {ab} \right) = \text{ln}\ a + \text{ln}\ b$

3. $\text{ln}\left( \frac{a}{b} \right) = \text{ln}\ a - \text{ln}\ b$

4. $\text{ln}\left( a^{r} \right) = r\ \text{ln}\ a$

Proof 证明

i\. By definition, $\text{ln}\ 1 = {\int_{1}^{1}\frac{1}{t}}dt = 0.$

i\. 由定义，$\text{ln}\ 1 = {\int_{1}^{1}\frac{1}{t}}dt = 0.$

ii\. We have

ii\. 我们有

$$\text{ln}\left( {ab} \right) = {\int_{1}^{ab}\frac{1}{t}}dt = {\int_{1}^{a}\frac{1}{t}}dt + {\int_{a}^{ab}\frac{1}{t}}dt.$$

Use $u\text{-substitution}$ on the last integral in this expression. Let $u = t\text{/}a.$ Then $du = \left( {1\text{/}a} \right)dt.$ Furthermore, when $t = a,u = 1,$ and when $t = ab,u = b.$ So we get

对上式中最后一个积分使用换元法。令 $u = t\text{/}a.$ 于是 $du = \left( {1\text{/}a} \right)dt.$ 并且，当 $t = a$ 时 $u = 1,$ 当 $t = ab$ 时 $u = b.$ 于是我们得到

$$\text{ln}\left( {ab} \right) = {\int_{1}^{a}\frac{1}{t}}dt + {\int_{a}^{ab}\frac{1}{t}}dt = {\int_{1}^{a}\frac{1}{t}}dt + {\int_{a}^{ab}{\frac{a}{t} \cdot \frac{1}{a}}}dt = {\int_{1}^{a}{\frac{1}{t}dt +}}{\int_{1}^{b}{\frac{1}{u}du = \text{ln}\ a + \text{ln}\ b.}}$$

iv\. Note that

iv\. 注意

$$\frac{d}{dx}\text{ln}\left( x^{r} \right) = \frac{rx^{r - 1}}{x^{r}} = \frac{r}{x}.$$

Furthermore,

此外，

$$\frac{d}{dx}\left( {r\ \text{ln}\ x} \right) = \frac{r}{x}.$$

Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have

由于这两个函数的导数相同，根据微积分基本定理，它们必定相差一个常数。于是我们有

$$\text{ln}\left( x^{r} \right) = r\ \text{ln}\ x + C$$

for some constant $C.$ Taking $x = 1,$ we get

其中 $C$ 为某常数。取 $x = 1,$ 我们得到

$$\begin{array}{rll} {\text{ln}\left( 1^{r} \right)} & = & {r\ \text{ln}(1) + C} \\ 0 & = & {r(0) + C} \\ C & = & {0.} \end{array}$$

Thus $\text{ln}\left( x^{r} \right) = r\ \text{ln}\ x$ and the proof is complete. Note that we can extend this property to irrational values of $r$ later in this section.

于是 $\text{ln}\left( x^{r} \right) = r\ \text{ln}\ x$，证明完毕。注意在本节后文我们可将该性质推广到 $r$ 的无理数值。

Part iii. follows from parts ii. and iv. and the proof is left to you.

第 iii. 部分可由第 ii. 与第 iv. 部分推出，证明留作练习。

□

Using Properties of Logarithms 利用对数性质

Use properties of logarithms to simplify the following expression into a single logarithm:

利用对数的性质，将下列表达式化简为单个对数：

$$\text{ln}\ 9 - 2\ \text{ln}\ 3 + \text{ln}\left( \frac{1}{3} \right).$$

Solution 解答

We have

我们有

$$\text{ln}\ 9 - 2\ \text{ln}\ 3 + \text{ln}\left( \frac{1}{3} \right) = \text{ln}\left( 3^{2} \right) - 2\ \text{ln}\ 3 + \text{ln}\left( 3^{-1} \right) = 2\ \text{ln}\ 3 - 2\ \text{ln}\ 3 - \text{ln}\ 3 = \text{−}\text{ln}\ 3.$$

Use properties of logarithms to simplify the following expression into a single logarithm:

利用对数的性质，将下列表达式化简为单个对数：

$$\text{ln}\ 8 - \text{ln}\ 2 - \text{ln}\left( \frac{1}{4} \right).$$

Defining the Number e 定义数 e

Now that we have the natural logarithm defined, we can use that function to define the number $e.$

既然已经定义了自然对数，我们便可以用该函数来定义数 $e.$

The number $e$ is defined to be the real number such that

数 $e$ 被定义为满足下式的实数

$$\text{ln}\ e = 1.$$

To put it another way, the area under the curve $y = 1\text{/}t$ between $t = 1$ and $t = e$ is $1$ (Figure 2.77). The proof that such a number exists and is unique is left to you. (*Hint*: Use the Intermediate Value Theorem to prove existence and the fact that $\text{ln}\ x$ is increasing to prove uniqueness.)

换言之，曲线 $y = 1\text{/}t$ 在 $t = 1$ 与 $t = e$ 之间的面积为 $1$（图 2.77）。该数存在且唯一的证明留作练习。（*提示*：用介值定理证明存在性，并用 $\text{ln}\ x$ 单调递增的事实证明唯一性。）

The number $e$ can be shown to be irrational, although we won’t do so here (see the Student Project in Taylor and Maclaurin Series). Its approximate value is given by

可以证明数 $e$ 是无理数，不过我们在此不作证明（见 Taylor 级数与 Maclaurin 级数中的学生项目）。它的近似值由下式给出

$$e \approx 2.71828182846.$$

The Exponential Function 指数函数

We now turn our attention to the function $e^{x}.$ Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by $\text{exp}\ x.$ Then,

现在我们把注意力转向函数 $e^{x}.$ 注意自然对数是一一对应的，因此具有反函数。目前，我们用 $\text{exp}\ x$ 表示这个反函数。于是，

$$\text{exp}\left( {\text{ln}\ x} \right) = x\ \text{for}\ x > 0\ \text{and}\ \text{ln}\left( {\text{exp}\ x} \right) = x\ \text{for all}\ x.$$

The following figure shows the graphs of $\text{exp}\ x$ and $\text{ln}\ x.$

下图展示了 $\text{exp}\ x$ 与 $\text{ln}\ x$ 的图像。

We hypothesize that $\text{exp}\ x = e^{x}.$ For rational values of $x,$ this is easy to show. If $x$ is rational, then we have $\text{ln}\left( e^{x} \right) = x\ \text{ln}\ e = x.$ Thus, when $x$ is rational, $e^{x} = \text{exp}\ x.$ For irrational values of $x,$ we simply define $e^{x}$ as the inverse function of $\text{ln}\ x.$

我们猜想 $\text{exp}\ x = e^{x}.$ 当 $x$ 为有理数时，这很容易证明。若 $x$ 为有理数，则有 $\text{ln}\left( e^{x} \right) = x\ \text{ln}\ e = x.$ 因此，当 $x$ 为有理数时 $e^{x} = \text{exp}\ x.$ 对于 $x$ 的无理数值，我们直接将 $e^{x}$ 定义为 $\text{ln}\ x$ 的反函数。

For any real number $x,$ define $y = e^{x}$ to be the number for which

对任意实数 $x,$ 定义 $y = e^{x}$ 为满足下式的数

$$\text{ln}\ y = \text{ln}\left( e^{x} \right) = x.$$ (2.25)

Then we have $e^{x} = \text{exp}(x)$ for all $x,$ and thus

于是对于任意 $x$ 都有 $e^{x} = \text{exp}(x)$，从而

$$e^{\text{ln}\ x} = x\ \text{for}\ x > 0\ \text{and}\ \text{ln}\left( e^{x} \right) = x$$ (2.26)

for all $x.$

对任意 $x$ 均成立。

Properties of the Exponential Function 指数函数的性质

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of $e,$ we must verify that the usual laws of exponents hold for the function $e^{x}.$

由于指数函数是通过反函数来定义的，而非通过 $e$ 的幂来定义，我们必须验证通常的指数律对函数 $e^{x}$ 成立。

Properties of the Exponential Function 指数函数的性质

If $p$ and $q$ are any real numbers and $r$ is a rational number, then

若 $p$ 与 $q$ 为任意实数，且 $r$ 为有理数，则

1. $e^{p}e^{q} = e^{p + q}$

2. $\frac{e^{p}}{e^{q}} = e^{p - q}$

3. $\left( e^{p} \right)^{r} = e^{pr}$

Proof 证明

Note that if $p$ and $q$ are rational, the properties hold. However, if $p$ or $q$ are irrational, we must apply the inverse function definition of $e^{x}$ and verify the properties. Only the first property is verified here; the other two are left to you. We have

注意，若 $p$ 与 $q$ 为有理数，则这些性质成立。然而，若 $p$ 或 $q$ 为无理数，我们必须运用 $e^{x}$ 的反函数定义来验证这些性质。此处只验证第一条性质；其余两条留作练习。我们有

$$\text{ln}\left( {e^{p}e^{q}} \right) = \text{ln}\left( e^{p} \right) + \text{ln}\left( e^{q} \right) = p + q = \text{ln}\left( e^{p + q} \right).$$

Since $\text{ln}\ x$ is one-to-one, then

由于 $\text{ln}\ x$ 是一一对应的，于是

$$e^{p}e^{q} = e^{p + q}.$$

□

As with part iv. of the logarithm properties, we can extend property iii. to irrational values of $r,$ and we do so by the end of the section.

与对数性质的第 iv. 部分类似，我们可以把第 iii. 条性质推广到 $r$ 的无理数值，并在本节结束时完成。

We also want to verify the differentiation formula for the function $y = e^{x}.$ To do this, we need to use implicit differentiation. Let $y = e^{x}.$ Then

我们还想验证函数 $y = e^{x}$ 的求导公式。为此，需要使用隐函数求导法。令 $y = e^{x}.$ 于是

$$\begin{array}{rll} {\text{ln}\ y} & = & x \\ {\frac{d}{dx}\text{ln}\ y} & = & {\frac{d}{dx}x} \\ {\frac{1}{y}\mspace{2mu}\frac{dy}{dx}} & = & 1 \\ \frac{dy}{dx} & = & {y.} \end{array}$$

Thus, we see

于是我们看到

$$\frac{d}{dx}e^{x} = e^{x}$$

as desired, which leads immediately to the integration formula

正如所愿，由此立刻得到积分公式

$${\int{e^{x}dx}} = e^{x} + C.$$

We apply these formulas in the following examples.

我们在下面的例子中应用这些公式。

Using Properties of Exponential Functions 利用指数函数的性质

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dt}e^{3t}e^{t^{2}}$

2. $\frac{d}{dx}e^{3x^{2}}$

Solution 解答

We apply the chain rule as necessary.

我们视需要使用链式法则。

1. $\frac{d}{dt}e^{3t}e^{t^{2}} = \frac{d}{dt}e^{3t + t^{2}} = e^{3t + t^{2}}\left( {3 + 2t} \right)$

2. $\frac{d}{dx}e^{3x^{2}} = e^{3x^{2}}6x$

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dx}\left( \frac{e^{x^{2}}}{e^{5x}} \right)$

2. $\frac{d}{dt}\left( e^{2t} \right)^{3}$

Using Properties of Exponential Functions 利用指数函数的性质

Evaluate the following integral: ${\int{2xe^{\text{−}x^{2}}dx}}.$

求下列积分：${\int{2xe^{\text{−}x^{2}}dx}}.$

Solution 解答

Using $u$-substitution, let $u = \text{−}x^{2}.$ Then $du = -2x\ dx,$ and we have

使用换元法，令 $u = \text{−}x^{2}.$ 于是 $du = -2x\ dx$，我们得到

$${\int{2xe^{\text{−}x^{2}}dx}} = \text{−}{\int{e^{u}du}} = \text{−}e^{u} + C = \text{−}e^{\text{−}x^{2}} + C.$$

Evaluate the following integral: ${\int\frac{4}{e^{3x}}}dx.$

求下列积分：${\int\frac{4}{e^{3x}}}dx.$

General Logarithmic and Exponential Functions 一般对数与指数函数

We close this section by looking at exponential functions and logarithms with bases other than $e.$ Exponential functions are functions of the form $f(x) = a^{x}.$ Note that unless $a = e,$ we still do not have a mathematically rigorous definition of these functions for irrational exponents. Let’s rectify that here by defining the function $f(x) = a^{x}$ in terms of the exponential function $e^{x}.$ We then examine logarithms with bases other than $e$ as inverse functions of exponential functions.

本节最后，我们考察以 $e$ 以外为底的指数函数与对数函数。指数函数是指形如 $f(x) = a^{x}$ 的函数。注意，除非 $a = e,$ 否则我们对这些函数在无理数指数下的定义仍不严格。我们在此通过将函数 $f(x) = a^{x}$ 用指数函数 $e^{x}$ 来定义，以弥补这一不足。然后，我们把以 $e$ 以外为底的对数函数作为指数函数的反函数加以研究。

For any $a > 0,$ and for any real number $x,$ define $y = a^{x}$ as follows:

对任意 $a > 0,$ 以及任意实数 $x,$ 定义 $y = a^{x}$ 如下：

$$y = a^{x} = e^{x\ \text{ln}\ a}.$$

Now $a^{x}$ is defined rigorously for all values of *x*. This definition also allows us to generalize property iv. of logarithms and property iii. of exponential functions to apply to both rational and irrational values of $r.$ It is straightforward to show that properties of exponents hold for general exponential functions defined in this way.

现在 $a^{x}$ 对所有 *x* 值都有了严格的定义。这个定义还使我们可以将对数的第 iv. 条性质与指数函数的第 iii. 条性质推广到 $r$ 的有理与无理数值。不难看出，如此定义的一般指数函数满足指数律。

Let’s now apply this definition to calculate a differentiation formula for $a^{x}.$ We have

现在应用这个定义来推导 $a^{x}$ 的求导公式。我们有

$$\frac{d}{dx}a^{x} = \frac{d}{dx}e^{x\ \text{ln}\ a} = e^{x\ \text{ln}\ a}\text{ln}\ a = a^{x}\text{ln}\ a.$$

The corresponding integration formula follows immediately.

相应的积分公式随即得出。

Derivatives and Integrals Involving General Exponential Functions 含一般指数函数的求导与积分

Let $a > 0.$ Then,

设 $a > 0.$ 则

$$\frac{d}{dx}a^{x} = a^{x}\text{ln}\ a$$

and

且

$${\int{a^{x}dx}} = \frac{1}{\text{ln}\ a}a^{x} + C.$$

If $a \neq 1,$ then the function $a^{x}$ is one-to-one and has a well-defined inverse. Its inverse is denoted by $\text{log}_{a}x.$ Then,

若 $a \neq 1,$ 则函数 $a^{x}$ 是一一对应的，并且具有定义良好的反函数。其反函数记为 $\text{log}_{a}x.$ 于是，

$$y = \text{log}_{a}x\ \text{if and only if}\ x = a^{y}.$$

Note that general logarithm functions can be written in terms of the natural logarithm. Let $y = \text{log}_{a}x.$ Then, $x = a^{y}.$ Taking the natural logarithm of both sides of this second equation, we get

注意一般对数函数可以用自然对数表示。令 $y = \text{log}_{a}x.$ 则 $x = a^{y}.$ 对后一方程两边取自然对数，我们得到

$$\begin{array}{rll} {\text{ln}\ x} & = & {\text{ln}\left( a^{y} \right)} \\ {\text{ln}\ x} & = & {y\ \text{ln}\ a} \\ y & = & \frac{\text{ln}\ x}{\text{ln}\ a} \\ {\text{log}_{a}x} & = & {\frac{\text{ln}\ x}{\text{ln}\ a}.} \end{array}$$

Thus, we see that all logarithmic functions are constant multiples of one another. Next, we use this formula to find a differentiation formula for a logarithm with base $a.$ Again, let $y = \text{log}_{a}x.$ Then,

于是我们看到，所有对数函数彼此之间都是常数倍的关系。接下来利用这个公式求以 $a$ 为底的对数的求导公式。再次令 $y = \text{log}_{a}x.$ 则

$$\begin{array}{cl} \frac{dy}{dx} & {= \frac{d}{dx}\left( {\text{log}_{a}x} \right)} \\ & {= \frac{d}{dx}\left( \frac{\text{ln}\ x}{\text{ln}\ a} \right)} \\ & {= \left( \frac{1}{\text{ln}\ a} \right)\frac{d}{dx}\left( {\text{ln}\ x} \right)} \\ & {= \frac{1}{\text{ln}\ a} \cdot \frac{1}{x}} \\ & {= \frac{1}{x\ \text{ln}\ a}.} \end{array}$$

Derivatives of General Logarithm Functions 一般对数函数求导

Let $a > 0.$ Then,

设 $a > 0.$ 则

$$\frac{d}{dx}\text{log}_{a}x = \frac{1}{x\ \text{ln}\ a}.$$

Calculating Derivatives of General Exponential and Logarithm Functions 一般指数与对数函数导数的计算

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dt}\left( {4^{t} \cdot 2^{t^{2}}} \right)$

2. $\frac{d}{dx}\text{log}_{8}\left( {7x^{2} + 4} \right)$

Solution 解答

We need to apply the chain rule as necessary.

我们视需要使用链式法则。

1. $\frac{d}{dt}\left( {4^{t} \cdot 2^{t^{2}}} \right) = \frac{d}{dt}\left( {2^{2t} \cdot 2^{t^{2}}} \right) = \frac{d}{dt}\left( 2^{2t + t^{2}} \right) = 2^{2t + t^{2}}\text{ln}(2)\left( {2 + 2t} \right)$

2. $\frac{d}{dx}\text{log}_{8}\left( {7x^{2} + 4} \right) = \frac{1}{\left( {7x^{2} + 4} \right)\left( {\text{ln}\ 8} \right)}\left( {14x} \right)$

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dt}\ 4^{t^{4}}$

2. $\frac{d}{dx}\text{log}_{3}\left( \sqrt{x^{2} + 1} \right)$

Integrating General Exponential Functions 一般指数函数的积分

Evaluate the following integral: ${\int\frac{3}{2^{3x}}}dx.$

求下列积分：${\int\frac{3}{2^{3x}}}dx.$

Solution 解答

Use $u\text{-substitution}$ and let $u = -3x.$ Then $du = -3dx$ and we have

使用换元法，令 $u = -3x.$ 于是 $du = -3dx$，我们得到

$${\int\frac{3}{2^{3x}}}dx = {\int{3 \cdot 2^{-3x}}}dx = \text{−}{\int{2^{u}du}} = - \frac{1}{\text{ln}\ 2}2^{u} + C = - \frac{1}{\text{ln}\ 2}2^{-3x} + C.$$

Evaluate the following integral: ${\int{x^{2}2^{x^{3}}dx}}.$

求下列积分：${\int{x^{2}2^{x^{3}}dx}}.$

Section 2.7 Exercises 2.7 节习题

For the following exercises, find the derivative $\frac{dy}{dx}.$

对下列习题，求导数 $\frac{dy}{dx}.$

295.

$y = \text{ln}\left( {2x} \right)$

296\.

$y = \text{ln}\left( {2x + 1} \right)$

297.

$y = \frac{1}{\text{ln}\ x}$

For the following exercises, find the indefinite integral.

对下列习题，求不定积分。

298\.

$\int\frac{dt}{3t}$

299.

$\int\frac{dx}{1 + x}$

For the following exercises, find the derivative $dy\text{/}dx.$ (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

对下列习题，求导数 $dy\text{/}dx.$（可以使用计算器绘制函数及其导数图像，以确认结果是否正确。）

300\.

\[T\] $y = \frac{\text{ln}(x)}{x}$

301.

\[T\] $y = x\ \text{ln}(x)$

302\.

\[T\] $y = \text{log}_{10}x$

303.

\[T\] $y = \text{ln}\left( {\text{sin}\ x} \right)$

304\.

\[T\] $y = \text{ln}\left( {\text{ln}\ x} \right)$

305.

\[T\] $y = 7\ \text{ln}\left( {4x} \right)$

306\.

\[T\] $y = \text{ln}\left( \left( {4x} \right)^{7} \right)$

307.

\[T\] $y = \text{ln}\left( {\text{tan}\ x} \right)$

308\.

\[T\] $y = \text{ln}\left( {\text{tan}\left( {3x} \right)} \right)$

309.

\[T\] $y = \text{ln}\left( {\text{cos}^{2}x} \right)$

For the following exercises, find the definite or indefinite integral.

对下列习题，求定积分或不定积分。

310\.

$\int_{0}^{1}\frac{dx}{3 + x}$

311.

$\int_{0}^{1}\frac{dt}{3 + 2t}$

312\.

$\int_{0}^{2}\frac{x\ dx}{x^{2} + 1}$

313.

$\int_{0}^{2}\frac{x^{3}dx}{x^{2} + 1}$

314\.

$\int_{2}^{e}\frac{dx}{x\ \text{ln}\ x}$

315.

$\int_{2}^{e}\frac{dx}{{x\ (\text{ln}{\ x)}}^{2}}$

316\.

$\int\frac{\text{cos}\ x\ dx}{\text{sin}\ x}$

317.

$\int_{0}^{\pi\text{/}4}{\text{tan}\ x\ dx}$

318\.

$\int{\text{cot}\left( {3x} \right)dx}$

319.

$\int\frac{\left( {\text{ln}\ x} \right)^{2}dx}{x}$

For the following exercises, compute $dy\text{/}dx$ by differentiating $\text{ln}\ y.$

对下列习题，通过对 $\text{ln}\ y$ 求导来计算 $dy\text{/}dx$。

320\.

$y = \sqrt{x^{2} + 1}$

321.

$y = \sqrt{x^{2} + 1}\sqrt{x^{2} - 1}$

322\.

$y = e^{\text{sin}\ x}$

323.

$y = x^{-1\text{/}x}$

324\.

$y = e^{({ex})}$

325.

$y = x^{e}$

326\.

$y = x^{({ex})}$

327.

$y = \sqrt{x}\ \sqrt[3]{x}\ \sqrt[6]{x}$

328\.

$y = x^{-1\text{/}\text{ln}\ x}$

329.

$y = e^{\text{−}\text{ln}\ x}$

For the following exercises, evaluate by any method.

对下列习题，用任意方法求值。

330\.

$\int_{5}^{10}{\frac{dt}{t} - {\int_{5x}^{10x}\frac{dt}{t}}}$

331.

${\int_{1}^{e^{\pi}}\frac{dx}{x}} + {\int_{-2}^{-1}\frac{dx}{x}}$

332\.

$\frac{d}{dx}{\int_{x}^{1}\frac{dt}{t}}$

333.

$\frac{d}{dx}{\int_{x}^{x^{2}}\frac{dt}{t}}$

334\.

$\frac{d}{dx}\text{ln}\left( {\text{sec}\ x + \text{tan}\ x} \right)$

For the following exercises, use the function $\text{ln}\ x.$ If you are unable to find intersection points analytically, use a calculator.

对下列习题，使用函数 $\text{ln}\ x.$ 如果无法用解析法求出交点，请使用计算器。

335.

Find the area of the region enclosed by $x = 1$ and $y = 5$ above $y = \text{ln}\ x.$

求由 $x = 1$ 与 $y = 5$ 所围成、且位于 $y = \text{ln}\ x$ 上方的区域面积。

336\.

\[T\] Find the arc length of $\text{ln}\ x$ from $x = 1$ to $x = 2.$

\[T\] 求 $\text{ln}\ x$ 从 $x = 1$ 到 $x = 2$ 的弧长。

337.

Find the area between $\text{ln}\ x$ and the *x*-axis from $x = 1\ \text{to}\ x = 2.$

求 $\text{ln}\ x$ 与 *x* 轴之间从 $x = 1\ \text{到}\ x = 2$ 的面积。

338\.

Find the volume of the shape created when rotating this curve from $x = 1\ \text{to}\ x = 2$ around the *x*-axis, as pictured here.

求将此曲线从 $x = 1\ \text{到}\ x = 2$ 绕 *x* 轴旋转所形成的几何体体积（如图所示）。

339.

\[T\] Find the surface area of the shape created when rotating the curve in the previous exercise from $x = 1$ to $x = 2$ around the *x*-axis.

\[T\] 求将上一题曲线从 $x = 1$ 到 $x = 2$ 绕 *x* 轴旋转所形成的旋转面的表面积。

If you are unable to find intersection points analytically in the following exercises, use a calculator.

如果在下列习题中无法用解析法求出交点，请使用计算器。

340\.

Find the area of the hyperbolic quarter-circle enclosed by $x = 2\ \text{and}\ y = 2$ above $y = 1\text{/}x.$

求由 $x = 2\ \text{与}\ y = 2$ 所围成、且位于 $y = 1\text{/}x$ 上方的双曲四分之一圆的面积。

341.

\[T\] Find the arc length of $y = 1\text{/}x$ from $x = 1\ \text{to}\ x = 4.$

\[T\] 求 $y = 1\text{/}x$ 从 $x = 1\ \text{到}\ x = 4$ 的弧长。

342\.

Find the area under $y = 1\text{/}x$ and above the *x*-axis from $x = 1\ \text{to}\ x = 4.$

求 $y = 1\text{/}x$ 下方、*x* 轴上方从 $x = 1\ \text{到}\ x = 4$ 的面积。

For the following exercises, verify the derivatives and antiderivatives.

对下列习题，验证导数与反导数。

343\.

$\frac{d}{dx}\text{ln}\left( {x + \sqrt{x^{2} + 1}} \right) = \frac{1}{\sqrt{1 + x^{2}}}$

344.

$\frac{d}{dx}\text{ln}\left( \frac{x - a}{x + a} \right) = \frac{2a}{\left( {x^{2} - a^{2}} \right)}$

345\.

$\frac{d}{dx}\text{ln}\left( \frac{1 + \sqrt{1 - x^{2}}}{x} \right) = - \frac{1}{x\sqrt{1 - x^{2}}}$

346.

$\frac{d}{dx}\text{ln}\left( {x + \sqrt{x^{2} - a^{2}}} \right) = \frac{1}{\sqrt{x^{2} - a^{2}}}$

347\.

${\int\frac{dx}{x\ \text{ln}(x)\text{ln}\left( {\text{ln}\ x} \right)}} = \text{ln}\left( {\text{ln}\left( {\text{ln}\ x} \right)} \right) + C$

2.8 Exponential Growth and Decay 2.8 指数增长与衰减

2.8.1 Use the exponential growth model in applications, including population growth and compound interest.
2.8.2 Explain the concept of doubling time.
2.8.3 Use the exponential decay model in applications, including radioactive decay and Newton's law of cooling.
2.8.4 Explain the concept of half-life.

2.8.1 在应用中使用指数增长模型，包括人口增长与复利。
2.8.2 解释倍增时间的概念。
2.8.3 在应用中使用指数衰减模型，包括放射性衰变与牛顿冷却定律。
2.8.4 解释半衰期的概念。

One of the most prevalent applications of exponential functions involves growth and decay models. Exponential growth and decay show up in a host of natural applications. From population growth and continuously compounded interest to radioactive decay and Newton's law of cooling, exponential functions are ubiquitous in nature. In this section, we examine exponential growth and decay in the context of some of these applications.

指数函数最普遍的应用之一涉及增长与衰减模型。指数增长与衰减出现在大量自然应用中。从人口增长与连续复利，到放射性衰变与牛顿冷却定律，指数函数在自然界中无处不在。本节中，我们在其中一些应用的背景下考察指数增长与衰减。

Exponential Growth Model 指数增长模型

Many systems exhibit exponential growth. These systems follow a model of the form $y = y_{0}e^{kt},$ where $y_{0}$ represents the initial state of the system and $k$ is a positive constant, called the *growth constant*. Notice that in an exponential growth model, we have

许多系统表现出指数增长。这些系统遵循形如 $y = y_{0}e^{kt}$ 的模型，其中 $y_{0}$ 表示系统的初始状态，$k$ 为一个正的常数，称为*增长常数*。注意，在指数增长模型中，我们有

$$y^{\prime} = ky_{0}e^{kt} = ky.$$ (2.27)

That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth. Equation 2.27 involves derivatives and is called a *differential equation.* We learn more about differential equations in Introduction to Differential Equations.

也就是说，增长率与函数的当前值成正比。这是指数增长的一个关键特征。方程 2.27 涉及导数，被称为*微分方程*。我们将在《微分方程导论》中更详细地学习微分方程。

Systems that exhibit exponential growth increase according to the mathematical model

表现出指数增长的系统按照如下数学模型增长

$$y = y_{0}e^{kt},$$

where $y_{0}$ represents the initial state of the system and $k > 0$ is a constant, called the *growth constant*.

其中 $y_{0}$ 表示系统的初始状态，$k > 0$ 为常数，称为*增长常数*。

Population growth is a common example of exponential growth. Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows. Figure 2.79 and Table 2.1 represent the growth of a population of bacteria with an initial population of $200$ bacteria and a growth constant of $0.02.$ Notice that after only $2$ hours $(120$ minutes), the population is $10$ times its original size!

人口增长是指数增长的一个常见例子。以细菌种群为例。人口增长率与种群规模成正比，这看来是合理的。毕竟，可繁殖的细菌越多，种群增长就越快。图 2.79 与表 2.1 展示了一个初始种群为 $200$ 个细菌、增长常数为 $0.02$ 的细菌种群的增长情况。注意，仅仅经过 $2$ 小时（即 $120$ 分钟），种群规模就达到原来规模的 $10$ 倍！

Table 2.1 Exponential Growth of a Bacterial Population

表 2.1 细菌种群的指数增长

Time (min)	Population Size (no. of bacteria)
$10$	$244$
$20$	$298$
$30$	$364$
$40$	$445$
$50$	$544$
$60$	$664$
$70$	$811$
$80$	$991$
$90$	$1210$
$100$	$1478$
$110$	$1805$
$120$	$2205$

时间（分钟）	种群规模（细菌数量）
$10$	$244$
$20$	$298$
$30$	$364$
$40$	$445$
$50$	$544$
$60$	$664$
$70$	$811$
$80$	$991$
$90$	$1210$
$100$	$1478$
$110$	$1805$
$120$	$2205$

Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.

注意，我们使用一个连续函数来建模本质上离散的行为。在任何给定时刻，现实世界中的种群都包含整数个细菌，尽管模型会取到非整数值。在使用指数增长模型时，我们必须始终谨慎，要在所建模现象的背景下解释函数值。

Population Growth 种群增长

Consider the population of bacteria described earlier. This population grows according to the function $f(t) = 200e^{0.02t},$ where *t* is measured in minutes. How many bacteria are present in the population after $5$ hours $(300$ minutes)? When does the population reach $100,000$ bacteria?

考虑前面描述的细菌种群。该种群按照函数 $f(t) = 200e^{0.02t}$ 增长，其中 *t* 以分钟为单位。经过 $5$ 小时（即 $300$ 分钟）后，种群中有多少个细菌？种群何时达到 $100,000$ 个细菌？

Solution 解答

We have $f(t) = 200e^{0.02t}.$ Then

我们有 $f(t) = 200e^{0.02t}.$ 于是

$$f(300) = 200e^{0.02{(300)}} \approx 80,686.$$

There are $80,686$ bacteria in the population after $5$ hours.

经过 $5$ 小时后，种群中有 $80,686$ 个细菌。

To find when the population reaches $100,000$ bacteria, we solve the equation

为了找出种群何时达到 $100,000$ 个细菌，我们解方程

$$\begin{array}{rll} 100,000 & = & {200e^{0.02t}} \\ 500 & = & e^{0.02t} \\ {\text{ln}\ 500} & = & {0.02t} \\ t & = & {\frac{\text{ln}\ 500}{0.02} \approx 310.73.} \end{array}$$

The population reaches $100,000$ bacteria after $310.73$ minutes.

经过 $310.73$ 分钟后，种群达到 $100,000$ 个细菌。

Consider a population of bacteria that grows according to the function $f(t) = 500e^{0.05t},$ where $t$ is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach $100$ million bacteria?

考虑一个按照函数 $f(t) = 500e^{0.05t}$ 增长的细菌种群，其中 $t$ 以分钟为单位。经过 4 小时后，种群中有多少个细菌？种群何时达到 $100$ 百万个细菌？

Let's now turn our attention to a financial application: compound interest. Interest that is not compounded is called *simple interest*. Simple interest is paid once, at the end of the specified time period (usually $1$ year). So, if we put $\text{\$}1000$ in a savings account earning $2\text{\%}$ simple interest per year, then at the end of the year we have

现在我们把注意力转向一个金融应用：复利。未被复利的利息称为*单利*。单利在指定期限（通常为 $1$ 年）结束时支付一次。因此，如果我们将 $\text{\$}1000$ 存入一个按每年 $2\text{\%}$ 支付单利的储蓄账户，那么到年底我们有

$$1000\left( {1 + 0.02} \right) = \text{\$}1020.$$

Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every $6$ months, it credits half of the year's interest to the account after $6$ months. During the second half of the year, the account earns interest not only on the initial $\text{\$}1000,$ but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have

复利每年支付多次，具体取决于复利周期。因此，如果银行每 $6$ 个月复利一次，它会在 $6$ 个月后将半年利息记入账户。在后半年中，账户不仅根据初始的 $\text{\$}1000$ 赚取利息，还根据前半年赚取的利息赚取利息。用数学语言来说，到年底我们有

$$1000\left( {1 + \frac{0.02}{2}} \right)^{2} = \text{\$}1020.10.$$

Similarly, if the interest is compounded every $4$ months, we have

类似地，如果利息每 $4$ 个月复利一次，我们有

$$1000\left( {1 + \frac{0.02}{3}} \right)^{3} = \text{\$}1020.13,$$

and if the interest is compounded daily $(365$ times per year), we have $\text{\$}1020.20.$ If we extend this concept, so that the interest is compounded continuously, after $t$ years we have

而如果利息每天复利（每年 $365$ 次），我们有 $\text{\$}1020.20.$ 如果我们延伸这一概念，使利息连续复利，那么在 $t$ 年后我们有

$$1000\underset{n\rightarrow\infty}{\text{lim}}\left( {1 + \frac{0.02}{n}} \right)^{nt}.$$

Now let's manipulate this expression so that we have an exponential growth function. Recall that the number $e$ can be expressed as a limit:

现在我们来变换这个表达式，使其成为指数增长函数。回顾数 $e$ 可以表示为一个极限：

$$e = \underset{m\rightarrow\infty}{\text{lim}}\left( {1 + \frac{1}{m}} \right)^{m}.$$

Based on this, we want the expression inside the parentheses to have the form $\left( {1 + {1\text{/}m}} \right).$ Let $n = 0.02m.$ Note that as $n\rightarrow\infty,$ $m\rightarrow\infty$ as well. Then we get

基于这一点，我们希望括号内的表达式具有 $\left( {1 + {1\text{/}m}} \right)$ 的形式。令 $n = 0.02m.$ 注意，当 $n\rightarrow\infty$ 时，$m\rightarrow\infty$ 同样成立。于是我们得到

$$1000\underset{n\rightarrow\infty}{\text{lim}}\left( {1 + \frac{0.02}{n}} \right)^{nt} = 1000\underset{m\rightarrow\infty}{\text{lim}}\left( {1 + \frac{0.02}{0.02m}} \right)^{0.02mt} = 1000\left\lbrack {\underset{m\rightarrow\infty}{\text{lim}}\left( {1 + \frac{1}{m}} \right)^{m}} \right\rbrack^{0.02t}.$$

We recognize the limit inside the brackets as the number $e.$ So, the balance in our bank account after $t$ years is given by $1000e^{0.02t}.$ Generalizing this concept, we see that if a bank account with an initial balance of $\text{\$}P$ earns interest at a rate of $r\text{\%},$ compounded continuously, then the balance of the account after $t$ years is

我们认出括号内的极限就是数 $e.$ 因此，我们的银行账户在 $t$ 年后的余额为 $1000e^{0.02t}.$ 推广这一概念，我们看到，如果一个初始余额为 $\text{\$}P$、按 $r\text{\%}$ 的利率连续复利的银行账户，那么其在 $t$ 年后的余额为

$$\text{Balance} = Pe^{rt}.$$

Compound Interest 复利

A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays $5\text{\%}$ annual interest compounded continuously. How much does the student need to invest today to have $\text{\$}1$ million when she retires at age $65?$ What if she could earn $6\text{\%}$ annual interest compounded continuously instead?

一位 25 岁的学生得到一个机会，将一笔钱投资于一个退休账户，该账户按每年 $5\text{\%}$ 的利率连续复利。该学生今天需要投资多少，才能在 65 岁退休时拥有 $\text{\$}1$ 百万？如果她改为能赚取每年 $6\text{\%}$ 的连续复利呢？

Solution 解答

We have

我们有

$$\begin{array}{rll} 1,000,000 & = & {Pe^{0.05{(40)}}} \\ P & = & 135,335.28. \end{array}$$

She must invest $\text{\$}135,335.28$ at $5\text{\%}$ interest.

她必须以 $5\text{\%}$ 的利率投资 $\text{\$}135,335.28$。

If, instead, she is able to earn $6\text{\%},$ then the equation becomes

如果她改为能赚取 $6\text{\%}$，则方程变为

$$\begin{array}{rll} 1,000,000 & = & {Pe^{0.06{(40)}}} \\ P & = & 90,717.95. \end{array}$$

In this case, she needs to invest only $\text{\$}90,717.95.$ This is roughly two-thirds the amount she needs to invest at $5\text{\%}.$ The fact that the interest is compounded continuously greatly magnifies the effect of the $1\text{\%}$ increase in interest rate.

在这种情况下，她只需投资 $\text{\$}90,717.95.$ 这大约是以 $5\text{\%}$ 投资所需金额的三分之二。利息连续复利这一事实极大地放大了利率提高 $1\text{\%}$ 所带来的效果。

Suppose instead of investing at age $25$, the student waits until age $35.$ How much would she have to invest at $5\text{\%}?$ At $6\text{\%}?$

假设该学生不在 25 岁投资，而是等到 35 岁。那么她在 $5\text{\%}$ 的利率下需要投资多少？在 $6\text{\%}$ 下又是多少？

If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from $100$ to $200$ bacteria as it does to grow from $10,000$ to $20,000$ bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have

如果一个量呈指数增长，则该量翻倍所需的时间是恒定的。换句话说，一个细菌种群从 $100$ 个增长到 $200$ 个所需的时间，与从 $10,000$ 个增长到 $20,000$ 个所需的时间是相同的。这一时间被称为倍增时间。为了计算倍增时间，我们想要求出该量达到其原始规模两倍时的时间。于是我们有

$$\begin{array}{rll} {2y_{0}} & = & {y_{0}e^{kt}} \\ 2 & = & e^{kt} \\ {\text{ln}\ 2} & = & {kt} \\ t & = & {\frac{\text{ln}\ 2}{k}.} \end{array}$$

If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by

如果一个量呈指数增长，倍增时间就是该量翻倍所需的时间。它由下式给出

$$\text{Doubling time}\ = \frac{\text{ln}\ 2}{k}.$$

Using the Doubling Time 应用倍增时间

Assume a population of fish grows exponentially. A pond is stocked initially with $500$ fish. After $6$ months, there are $1000$ fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches $10,000.$ When will the owner's friends be allowed to fish?

假设一个鱼群呈指数增长。一个池塘最初放养了 $500$ 条鱼。经过 $6$ 个月后，池塘中有 $1000$ 条鱼。当鱼群数量达到 $10,000$ 条后，池塘主人将允许他的朋友和邻居在他的池塘中钓鱼。主人的朋友们何时才能被允许钓鱼？

Solution 解答

We know it takes the population of fish $6$ months to double in size. So, if *t* represents time in months, by the doubling-time formula, we have $6 = {\left( {\text{ln}\ 2} \right)\text{/}k}.$ Then, $k = {\left( {\text{ln}\ 2} \right)\text{/}6}.$ Thus, the population is given by $y = 500e^{{({{({\text{ln}\ 2})}\text{/}6})}t}.$ To figure out when the population reaches $10,000$ fish, we must solve the following equation:

我们知道该鱼群规模翻倍需要 $6$ 个月。因此，如果 *t* 表示以月为单位的时间，由倍增时间公式，我们有 $6 = {\left( {\text{ln}\ 2} \right)\text{/}k}.$ 于是 $k = {\left( {\text{ln}\ 2} \right)\text{/}6}.$ 因此，该种群由 $y = 500e^{{({{({\text{ln}\ 2})}\text{/}6})}t}$ 给出。为了计算种群何时达到 $10,000$ 条鱼，我们必须解下面的方程：

$$\begin{array}{rll} 10,000 & = & {500e^{(\text{ln}\ 2\text{/}6)t}} \\ 20 & = & e^{(\text{ln}\ 2\text{/}6)t} \\ {\text{ln}\ 20} & = & {\left( \frac{\text{ln}\ 2}{6} \right)t} \\ t & = & {\frac{6\left( {\text{ln}\ 20} \right)}{\text{ln}\ 2} \approx 25.93.} \end{array}$$

The owner's friends have to wait $25.93$ months (a little more than $2$ years) to fish in the pond.

主人的朋友们必须等待 $25.93$ 个月（略多于 $2$ 年）才能在池塘中钓鱼。

Suppose it takes $9$ months for the fish population in Example 2.44 to reach $1000$ fish. Under these circumstances, how long do the owner's friends have to wait?

假设示例 2.44 中的鱼群需要 $9$ 个月才能达到 $1000$ 条鱼。在这种情况下，主人的朋友们要等多久？

Exponential Decay Model 指数衰减模型

Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant $k,$ we have $y = y_{0}e^{\text{−}kt}.$

指数函数也可用来建模不断缩减的种群（例如因疾病），或随时间分解的化学物质。我们说此类系统呈现指数衰减，而非指数增长。模型几乎相同，只是指数中出现负号。于是，对于某个正常数 $k,$ 我们有 $y = y_{0}e^{\text{−}kt}.$

As with exponential growth, there is a differential equation associated with exponential decay. We have

与指数增长一样，指数衰减也对应一个微分方程。我们有

$$y^{\prime} = \text{−}ky_{0}e^{\text{−}kt} = \text{−}ky.$$

Systems that exhibit exponential decay behave according to the model

呈现指数衰减的系统按如下模型演化

$$y = y_{0}e^{\text{−}kt},$$

where $y_{0}$ represents the initial state of the system and $k > 0$ is a constant, called the *decay constant*.

其中 $y_{0}$ 表示系统的初始状态，$k > 0$ 为一常数，称为*衰减常数*。

The following figure shows a graph of a representative exponential decay function.

下图展示了一个具代表性的指数衰减函数的图像。

Let’s look at a physical application of exponential decay. Newton’s law of cooling says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if $T$ represents the temperature of the object and $T_{a}$ represents the ambient temperature in a room, then

我们来看指数衰减的一个物理应用。牛顿冷却定律说，物体的冷却速率与物体温度和环境温度之差成正比。换言之，若 $T$ 表示物体温度，$T_{a}$ 表示室内环境温度，则

$$T^{\prime} = \text{−}k\left( {T - T_{a}} \right).$$

Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional $T_{a}$ term. Fortunately, we can make a change of variables that resolves this issue. Let $y(t) = T(t) - T_{a}.$ Then $y^{\prime}(t) = T^{\prime}(t) - 0 = T^{\prime}(t),$ and our equation becomes

注意，这并不完全是指数衰减的正确模型。我们希望导数与函数本身成正比，而这个表达式多出了一个 $T_{a}$ 项。幸运的是，我们可以通过变量代换解决这个问题。令 $y(t) = T(t) - T_{a}.$ 于是 $y^{\prime}(t) = T^{\prime}(t) - 0 = T^{\prime}(t),$ 方程化为

$$y^{\prime} = \text{−}ky.$$

From our previous work, we know this relationship between *y* and its derivative leads to exponential decay. Thus,

由前面的讨论可知，*y* 与其导数之间的这一关系将导致指数衰减。于是

$$y = y_{0}e^{\text{−}kt},$$

and we see that

于是我们看到

$$\begin{array}{rll} {T - T_{a}} & = & {\left( {T_{0} - T_{a}} \right)e^{\text{−}kt}} \\ T & = & {\left( {T_{0} - T_{a}} \right)e^{\text{−}kt} + T_{a}} \end{array}$$

where $T_{0}$ represents the initial temperature. Let’s apply this formula in the following example.

其中 $T_{0}$ 表示初始温度。我们在下面的示例中应用这一公式。

Newton’s Law of Cooling 牛顿冷却定律

According to experienced baristas, the optimal temperature to serve coffee is between $155\text{°}\text{F}$ and $175\text{°}\text{F}.$ Suppose coffee is poured at a temperature of $200\text{°}\text{F},$ and after $2$ minutes in a $70\text{°}\text{F}$ room it has cooled to $180\text{°}\text{F}.$ When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.

据有经验的咖啡师说，供应咖啡的最佳温度在 $155\text{°}\text{F}$ 与 $175\text{°}\text{F}$ 之间。假定咖啡倒出时温度为 $200\text{°}\text{F}$，在 $70\text{°}\text{F}$ 的房间里放置 $2$ 分钟后冷却到 $180\text{°}\text{F}$。咖啡何时首次降到可供应温度？咖啡何时变得太冷而无法供应？答案四舍五入到最近的半分钟。

Solution 解答

We have

我们有

$$\begin{array}{rll} T & = & {\left( {T_{0} - T_{a}} \right)e^{\text{−}kt} + T_{a}} \\ 180 & = & {(200 - 70)e^{\text{−}k{(2)}} + 70} \\ 110 & = & {130e^{-2k}} \\ \frac{11}{13} & = & e^{-2k} \\ {\text{ln}\ \frac{11}{13}} & = & {-2k} \\ {\text{ln}\ 11 - \text{ln}\ 13} & = & {-2k} \\ k & = & {\frac{\text{ln}\ 13 - \text{ln}\ 11}{2}.} \end{array}$$

Then, the model is

于是模型为

$$T = 130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} + 70.$$

The coffee reaches $175\text{°}\text{F}$ when

当咖啡达到 $175\text{°}\text{F}$ 时

$$\begin{array}{rll} 175 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} + 70} \\ 105 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t}} \\ \frac{21}{26} & = & e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} \\ {\text{ln}\ \frac{21}{26}} & = & {\frac{\text{ln}\ 11 - \text{ln}\ 13}{2}t} \\ {\text{ln}\ 21 - \text{ln}\ 26} & = & {\frac{\text{ln}\ 11 - \text{ln}\ 13}{2}t} \\ t & = & {\frac{2\left( {\text{ln}\ 21 - \text{ln}\ 26} \right)}{\text{ln}\ 11 - \text{ln}\ 13} \approx 2.56.} \end{array}$$

The coffee can be served about $2.5$ minutes after it is poured. The coffee reaches $155\text{°}\text{F}$ at

咖啡在倒出后约 $2.5$ 分钟即可供应。咖啡达到 $155\text{°}\text{F}$ 时

$$\begin{array}{rll} 155 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} + 70} \\ 85 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t}} \\ \frac{17}{26} & = & e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} \\ {\text{ln}\ 17 - \text{ln}\ 26} & = & {\left( \frac{\text{ln}\ 11 - \text{ln}\ 13}{2} \right)t} \\ t & = & {\frac{2\left( {\text{ln}\ 17 - \text{ln}\ 26} \right)}{\text{ln}\ 11 - \text{ln}\ 13} \approx 5.09.} \end{array}$$

The coffee is too cold to be served about $5$ minutes after it is poured.

咖啡在倒出后约 $5$ 分钟变得太冷而无法供应。

Suppose the room is warmer $(75\text{°}\text{F})$ and, after $2$ minutes, the coffee has cooled only to $185\text{°}\text{F}.$ When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.

假定房间更暖，为 $(75\text{°}\text{F})$，且 $2$ 分钟后咖啡仅冷却到 $185\text{°}\text{F}$。咖啡何时首次降到可供应温度？咖啡何时变得太冷而无法供应？答案四舍五入到最近的半分钟。

Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant half-life. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have

正如呈现指数增长的系统具有恒定的倍增时间，呈现指数衰减的系统具有恒定的半衰期。要计算半衰期，我们想知道该量何时减少到原来大小的一半。于是

$$\begin{array}{rll} \frac{y_{0}}{2} & = & {y_{0}e^{\text{−}kt}} \\ \frac{1}{2} & = & e^{\text{−}kt} \\ {- \text{ln}\ 2} & = & {\text{−}kt} \\ t & = & {\frac{\text{ln}\ 2}{k}.} \end{array}$$

*Note*: This is the same expression we came up with for doubling time.

*注*：这与我们为倍增时间推出的表达式相同。

If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by

若一个量呈指数衰减，半衰期就是该量减少一半所需的时间，由下式给出

$$\text{Half-life} = \frac{\text{ln}\ 2}{k}.$$

Radiocarbon Dating 放射性碳定年

One of the most common applications of an exponential decay model is carbon dating. $\text{Carbon-}14$ decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon was originally present in an object and how much carbon remains, we can determine the age of the object. The half-life of $\text{carbon-}14$ is approximately $5730$ years—meaning, after that many years, half the material has converted from the original $\text{carbon-}14$ to the new nonradioactive $\text{nitrogen-}14.$ If we have $100$ g $\text{carbon-}14$ today, how much is left in $50$ years? If an artifact that originally contained $100$ g of carbon now contains $10$ g of carbon, how old is it? Round the answer to the nearest hundred years.

指数衰减模型最常见的应用之一是碳定年。$\text{Carbon-}14$ 以规则而稳定的指数速率衰变（释放放射性粒子）。因此，如果我们知道一个物体中原本含有多少碳、现在还剩多少碳，就能确定该物体的年代。$\text{carbon-}14$ 的半衰期约为 $5730$ 年——也就是说，经过这么多年后，一半的物质已从原有的 $\text{carbon-}14$ 转变为新的非放射性 $\text{nitrogen-}14$。如果今天有 $100$ g 的 $\text{carbon-}14$，50 年后还剩多少？如果一件原本含 $100$ g 碳的文物现在只含 $10$ g 碳，它的年代是多少？答案四舍五入到最近的百年。

Solution 解答

We have

我们有

$$\begin{array}{rll} 5730 & = & \frac{\text{ln}\ 2}{k} \\ k & = & {\frac{\text{ln}\ 2}{5730}.} \end{array}$$

So, the model says

于是模型为

$$y = 100e^{\text{−}{({\text{ln}\ 2\text{/}5730})}t}.$$

In $50$ years, we have

在 $50$ 年后，我们有

$$\begin{array}{cll} y & = & {100e^{\text{−}{({\text{ln}\ 2\text{/}5730})}{(50)}}} \\ & \approx & {99.40.} \end{array}$$

Therefore, in $50$ years, $99.40$ g of $\text{carbon-}14$ remains.

因此，50 年后，剩余 $99.40$ g 的 $\text{carbon-}14$。

To determine the age of the artifact, we must solve

要确定该文物的年代，我们必须解

$$\begin{array}{rll} 10 & = & {100e^{\text{−}{({\text{ln}\ 2\text{/}5730})}t}} \\ \frac{1}{10} & = & e^{\text{−}{({\text{ln}\ 2\text{/}5730})}t} \\ t & \approx & 19035. \end{array}$$

The artifact is about $19,000$ years old.

该文物的年代约为 $19,000$ 年。

If we have $100$ g of $\text{carbon-}14,$ how much is left after $500$ years? If an artifact that originally contained $100$ g of carbon now contains $20g$ of carbon, how old is it? Round the answer to the nearest hundred years.

如果我们有 $100$ g 的 $\text{carbon-}14$，500 年后还剩多少？如果一件原本含 $100$ g 碳的文物现在只含 $20g$ 碳，它的年代是多少？答案四舍五入到最近的百年。

Section 2.8 Exercises 2.8 节习题

*True or False*? If true, prove it. If false, find the true answer.

*判断正误*？若正确，请证明；若错误，请找出正确答案。

348\.

The doubling time for $y = e^{ct}$ is $\left( {\text{ln}\ (2)} \right)\text{/}\left( {\text{ln}\ (c)} \right).$

$y = e^{ct}$ 的倍增时间为 $\left( {\text{ln}\ (2)} \right)\text{/}\left( {\text{ln}\ (c)} \right).$

349.

If you invest $\text{\$}500,$ an annual rate of interest of $3\text{\%}$ yields more money in the first year than a $2.5\text{\%}$ continuous rate of interest.

若你投资 $\text{\$}500,$ 那么 $3\text{\%}$ 的年利率在第一年产生的收益多于 $2.5\text{\%}$ 的连续利率。

350\.

If you leave a $100\text{°}\text{C}$ pot of tea at room temperature $(25\text{°}\text{C})$ and an identical pot in the refrigerator $(5\text{°}\text{C}),$ with $k = 0.02,$ the tea in the refrigerator reaches a drinkable temperature $(70\text{°}\text{C})$ more than $5$ minutes before the tea at room temperature.

若把一壶 $100\text{°}\text{C}$ 的茶放在室温 $(25\text{°}\text{C})$ 下，另一壶相同的茶放在冰箱 $(5\text{°}\text{C})$ 中，取 $k = 0.02$，则冰箱中的茶比室温下的茶提前 5 分钟以上达到可饮温度 $(70\text{°}\text{C})$。

351.

If given a half-life of *t* years, the constant $k$ for $y = e^{kt}$ is calculated by $k = {{\text{ln}\ \left( {1\text{/}2} \right)}\text{/}t}.$

若给定半衰期为 *t* 年，则 $y = e^{kt}$ 的常数 $k$ 由 $k = {{\text{ln}\ \left( {1\text{/}2} \right)}\text{/}t}.$ 算出。

For the following exercises, use $y = y_{0}e^{kt}.$

在以下习题中，使用 $y = y_{0}e^{kt}.$

352\.

If a culture of bacteria doubles in $3$ hours, how many hours does it take to multiply by $10?$

若一培养皿中的细菌在 $3$ 小时内翻倍，则要增殖到 $10$ 倍需要多少小时？

353.

If bacteria increase by a factor of $10$ in $10$ hours, how many hours does it take to increase by $100?$

若细菌在 $10$ 小时内增加到 $10$ 倍，则增加到 $100$ 倍需要多少小时？

354\.

How old is a skull that contains one-fifth as much radiocarbon as a modern skull? Note that the half-life of radiocarbon is $5730$ years.

一个头骨所含的放射性碳只有现代头骨的五分之一，它的年代是多少？注意放射性碳的半衰期为 $5730$ 年。

355.

If a relic contains $90\text{\%}$ as much radiocarbon as new material, can it have come from the time of Christ (approximately $2000$ years ago)? Note that the half-life of radiocarbon is $5730$ years.

若一件文物所含的放射性碳为新材料的 $90\text{\%}$，它可能来自基督时代（约 $2000$ 年前）吗？注意放射性碳的半衰期为 $5730$ 年。

356\.

The population of Cairo grew from $5$ million to $10$ million in $20$ years. Use an exponential model to find when the population was $8$ million.

开罗人口在 $20$ 年内从 $500$ 万增长到 $1000$ 万。用指数模型求出人口达到 $800$ 万的时间。

357.

The populations of New York and Los Angeles are growing at $1\text{\%}$ and $1.4\text{\%}$ a year, respectively. Starting from $8$ million (New York) and $6$ million (Los Angeles), when are the populations equal? Round your answer to a whole number of years.

纽约与洛杉矶的人口分别以每年 $1\text{\%}$ 和 $1.4\text{\%}$ 的速率增长。从 $800$ 万（纽约）和 $600$ 万（洛杉矶）起步，两市人口何时相等？答案四舍五入到整年。

358\.

Suppose the value of $\text{\$}1$ in Japanese yen decreases at $2\text{\%}$ per year. Starting from $\text{\$}1 = \text{¥}250,$ when will $\text{\$}1 = \text{¥}1?$

假定 $\text{\$}1$ 美元（以日元计）的价值每年下降 $2\text{\%}$。从 $\text{\$}1 = \text{¥}250$ 起步，何时 $\text{\$}1 = \text{¥}1?$

359.

The effect of advertising decays exponentially. If $40\text{\%}$ of the population remembers a new product after $3$ days, how long will $20\text{\%}$ remember it?

广告的效果呈指数衰减。若 $40\text{\%}$ 的人口在 $3$ 天后还记得某新产品，那么 $20\text{\%}$ 的人口会记得多久？

360\.

If $y = 1000$ at $t = 3$ and $y = 3000$ at $t = 4,$ what was $y_{0}$ at $t = 0?$

若 $y = 1000$ 于 $t = 3$，$y = 3000$ 于 $t = 4$，则 $t = 0$ 时的 $y_{0}$ 是多少？

361.

If $y = 100$ at $t = 4$ and $y = 10$ at $t = 8,$ when does $y = 1?$

若 $y = 100$ 于 $t = 4$，$y = 10$ 于 $t = 8$，则何时 $y = 1?$

362\.

If a bank offers annual interest of $7.5\text{\%}$ or continuous interest of $7.25\text{\%},$ which has a better annual yield?

若某银行提供 $7.5\text{\%}$ 的年利率或 $7.25\text{\%}$ 的连续利率，哪一种年化收益更高？

363.

What continuous interest rate has the same yield as an annual rate of $9\text{\%}?$

何种连续利率与 $9\text{\%}$ 的年利率收益相同？

364\.

If you deposit $\text{\$}5000$ at $8\text{\%}$ annual interest, how many years can you withdraw $\text{\$}500$ (starting after the first year) without running out of money?

若你在 $8\text{\%}$ 的年利率下存入 $\text{\$}5000$，在不耗尽资金的前提下，从第一年后开始每年取出 $\text{\$}500$，可以持续多少年？

365.

You are trying to save $\text{\$}50,000$ in $20$ years for college tuition for your child. If interest is a continuous $10\text{\%},$ how much do you need to invest initially?

你打算在 $20$ 年内为孩子攒下 $\text{\$}50,000$ 大学学费。若利率为连续 $10\text{\%}$，你最初需要投资多少？

366\.

You are cooling a turkey that was taken out of the oven with an internal temperature of $165\text{°}\text{F}.$ After $10$ minutes of resting the turkey in a $70\text{°}\text{F}$ apartment, the temperature has reached $155\text{°}\text{F}\text{.}$ What is the temperature of the turkey $20$ minutes after taking it out of the oven?

你在冷却一只火鸡，它刚从烤箱取出时内部温度为 $165\text{°}\text{F}$。在 $70\text{°}\text{F}$ 的公寓中静置 $10$ 分钟后，温度降到了 $155\text{°}\text{F}\text{.}$ 火鸡取出 $20$ 分钟后温度是多少？

367.

You are trying to thaw some vegetables that are at a temperature of $1\text{°}\text{F}\text{.}$ To thaw vegetables safely, you must put them in the refrigerator, which has an ambient temperature of $44\text{°}\text{F}.$ You check on your vegetables $2$ hours after putting them in the refrigerator to find that they are now $12\text{°}\text{F}\text{.}$ Plot the resulting temperature curve and use it to determine when the vegetables reach $33\text{°}\text{F}\text{.}$

你想解冻一些温度为 $1\text{°}\text{F}\text{.}$ 的蔬菜。为安全解冻，必须把它们放进环境温度为 $44\text{°}\text{F}$ 的冰箱。放入冰箱 $2$ 小时后你查看蔬菜，发现它们现在是 $12\text{°}\text{F}\text{.}$ 绘出所得的温度曲线，并用它确定蔬菜何时达到 $33\text{°}\text{F}\text{.}$

368\.

You are an archaeologist and are given a bone that is claimed to be from a Tyrannosaurus Rex. You know these dinosaurs lived during the Cretaceous Era $(146$ million years to $65$ million years ago), and you find by radiocarbon dating that there is $0.000001\text{\%}$ the amount of radiocarbon. Is this bone from the Cretaceous?

你是一位考古学家，得到一根据称来自霸王龙（Tyrannosaurus Rex）的骨头。你知道这些恐龙生活在白垩纪（$(146$ 百万年前至 $65$ 百万年前），并通过放射性碳定年发现其中放射性碳的含量仅为 $0.000001\text{\%}$。这根骨头来自白垩纪吗？

369.

The spent fuel of a nuclear reactor contains plutonium-239, which has a half-life of $24,000$ years. If $1$ barrel containing $10\ \text{kg}$ of plutonium-239 is sealed, how many years must pass until only $10g$ of plutonium-239 is left?

核反应堆的乏燃料含有钚-239，其半衰期为 $24,000$ 年。若密封 $1$ 桶含 $10\ \text{kg}$ 钚-239 的燃料，要经过多少年才会只剩 $10g$ 钚-239？

For the next set of exercises, use the following table, which features the world population by decade.

在接下来的这组习题中，使用下表，表中给出了按十年计的世界人口。

Years since 1950	Population (millions)
$0$	$2,556$
$10$	$3,039$
$20$	$3,706$
$30$	$4,453$
$40$	$5,279$
$50$	$6,083$
$60$	$6,849$

自 1950 年起的年数	人口（百万）
$0$	$2,556$
$10$	$3,039$
$20$	$3,706$
$30$	$4,453$
$40$	$5,279$
$50$	$6,083$
$60$	$6,849$

370.

\[T\] The best-fit exponential curve to the data of the form $P(t) = ae^{bt}$ is given by $P(t) = 2686e^{0.01604t}.$ Use a graphing calculator to graph the data and the exponential curve together.

\[T\] 对形如 $P(t) = ae^{bt}$ 的数据的最佳拟合指数曲线为 $P(t) = 2686e^{0.01604t}$。使用绘图计算器将数据与指数曲线一起绘图。

371.

\[T\] Find and graph the derivative $y^{\prime}$ of your equation. Where is it increasing and what is the meaning of this increase?

\[T\] 求并绘出你所得方程的导数 $y^{\prime}$。它在何处递增？这一递增意味着什么？

372\.

\[T\] Find and graph the second derivative of your equation. Where is it increasing and what is the meaning of this increase?

\[T\] 求并绘出你所得方程的二阶导数。它在何处递增？这一递增意味着什么？

373.

\[T\] Find the predicted date when the population reaches $10$ billion. Using your previous answers about the first and second derivatives, explain why exponential growth is unsuccessful in predicting the future.

\[T\] 求出人口达到 $100$ 亿的预测年份。利用你关于一阶、二阶导数的前述答案，解释为什么指数增长无法成功预测未来。

For the next set of exercises, use the following table, which shows the population of San Francisco during the 19th century.

在接下来的这组习题中，使用下表，表中给出了 19 世纪旧金山的人口。

Years since 1850	Population (thousands)
$0$	$21.00$
$10$	$56.80$
$20$	$149.5$
$30$	$234.0$

自 1850 年起的年数	人口（千）
$0$	$21.00$
$10$	$56.80$
$20$	$149.5$
$30$	$234.0$

374.

\[T\] The best-fit exponential curve to the data of the form $P(t) = ae^{bt}$ is given by $P(t) = 35.26e^{0.06407t}.$ Use a graphing calculator to graph the data and the exponential curve together.

\[T\] 对形如 $P(t) = ae^{bt}$ 的数据的最佳拟合指数曲线为 $P(t) = 35.26e^{0.06407t}$。使用绘图计算器将数据与指数曲线一起绘图。

375.

\[T\] Find and graph the derivative $y^{\prime}$ of your equation. Where is it increasing? What is the meaning of this increase? Is there a value where the increase is maximal?

\[T\] 求并绘出你所得方程的导数 $y^{\prime}$。它在何处递增？这一递增意味着什么？是否存在使递增达到最大的值？

376\.

\[T\] Find and graph the second derivative of your equation. Where is it increasing? What is the meaning of this increase?

\[T\] 求并绘出你所得方程的二阶导数。它在何处递增？这一递增意味着什么？

2.9 Calculus of the Hyperbolic Functions 2.9 双曲函数的微积分

2.9.1 Apply the formulas for derivatives and integrals of the hyperbolic functions.
2.9.2 Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals.
2.9.3 Describe the common applied conditions of a catenary curve.

2.9.1 应用双曲函数的导数与积分公式。
2.9.2 应用反双曲函数的导数公式及其相关积分。
2.9.3 描述悬链线曲线的常见应用条件。

We were introduced to hyperbolic functions in Introduction to Functions and Graphs, along with some of their basic properties. In this section, we look at differentiation and integration formulas for the hyperbolic functions and their inverses.

我们在《函数与图像导论》中介绍了双曲函数及其一些基本性质。在本节中，我们考察双曲函数及其反函数的求导与积分公式。

Derivatives and Integrals of the Hyperbolic Functions 双曲函数的导数与积分

Recall that the hyperbolic sine and hyperbolic cosine are defined as

回顾一下，双曲正弦与双曲余弦定义为

$$\text{sinh}\ x = \frac{e^{x} - e^{\text{−}x}}{2}\ \text{and}\ \text{cosh}\ x = \frac{e^{x} + e^{\text{−}x}}{2}.$$

The other hyperbolic functions are then defined in terms of $\text{sinh}\ x$ and $\text{cosh}\ x.$ The graphs of the hyperbolic functions are shown in the following figure.

其余双曲函数则由 $\text{sinh}\ x$ 与 $\text{cosh}\ x$ 定义。双曲函数的图像如下图所示。

It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at $\text{sinh}\ x$ we have

推导双曲函数的求导公式很容易。例如，考察 $\text{sinh}\ x$，我们有

$$\begin{array}{cl} {\frac{d}{dx}\left( {\text{sinh}\ x} \right)} & {= \frac{d}{dx}\left( \frac{e^{x} - e^{\text{−}x}}{2} \right)} \\ & {= \frac{1}{2}\left\lbrack {\frac{d}{dx}\left( e^{x} \right) - \frac{d}{dx}\left( e^{\text{−}x} \right)} \right\rbrack} \\ & {= \frac{1}{2}\left\lbrack {e^{x} + e^{\text{−}x}} \right\rbrack = \text{cosh}\ x.} \end{array}$$

Similarly, $\left( {d\text{/}{dx}} \right)\text{cosh}\ x = \text{sinh}\ x.$ We summarize the differentiation formulas for the hyperbolic functions in the following table.

类似地，$\left( {d\text{/}{dx}} \right)\text{cosh}\ x = \text{sinh}\ x.$ 我们将双曲函数的求导公式总结于下表。

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}\ x$	$\text{cosh}\ x$
$\text{cosh}\ x$	$\text{sinh}\ x$
$\text{tanh}\ x$	$\text{sech}^{2}\ x$
$\text{coth}\ x$	$\text{−}\text{csch}^{2}\ x$
$\text{sech}\ x$	$\text{−}\text{sech}\ x\ \text{tanh}\ x$
$\text{csch}\ x$	$\text{−}\text{csch}\ x\ \text{coth}\ x$

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}\ x$	$\text{cosh}\ x$
$\text{cosh}\ x$	$\text{sinh}\ x$
$\text{tanh}\ x$	$\text{sech}^{2}\ x$
$\text{coth}\ x$	$\text{−}\text{csch}^{2}\ x$
$\text{sech}\ x$	$\text{−}\text{sech}\ x\ \text{tanh}\ x$
$\text{csch}\ x$	$\text{−}\text{csch}\ x\ \text{coth}\ x$

Table 2.2 Derivatives of the Hyperbolic Functions

表 2.2 双曲函数的导数

Let’s take a moment to compare the derivatives of the hyperbolic functions with the derivatives of the standard trigonometric functions. There are a lot of similarities, but differences as well. For example, the derivatives of the sine functions match: $\left( {d\text{/}{dx}} \right)\text{sin}\ x = \text{cos}\ x$ and $\left( {d\text{/}{dx}} \right)\text{sinh}\ x = \text{cosh}\ x.$ The derivatives of the cosine functions, however, differ in sign: $\left( {d\text{/}{dx}} \right)\text{cos}\ x = \text{−}\text{sin}\ x,$ but $\left( {d\text{/}{dx}} \right)\text{cosh}\ x = \text{sinh}\ x.$ As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions.

我们花点时间，将双曲函数的导数与标准三角函数的导数作一比较。二者有许多相似之处，但也存在差异。例如，正弦函数的导数相符：$\left( {d\text{/}{dx}} \right)\text{sin}\ x = \text{cos}\ x$，且 $\left( {d\text{/}{dx}} \right)\text{sinh}\ x = \text{cosh}\ x.$ 然而，余弦函数的导数符号不同：$\left( {d\text{/}{dx}} \right)\text{cos}\ x = \text{−}\text{sin}\ x$，但 $\left( {d\text{/}{dx}} \right)\text{cosh}\ x = \text{sinh}\ x.$ 在继续研究双曲函数时，我们必须留意它们与标准三角函数之间的相似与差异。

These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.

双曲函数的这些求导公式直接导出下列积分公式。

$$\begin{array}{rllccrll} {{\int{\text{sinh}\ u\ d}}u} & = & {\text{cosh}\ u + C} & & & {{\int{\text{csch}^{2}\ u\ d}}u} & = & {\text{−}\text{coth}\ u + C} \\ {{\int{\text{cosh}\ u\ d}}u} & = & {\text{sinh}\ u + C} & & & {{\int{\text{sech}\ u\ \text{tanh}\ u\ d}}u} & = & {\text{−}\text{sech}\ u + C} \\ {{\int{\text{sech}^{2}u\ d}}u} & = & {\text{tanh}\ u + C} & & & {{\int{\text{csch}\ u\ \text{coth}\ u\ d}}u} & = & {\text{−}\text{csch}\ u + C} \end{array}$$

Differentiating Hyperbolic Functions 求双曲函数的导数

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dx}\left( {\text{sinh}\left( x^{2} \right)} \right)$

2. $\frac{d}{dx}\left( {\text{cosh}\ x} \right)^{2}$

Solution 解答

Using the formulas in Table 2.2 and the chain rule, we get

利用表 2.2 中的公式与链式法则，我们得到

1. $\frac{d}{dx}\left( {\text{sinh}\left( x^{2} \right)} \right) = \text{cosh}\left( x^{2} \right) \cdot 2x$

2. $\frac{d}{dx}\left( {\text{cosh}\ x} \right)^{2} = 2\ \text{cosh}\ x\ \text{sinh}\ x$

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dx}\left( {\text{tanh}\left( {x^{2} + 3x} \right)} \right)$

2. $\frac{d}{dx}\left( \frac{1}{\left( {\text{sinh}\ x} \right)^{2}} \right)$

Integrals Involving Hyperbolic Functions 涉及双曲函数的积分

Evaluate the following integrals:

求下列积分：

1. ${\int{x\ \text{cosh}\left( x^{2} \right)d}}x$

2. ${\int{\text{tanh}\ x\ d}}x$

Solution 解答

We can use *u*-substitution in both cases.

我们在两种情形下都可以使用 *u*-换元（法）。

1. Let $u = x^{2}.$ Then, $du = 2x\ dx$ and

1. 令 $u = x^{2}.$ 则 $du = 2x\ dx$，且

$${\int{x\ \text{cosh}\left( x^{2} \right)d}}x = {\int\frac{1}{2}}\text{cosh}\ u\ du = \frac{1}{2}\text{sinh}\ u + C = \frac{1}{2}\text{sinh}\left( x^{2} \right) + C.$$

2. Let $u = \text{cosh}\ x.$ Then, $du = \text{sinh}\ x\ dx$ and

2. 令 $u = \text{cosh}\ x.$ 则 $du = \text{sinh}\ x\ dx$，且

$${\int{\text{tanh}\ x\ d}}x = {\int{\frac{\text{sinh}\ x}{\text{cosh}\ x}d}}x = {\int{\frac{1}{u}d}}u = \text{ln}|u| + C = \text{ln}\left| {\text{cosh}\ x} \right| + C.$$

Note that $\text{cosh}\ x > 0$ for all $x,$ so we can eliminate the absolute value signs and obtain

注意，$\text{cosh}\ x > 0$ 对所有 $x,$ 都成立，因此我们可以去掉绝对值符号，得到

$${\int{\text{tanh}\ x\ d}}x = \text{ln}\left( {\text{cosh}\ x} \right) + C.$$

Evaluate the following integrals:

求下列积分：

1. ${\int{\text{sinh}^{3}x\ \text{cosh}\ x\ d}}x$

2. ${\int{\text{sech}^{2}\left( {3x} \right)d}}x$

Calculus of Inverse Hyperbolic Functions 反双曲函数的微积分

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.

观察双曲函数的图像可知，在适当的取值限制下，它们都有反函数。多数所需的取值范围限制可通过仔细观察图像来辨别。反双曲函数的定义域与值域总结于下表。

Function	Domain	Range
$\text{sinh}^{-1}x$	$\left( {\text{−}\infty,\infty} \right)$	$\left( {\text{−}\infty,\infty} \right)$
$\text{cosh}^{-1}x$	$\left\lbrack {1,\infty} \right)$	$\left\lbrack \left. {0,\infty} \right) \right.$
$\text{tanh}^{-1}x$	$\left( {-1,1} \right)$	$\left( {\text{−}\infty,\infty} \right)$
$\text{coth}^{-1}x$	$\left( {\text{−}\infty,-1} \right) \cup \left( {1,\infty} \right)$	$\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right)$
$\text{sech}^{-1}x$	$\left( {0\text{, 1}} \right\rbrack$	$\left\lbrack \left. {0,\infty} \right) \right.$
$\text{csch}^{-1}x$	$\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right)$	$\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right)$

函数	定义域	值域
$\text{sinh}^{-1}x$	$\left( {\text{−}\infty,\infty} \right)$	$\left( {\text{−}\infty,\infty} \right)$
$\text{cosh}^{-1}x$	$\left\lbrack {1,\infty} \right)$	$\left\lbrack \left. {0,\infty} \right) \right.$
$\text{tanh}^{-1}x$	$\left( {-1,1} \right)$	$\left( {\text{−}\infty,\infty} \right)$
$\text{coth}^{-1}x$	$\left( {\text{−}\infty,-1} \right) \cup \left( {1,\infty} \right)$	$\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right)$
$\text{sech}^{-1}x$	$\left( {0\text{, 1}} \right\rbrack$	$\left\lbrack \left. {0,\infty} \right) \right.$
$\text{csch}^{-1}x$	$\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right)$	$\left( {\text{−}\infty,0} \right) \cup \left( {0,\infty} \right)$

Table 2.3 Domains and Ranges of the Inverse Hyperbolic Functions

表 2.3 反双曲函数的定义域与值域

The graphs of the inverse hyperbolic functions are shown in the following figure.

反双曲函数的图像如下图所示。

To find the derivatives of the inverse functions, we use implicit differentiation. We have

为求反函数的导数，我们使用隐函数求导法。我们有

$$\begin{array}{rll} y & = & {\text{sinh}^{-1}\ x} \\ {\text{sinh}\ y} & = & x \\ {\frac{d}{dx}\text{sinh}\ y} & = & {\frac{d}{dx}x} \\ {\text{cosh}\ y\frac{dy}{dx}} & = & {1.} \end{array}$$

Recall that $\text{cosh}^{2}y - \text{sinh}^{2}y = 1,$ so $\text{cosh}\ y = \sqrt{1 + \text{sinh}^{2}y}.$ Then,

回顾 $\text{cosh}^{2}y - \text{sinh}^{2}y = 1,$ 可知 $\text{cosh}\ y = \sqrt{1 + \text{sinh}^{2}y}.$ 于是，

$$\frac{dy}{dx} = \frac{1}{\text{cosh}\ y} = \frac{1}{\sqrt{1 + \text{sinh}^{2}y}} = \frac{1}{\sqrt{1 + x^{2}}}.$$

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

我们可以用类似的方法导出其余反双曲函数的求导公式。这些求导公式总结于下表。

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}^{-1}x$	$\frac{1}{\sqrt{1 + x^{2}}}$
$\text{cosh}^{-1}x$	$\frac{1}{\sqrt{x^{2} - 1}}$
$\text{tanh}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{coth}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{sech}^{-1}x$	$\frac{-1}{x\sqrt{1 - x^{2}}}$
$\text{csch}^{-1}x$	$\frac{-1}{\|x\|\sqrt{1 + x^{2}}}$

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}^{-1}x$	$\frac{1}{\sqrt{1 + x^{2}}}$
$\text{cosh}^{-1}x$	$\frac{1}{\sqrt{x^{2} - 1}}$
$\text{tanh}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{coth}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{sech}^{-1}x$	$\frac{-1}{x\sqrt{1 - x^{2}}}$
$\text{csch}^{-1}x$	$\frac{-1}{\|x\|\sqrt{1 + x^{2}}}$

Table 2.4 Derivatives of the Inverse Hyperbolic Functions

表 2.4 反双曲函数的导数

Note that the derivatives of $\text{tanh}^{-1}\ x$ and $\text{coth}^{-1}\ x$ are the same. Thus, when we integrate ${1\text{/}\left( {1 - x^{2}} \right)},$ we need to select the proper antiderivative based on the domain of the functions and the values of $x.$ Integration formulas involving the inverse hyperbolic functions are summarized as follows.

注意，$\text{tanh}^{-1}\ x$ 与 $\text{coth}^{-1}\ x$ 的导数相同。因此，当我们计算 ${1\text{/}\left( {1 - x^{2}} \right)}$ 的积分时，需要根据函数的定义域以及 $x$ 的取值来选取恰当的反导数。涉及反双曲函数的积分公式总结如下。

$$\begin{array}{rllccccc} {{\int{\frac{1}{\sqrt{1 + u^{2}}}d}}u} & = & {\text{sinh}^{-1}u + C} & & & {{\int\frac{1}{u\sqrt{1 - u^{2}}}}du} & = & {\text{−}\text{sech}^{-1}|u| + C} \\ {{\int\frac{1}{\sqrt{u^{2} - 1}}}du} & = & {\text{cosh}^{-1}u + C} & & & {{\int\frac{1}{u\sqrt{1 + u^{2}}}}du} & = & {\text{−}\text{csch}^{-1}|u| + C} \\ {{\int\frac{1}{1 - u^{2}}}du} & = & \left\{ \begin{array}{l} {\text{tanh}^{-1}u + C\ \text{if}\ |u| < 1} \\ {\text{coth}^{-1}u + C\ \text{if}\ |u| > 1} \end{array} \right. & & & & & \end{array}$$

Differentiating Inverse Hyperbolic Functions 求反双曲函数的导数

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dx}\left( {\text{sinh}^{-1}\left( \frac{x}{3} \right)} \right)$

2. $\frac{d}{dx}\left( {\text{tanh}^{-1}x} \right)^{2}$

Solution 解答

Using the formulas in Table 2.4 and the chain rule, we obtain the following results:

利用表 2.4 中的公式与链式法则，我们得到如下结果：

1. $\frac{d}{dx}\left( {\text{sinh}^{-1}\left( \frac{x}{3} \right)} \right) = \frac{1}{3\sqrt{1 + \frac{x^{2}}{9}}} = \frac{1}{\sqrt{9 + x^{2}}}$

2. $\frac{d}{dx}\left( {\text{tanh}^{-1}x} \right)^{2} = \frac{2\left( {\text{tanh}^{-1}x} \right)}{1 - x^{2}}$

Evaluate the following derivatives:

求下列导数：

1. $\frac{d}{dx}\left( {\text{cosh}^{-1}\left( {3x} \right)} \right)$

2. $\frac{d}{dx}\left( {\text{coth}^{-1}x} \right)^{3}$

Integrals Involving Inverse Hyperbolic Functions 涉及反双曲函数的积分

Evaluate the following integrals:

求下列积分：

1. ${\int{\frac{1}{\sqrt{4x^{2} - 1}}d}}x$

2. ${\int{\frac{1}{2x\sqrt{1 - 9x^{2}}}d}}x$

Solution 解答

We can use $u\text{-substitution}$ in both cases.

我们在两种情形下都可以使用 $u\text{-substitution}$。

1. Let $u = 2x.$ Then, $du = 2dx$ and we have

1. 令 $u = 2x.$ 则 $du = 2dx$，且我们有

$${\int{\frac{1}{\sqrt{4x^{2} - 1}}d}}x = {\int\frac{1}{2\sqrt{u^{2} - 1}}}du = \frac{1}{2}\text{cosh}^{-1}u + C = \frac{1}{2}\text{cosh}^{-1}\left( {2x} \right) + C.$$

2. Let $u = 3x.$ Then, $du = 3dx$ and we obtain

2. 令 $u = 3x.$ 则 $du = 3dx$，且我们得到

$${\int{\frac{1}{2x\sqrt{1 - 9x^{2}}}dx = \frac{1}{2}{\int{\frac{1}{u\sqrt{1 - u^{2}}}du = - \frac{1}{2}\text{sech}^{-1}|u| + C = - \frac{1}{2}\text{sech}^{-1}\left| {3x} \right| + C}}}}.$$

Evaluate the following integrals:

求下列积分：

1. ${\int{\frac{1}{\sqrt{x^{2} - 4}}d}}x,\ \ x > 2$

2. ${\int{\frac{1}{\sqrt{1 - e^{2x}}}d}}x$

Applications 应用

One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

双曲函数的一个物理应用涉及悬挂的缆索。若一根密度均匀的缆索悬挂于两个支点之间，除自身重量外不受其他载荷，则缆索形成一条称为悬链线（catenary）的曲线。高压输电线、悬挂于两柱之间的链条，以及蜘蛛网丝，都形成悬链线。下图展示了悬挂在一排柱子上的链条。

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form $y = a\ \text{cosh}\left( {x\text{/}a} \right)$ are catenaries. Figure 2.84 shows the graph of $y = 2\ \text{cosh}\left( {x\text{/}2} \right).$

双曲函数可用于为悬链线建模。具体地，形如 $y = a\ \text{cosh}\left( {x\text{/}a} \right)$ 的函数就是悬链线。图 2.84 展示了 $y = 2\ \text{cosh}\left( {x\text{/}2} \right)$ 的图像。

Using a Catenary to Find the Length of a Cable 利用悬链线求电缆长度

Assume a hanging cable has the shape $10\ \text{cosh}\left( {x\text{/}10} \right)$ for $-15 \leq x \leq 15,$ where $x$ is measured in feet. Determine the length of the cable (in feet).

设一根悬挂的电缆在区间 $-15 \leq x \leq 15$ 上呈 $10\ \text{cosh}\left( {x\text{/}10} \right)$ 的形状，其中 $x$ 以英尺为单位。求该电缆的长度（以英尺计）。

Solution 解答

Recall from Section $2.4$ that the formula for arc length is

由第 $2.4$ 节回顾可知，弧长公式为

$$\text{Arc Length} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx.$$

We have $f(x) = 10\ \text{cosh}\left( {x\text{/}10} \right),$ so $f^{\prime}(x) = \text{sinh}\left( {x\text{/}10} \right).$ Then

我们有 $f(x) = 10\ \text{cosh}\left( {x\text{/}10} \right),$ 故 $f^{\prime}(x) = \text{sinh}\left( {x\text{/}10} \right).$ 于是

$$\begin{array}{cl} \text{Arc Length} & {= {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx} \\ & {= {\int_{-15}^{15}\sqrt{1 + \text{sinh}^{2}\left( \frac{x}{10} \right)}}\ dx.} \end{array}$$

Now recall that $1 + \text{sinh}^{2}x = \text{cosh}^{2}x,$ so we have

现在回顾 $1 + \text{sinh}^{2}x = \text{cosh}^{2}x,$ 于是我们有

$$\begin{array}{cl} \text{Arc Length} & {= {\int_{-15}^{15}\sqrt{1 + \text{sinh}^{2}\left( \frac{x}{10} \right)}}\ dx} \\ & {= \int_{-15}^{15}\text{cosh}\left( \frac{x}{10} \right)dx} \\ & {= 10\ \text{sinh}\left. \left( \frac{x}{10} \right) \right|_{-15}^{15} = 10\left\lbrack {\text{sinh}\left( \frac{3}{2} \right) - \text{sinh}\left( {- \frac{3}{2}} \right)} \right\rbrack = 20\ \text{sinh}\left( \frac{3}{2} \right)} \\ & {\approx 42.586\ \text{ft}\text{.}} \end{array}$$

Assume a hanging cable has the shape $15\ \text{cosh}\left( {x\text{/}15} \right)$ for $-20 \leq x \leq 20.$ Determine the length of the cable (in feet).

设一根悬挂的电缆在区间 $-20 \leq x \leq 20$ 上呈 $15\ \text{cosh}\left( {x\text{/}15} \right)$ 的形状。求该电缆的长度（以英尺计）。

Section 2.9 Exercises 2.9 节习题

377.

\[T\] Find expressions for $\text{cosh}\ x + \text{sinh}\ x$ and $\text{cosh}\ x - \text{sinh}\ x.$ Use a calculator to graph these functions and ensure your expression is correct.

\[T\] 求 $\text{cosh}\ x + \text{sinh}\ x$ 与 $\text{cosh}\ x - \text{sinh}\ x$ 的表达式。用计算器画出这些函数的图像，确认你的表达式正确。

378\.

From the definitions of $\text{cosh}(x)$ and $\text{sinh}(x),$ find their antiderivatives.

由 $\text{cosh}(x)$ 与 $\text{sinh}(x)$ 的定义，求其原函数。

379.

Show that $\text{cosh}(x)$ and $\text{sinh}(x)$ satisfy $y^{''} = y.$

证明 $\text{cosh}(x)$ 与 $\text{sinh}(x)$ 满足 $y^{''} = y.$

380\.

Use the quotient rule to verify that $\text{tanh}(x)\prime = \text{sech}^{2}(x).$

用商的求导法则验证 $\text{tanh}(x)\prime = \text{sech}^{2}(x).$

381.

Derive $\text{cosh}^{2}(x) + \text{sinh}^{2}(x) = \text{cosh}\left( {2x} \right)$ from the definition.

由定义推导 $\text{cosh}^{2}(x) + \text{sinh}^{2}(x) = \text{cosh}\left( {2x} \right).$

382\.

Take the derivative of the previous expression to find an expression for $\text{sinh}\left( {2x} \right).$

对前一表达式求导，得到 $\text{sinh}\left( {2x} \right)$ 的表达式。

383.

Prove $\text{sinh}\left( {x + y} \right) = \text{sinh}(x)\text{cosh}(y) + \text{cosh}(x)\text{sinh}(y)$ by changing the expression to exponentials.

将表达式化为指数形式，证明 $\text{sinh}\left( {x + y} \right) = \text{sinh}(x)\text{cosh}(y) + \text{cosh}(x)\text{sinh}(y).$

384\.

Take the derivative of the previous expression to find an expression for $\text{cosh}\left( {x + y} \right).$

对前一表达式求导，得到 $\text{cosh}\left( {x + y} \right)$ 的表达式。

For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.

对以下习题，求所给函数的导数，并与函数一起作图，以确认你的答案正确。

385.

\[T\] $\text{cosh}\left( {3x + 1} \right)$

386\.

\[T\] $\text{sinh}\left( x^{2} \right)$

387.

\[T\] $\frac{1}{\text{cosh}(x)}$

388\.

\[T\] $\text{sinh}\left( {\text{ln}(x)} \right)$

389.

\[T\] $\text{cosh}^{2}(x) + \text{sinh}^{2}(x)$

390\.

\[T\] $\text{cosh}^{2}(x) - \text{sinh}^{2}(x)$

391.

\[T\] $\text{tanh}\left( \sqrt{x^{2} + 1} \right)$

392\.

\[T\] $\frac{1 + \text{tanh}(x)}{1 - \text{tanh}(x)}$

393.

\[T\] $\text{sinh}^{6}(x)$

394\.

\[T\] $\text{ln}\left( {\text{sech}(x) + \text{tanh}(x)} \right)$

For the following exercises, find the antiderivatives for the given functions.

对以下习题，求所给函数的原函数。

395.

$\text{cosh}\left( {2x + 1} \right)$

396\.

$\text{tanh}\left( {3x + 2} \right)$

397.

$x\ \text{cosh}\left( x^{2} \right)$

398\.

$3x^{3}\text{tanh}\left( x^{4} \right)$

399.

$\text{cosh}^{2}(x)\text{sinh}(x)$

400\.

$\text{tanh}^{2}(x)\text{sech}^{2}(x)$

401.

$\frac{\text{sinh}(x)}{1 + \text{cosh}(x)}$

402\.

$\text{coth}(x)$

403.

$\text{cosh}(x) + \text{sinh}(x)$

404\.

$\left( {\text{cosh}(x) + \text{sinh}(x)} \right)^{n}$

For the following exercises, find the derivatives for the functions.

对以下习题，求各函数的导数。

405.

$\text{tanh}^{-1}\left( {4x} \right)$

406\.

$\text{sinh}^{-1}\left( x^{2} \right)$

407.

$\text{sinh}^{-1}\left( {\text{cosh}(x)} \right)$

408\.

$\text{cosh}^{-1}\left( x^{3} \right)$

409.

$\text{tanh}^{-1}\left( {\text{cos}(x)} \right)$

410\.

$e^{\text{sinh}^{-1}{(x)}}$

411.

$\text{ln}\left( {\text{tanh}^{-1}(x)} \right)$

For the following exercises, find the antiderivatives for the functions.

对以下习题，求各函数的原函数。

412\.

$\int\frac{dx}{4 - x^{2}}$

413.

$\int\frac{dx}{a^{2} - x^{2}}$

414\.

$\int\frac{dx}{\sqrt{x^{2} + 1}}$

415.

$\int\frac{x\ dx}{\sqrt{x^{2} + 1}}$

416\.

$\int{- \frac{dx}{x\sqrt{1 - x^{2}}}}$

417.

${\int\frac{e^{x}}{\sqrt{e^{2x} - 1}}}{dx}$

418\.

${\int{- \frac{2x}{x^{4} - 1}}}{dx}$

For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation ${{dv}\text{/}{dt}} = g - v^{2}.$

对以下习题，利用下落物体所受摩擦阻力等于速度平方这一事实，其满足方程 ${{dv}\text{/}{dt}} = g - v^{2}.$

419.

Show that $v(t) = \sqrt{g}\ \text{tanh}\left( \left( \sqrt{g} \right)t \right)$ satisfies this equation.

证明 $v(t) = \sqrt{g}\ \text{tanh}\left( \left( \sqrt{g} \right)t \right)$ 满足该方程。

420\.

Derive the previous expression for $v(t)$ by integrating $\frac{dv}{g - v^{2}} = dt.$

通过对 $\frac{dv}{g - v^{2}} = dt$ 积分，推导 $v(t)$ 的前一表达式。

421.

\[T\] Estimate how far a body has fallen in $12$ seconds by finding the area underneath the curve of $v(t).$

\[T\] 通过求 $v(t)$ 曲线下方的面积，估计物体在 $12$ 秒内下落的距离。

For the following exercises, use this scenario: A cable hanging under its own weight has a slope $S = {{dy}\text{/}{dx}}$ that satisfies ${{dS}\text{/}{dx}} = c\sqrt{1 + S^{2}}.$ The constant $c$ is the ratio of cable density to tension.

对以下习题，使用如下场景：一根缆索在自重作用下悬挂，其斜率 $S = {{dy}\text{/}{dx}}$ 满足 ${{dS}\text{/}{dx}} = c\sqrt{1 + S^{2}}.$ 常数 $c$ 是缆索密度与张力的比值。

422\.

Show that $S = \text{sinh}(cx)$ satisfies this equation.

证明 $S = \text{sinh}(cx)$ 满足该方程。

423.

Integrate ${{dy}\text{/}{dx}} = \text{sinh}(cx)$ to find the cable height $y(x)$ if $y(0) = {1\text{/}c}.$

对 ${{dy}\text{/}{dx}} = \text{sinh}(cx)$ 积分，在 $y(0) = {1\text{/}c}$ 的条件下求电缆高度 $y(x)$。

424\.

Sketch the cable and determine how far down it sags at $x = 0.$

画出电缆草图，并确定其在 $x = 0$ 处下垂的距离。

For the following exercises, solve each problem.

对以下习题，求解各问题。

425.

\[T\] A chain hangs from two posts $2$ m apart to form a catenary described by the equation $y = 2\ \text{cosh}\left( {x\text{/}2} \right) - 1.$ Find the slope of the catenary at the left fence post.

\[T\] 一条链子悬挂在相距 $2$ 米的两根柱子上，形成由方程 $y = 2\ \text{cosh}\left( {x\text{/}2} \right) - 1$ 描述的悬链线。求该悬链线在左侧柱处的斜率。

426\.

\[T\] A chain hangs from two posts four meters apart to form a catenary described by the equation $y = 4\ \text{cosh}\left( {x\text{/}4} \right) - 3.$ Find the total length of the catenary (arc length).

\[T\] 一条链子悬挂在相距四米的两根柱子上，形成由方程 $y = 4\ \text{cosh}\left( {x\text{/}4} \right) - 3$ 描述的悬链线。求该悬链线的总长度（弧长）。

427.

\[T\] A high-voltage power line is a catenary described by $y = 10\ \text{cosh}\left( {x\text{/}10} \right).$ Find the ratio of the area under the catenary to its arc length. What do you notice?

\[T\] 一条高压电线是悬链线，由 $y = 10\ \text{cosh}\left( {x\text{/}10} \right)$ 描述。求悬链线下方面积与其弧长之比。你注意到了什么？

428.

A telephone line is a catenary described by $y = a\ \text{cosh}\left( {x\text{/}a} \right).$ Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?

一条电话线是悬链线，由 $y = a\ \text{cosh}\left( {x\text{/}a} \right)$ 描述。求悬链线下方面积与其弧长之比。这是否印证了你上一问的答案？

429.

Prove the formula for the derivative of $y = \text{sinh}^{-1}(x)$ by differentiating $x = \text{sinh}(y).$ (*Hint:* Use hyperbolic trigonometric identities.)

通过对 $x = \text{sinh}(y)$ 求导，证明 $y = \text{sinh}^{-1}(x)$ 的导数公式。（*Hint:* 使用双曲三角恒等式。）

430\.

Prove the formula for the derivative of $y = \text{cosh}^{-1}(x)$ by differentiating $x = \text{cosh}(y).$

通过对 $x = \text{cosh}(y)$ 求导，证明 $y = \text{cosh}^{-1}(x)$ 的导数公式。

(*Hint:* Use hyperbolic trigonometric identities.)

（*Hint:* 使用双曲三角恒等式。）

431.

Prove the formula for the derivative of $y = \text{sech}^{-1}(x)$ by differentiating $x = \text{sech}(y).$ (*Hint:* Use hyperbolic trigonometric identities.)

通过对 $x = \text{sech}(y)$ 求导，证明 $y = \text{sech}^{-1}(x)$ 的导数公式。（*Hint:* 使用双曲三角恒等式。）

432\.

Prove that $\left( {\text{cosh}(x) + \text{sinh}(x)} \right)^{n} = \text{cosh}(nx) + \text{sinh}(nx).$

证明 $\left( {\text{cosh}(x) + \text{sinh}(x)} \right)^{n} = \text{cosh}(nx) + \text{sinh}(nx).$

433.

Prove the expression for $\text{sinh}^{-1}(x).$ Multiply $x = \text{sinh}(y) = \left( {1\text{/}2} \right)\left( {e^{y}–e^{\text{−}y}} \right)$ by $2e^{y}$ and solve for $y.$ Does your expression match the textbook?

证明 $\text{sinh}^{-1}(x)$ 的表达式。将 $x = \text{sinh}(y) = \left( {1\text{/}2} \right)\left( {e^{y}–e^{\text{−}y}} \right)$ 乘以 $2e^{y}$ 并解出 $y$。你的表达式与教材一致吗？

434\.

Prove the expression for $\text{cosh}^{-1}(x).$ Multiply $x = \text{cosh}(y) = \left( {1\text{/}2} \right)\left( {e^{y} + e^{\text{−}y}} \right)$ by $2e^{y}$ and solve for $y.$ Does your expression match the textbook?

证明 $\text{cosh}^{-1}(x)$ 的表达式。将 $x = \text{cosh}(y) = \left( {1\text{/}2} \right)\left( {e^{y} + e^{\text{−}y}} \right)$ 乘以 $2e^{y}$ 并解出 $y$。你的表达式与教材一致吗？

Key Terms 关键术语

arc length

弧长

the arc length of a curve can be thought of as the distance a person would travel along the path of the curve

曲线的弧长可以理解为一个人沿曲线路径行进所经过的距离。

catenary

悬链线

a curve in the shape of the function $y = a\ \text{cosh}\left( {x\text{/}a} \right)$ is a catenary; a cable of uniform density suspended between two supports assumes the shape of a catenary

形如函数 $y = a\ \text{cosh}\left( {x\text{/}a} \right)$ 的曲线是悬链线；密度均匀的缆索悬挂于两个支点之间时呈现悬链线的形状。

center of mass

质心

the point at which the total mass of the system could be concentrated without changing the moment

系统的总质量可集中于该点而不改变其力矩。

centroid

形心

the centroid of a region is the geometric center of the region; laminas are often represented by regions in the plane; if the lamina has a constant density, the center of mass of the lamina depends only on the shape of the corresponding planar region; in this case, the center of mass of the lamina corresponds to the centroid of the representative region

区域的形心是该区域的几何中心；薄片常用平面区域表示；若薄片的密度恒定，则其质心仅取决于相应平面区域的形状；此时薄片的质心对应于该代表区域的形心。

cross-section

横截面

the intersection of a plane and a solid object

平面与立体物体的交截部分。

density function

密度函数

a density function describes how mass is distributed throughout an object; it can be a linear density, expressed in terms of mass per unit length; an area density, expressed in terms of mass per unit area; or a volume density, expressed in terms of mass per unit volume; weight-density is also used to describe weight (rather than mass) per unit volume

密度函数描述质量在物体中的分布方式；它可以是线密度（以单位长度的质量表示）、面密度（以单位面积的质量表示）或体密度（以单位体积的质量表示）；重度也用于表示单位体积的重量（而非质量）。

disk method

圆盘法

a special case of the slicing method used with solids of revolution when the slices are disks

切片法在旋转体且切片为圆盘时的特例。

doubling time

倍增时间

if a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double, and is given by $\left( {\text{ln}\ 2} \right)\text{/}k$

若某量呈指数增长，倍增时间即为该量翻倍所需的时间，由 $\left( {\text{ln}\ 2} \right)\text{/}k$ 给出。

exponential decay

指数衰减

systems that exhibit exponential decay follow a model of the form $y = y_{0}e^{\text{−}kt}$

呈指数衰减的系统遵循形式为 $y = y_{0}e^{\text{−}kt}$ 的模型。

exponential growth

指数增长

systems that exhibit exponential growth follow a model of the form $y = y_{0}e^{kt}$

呈指数增长的系统遵循形式为 $y = y_{0}e^{kt}$ 的模型。

frustum

圆台（截头圆锥）

a portion of a cone; a frustum is constructed by cutting the cone with a plane parallel to the base

圆锥的一部分；圆台是用一个平行于底面的平面截圆锥而得到的。

half-life

半衰期

if a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by $\left( {\text{ln}\ 2} \right)\text{/}k$

若某量呈指数衰减，半衰期即为该量减半所需的时间，由 $\left( {\text{ln}\ 2} \right)\text{/}k$ 给出。

Hooke’s law

胡克定律

this law states that the force required to compress (or elongate) a spring is proportional to the distance the spring has been compressed (or stretched) from equilibrium; in other words, $F = kx,$ where $k$ is a constant

该定律指出，压缩（或拉伸）弹簧所需之力与弹簧偏离平衡位置被压缩（或拉伸）的距离成正比；换言之，$F = kx,$ 其中 $k$ 为常数。

hydrostatic pressure

静水压强

the pressure exerted by water on a submerged object

水对浸没物体施加的压强。

lamina

薄片（薄板）

a thin sheet of material; laminas are thin enough that, for mathematical purposes, they can be treated as if they are two-dimensional

一层薄的材料；薄片足够薄，在数学处理上可视为二维的。

method of cylindrical shells

圆柱壳法

a method of calculating the volume of a solid of revolution by dividing the solid into nested cylindrical shells; this method is different from the methods of disks or washers in that we integrate with respect to the opposite variable

将旋转体划分为嵌套圆柱壳以计算其体积的方法；该方法不同于圆盘法或垫圈法，因为它对相反的变量积分。

moment

力矩（矩）

if *n* masses are arranged on a number line, the moment of the system with respect to the origin is given by $M = \sum\limits_{i = 1}^{n}m_{i}x_{i};$ if, instead, we consider a region in the plane, bounded above by a function $f(x)$ over an interval $\left\lbrack {a,b} \right\rbrack,$ then the moments of the region with respect to the *x*- and *y*-axes are given by $M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}$ and $M_{y} = \rho{\int_{a}^{b}{xf(x)dx}},$ respectively

若 *n* 个质点在数轴上排列，系统关于原点的力矩由 $M = \sum\limits_{i = 1}^{n}m_{i}x_{i};$ 给出；若考虑平面中由函数 $f(x)$ 在区间 $\left\lbrack {a,b} \right\rbrack$ 上界定的区域，则该区域关于 *x* 轴与 *y* 轴的力矩分别为 $M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}$ 与 $M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}.$

slicing method

切片法

a method of calculating the volume of a solid that involves cutting the solid into pieces, estimating the volume of each piece, then adding these estimates to arrive at an estimate of the total volume; as the number of slices goes to infinity, this estimate becomes an integral that gives the exact value of the volume

通过将立体切成若干块、估计每块的体积、再将这些估计相加以得到总体积估计值来计算立体体积的方法；当切片数趋于无穷时，该估计变为给出精确体积的积分。

solid of revolution

旋转体

a solid generated by revolving a region in a plane around a line in that plane

平面中某区域绕该平面内一条直线旋转而生成的立体。

surface area

表面积

the surface area of a solid is the total area of the outer layer of the object; for objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces

立体的表面积是其外层的总面积；对于立方体或砖块等物体，其表面积等于所有面面积之和。

symmetry principle

对称原理

the symmetry principle states that if a region *R* is symmetric about a line *l*, then the centroid of *R* lies on *l*

对称原理指出：若区域 *R* 关于某直线 *l* 对称，则 *R* 的形心位于 *l* 上。

theorem of Pappus for volume

帕普斯体积定理

this theorem states that the volume of a solid of revolution formed by revolving a region around an external axis is equal to the area of the region multiplied by the distance traveled by the centroid of the region

该定理指出，区域绕一条外部轴旋转所形成的旋转体体积，等于该区域的面积乘以该区域形心所经过的路程。

washer method

垫圈法（圆环法）

a special case of the slicing method used with solids of revolution when the slices are washers

切片法在旋转体且切片为垫圈时的特例。

work

功

the amount of energy it takes to move an object; in physics, when a force is constant, work is expressed as the product of force and distance

移动物体所需的能量；在物理学中，当力为常量时，功表示为力与距离的乘积。

Key Equations 重要公式

Formula Name	Formula
*Area between two curves, integrating on the x-axis*	$A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx$
*Area between two curves, integrating on the y-axis*	$A = \int_{c}^{d}\left\lbrack {u(y) - v(y)} \right\rbrack dy$
*Disk Method along the x-axis*	$V = {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}$
*Disk Method along the y-axis*	$V = {\int_{c}^{d}{\pi\left\lbrack {g(y)} \right\rbrack^{2}dy}}$
Washer Method	$V = {\int_{a}^{b}{\pi\left\lbrack {\left( {f(x)} \right)^{2} - \left( {g(x)} \right)^{2}} \right\rbrack}}dx$
Method of Cylindrical Shells	$V = {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx$
Arc Length of a Function of x	$\text{Arc Length} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx$
Arc Length of a Function of y	$\text{Arc Length} = {\int_{c}^{d}\sqrt{1 + \left\lbrack {g^{\prime}(y)} \right\rbrack^{2}}}\ dy$
Surface Area of a Function of x	$\text{Surface Area} = {\int_{a}^{b}{\left( {2\pi f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}} \right)dx}}$
Mass of a one-dimensional object	$m = {\int_{a}^{b}\rho}(x)dx$
Mass of a circular object	$m = {\int_{0}^{r}2}\pi x\rho(x)dx$
Work done on an object	$W = {\int_{a}^{b}F}(x)dx$
Hydrostatic force on a plate	$F = {\int_{a}^{b}\rho}w(x)s(x)dx$
Mass of a lamina	$m = \rho\int_{a}^{b}f(x)dx$
Moments of a lamina	$M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}$
Center of mass of a lamina	$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}$
Natural logarithm function	$\text{ln}\ x = {\int_{1}^{x}\frac{1}{t}}dt$ Z
Exponential function $y = e^{x}$	$\text{ln}\ y = \text{ln}\left( e^{x} \right) = x$ Z

公式名称	公式
*曲线之间的面积，关于 x* 轴积分**	$A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx$
*曲线之间的面积，关于 y* 轴积分**	$A = \int_{c}^{d}\left\lbrack {u(y) - v(y)} \right\rbrack dy$
*沿 x* 轴的圆盘法**	$V = {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}$
*沿 y* 轴的圆盘法**	$V = {\int_{c}^{d}{\pi\left\lbrack {g(y)} \right\rbrack^{2}dy}}$
垫圈法	$V = {\int_{a}^{b}{\pi\left\lbrack {\left( {f(x)} \right)^{2} - \left( {g(x)} \right)^{2}} \right\rbrack}}dx$
圆柱壳法	$V = {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx$
*关于 x* 的函数的弧长**	$\text{Arc Length} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx$
*关于 y* 的函数的弧长**	$\text{Arc Length} = {\int_{c}^{d}\sqrt{1 + \left\lbrack {g^{\prime}(y)} \right\rbrack^{2}}}\ dy$
*关于 x* 的函数的表面积**	$\text{Surface Area} = {\int_{a}^{b}{\left( {2\pi f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}} \right)dx}}$
一维物体的质量	$m = {\int_{a}^{b}\rho}(x)dx$
圆形物体的质量	$m = {\int_{0}^{r}2}\pi x\rho(x)dx$
对物体所做的功	$W = {\int_{a}^{b}F}(x)dx$
板上的静水压力	$F = {\int_{a}^{b}\rho}w(x)s(x)dx$
薄片的质量	$m = \rho\int_{a}^{b}f(x)dx$
薄片的力矩	$M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}$
薄片的质心	$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}$
自然对数函数	$\text{ln}\ x = {\int_{1}^{x}\frac{1}{t}}dt$ Z
指数函数 $y = e^{x}$	$\text{ln}\ y = \text{ln}\left( e^{x} \right) = x$ Z

2.1 Areas between Curves 2.1 曲线之间的面积

Just as definite integrals can be used to find the area under a curve, they can also be used to find the area between two curves.
To find the area between two curves defined by functions, integrate the difference of the functions.
If the graphs of the functions cross, or if the region is complex, use the absolute value of the difference of the functions. In this case, it may be necessary to evaluate two or more integrals and add the results to find the area of the region.
Sometimes it can be easier to integrate with respect to *y* to find the area. The principles are the same regardless of which variable is used as the variable of integration.

正如定积分可用来求曲线下方的面积，它们也可用来求两条曲线之间的面积。
为了求由两个函数所定义的两曲线之间的面积，对被积函数的差进行积分。
若函数图像相交，或者区域较为复杂，则使用函数之差的绝对值。此时，可能需要计算两个或更多个积分并将结果相加，以求得该区域的面积。
有时关于 *y* 积分来求面积会更加容易。无论以哪个变量作为积分变量，原理都是相同的。

2.2 Determining Volumes by Slicing 2.2 用切片法确定体积

Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.

定积分可用来求立体的体积。利用切片法，我们可以通过积分横截面积来求体积。
对于旋转体，体积切片常是圆盘，横截面是圆。圆盘法的实质是在横截面为圆的特殊情形下应用切片法，并使用圆的面积公式。
若旋转体中心有空腔，则体积切片为垫圈（圆环）。采用垫圈法时，在积分之前，先以内圆的面积从外圆的面积中减去。

2.3 Volumes of Revolution: Cylindrical Shells 2.3 旋转体体积：圆柱壳

The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.

圆柱壳法是另一种利用定积分计算旋转体体积的方法。由于我们是就另一个变量积分，这种方法有时优于圆盘法或垫圈法。在某些情况下，一个积分会比另一个积分复杂得多。
函数的几何形状与积分的难度，是选择使用哪种积分方法的主要考量因素。

2.4 Arc Length of a Curve and Surface Area 2.4 曲线的弧长与表面积

The arc length of a curve can be calculated using a definite integral.
The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of $y.$
The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may be necessary to use a computer or calculator to approximate the values of the integrals.

曲线的弧长可用定积分计算。
弧长先用线段近似，由此产生一个黎曼和。再取极限，便得到定积分公式。同样的过程也可应用于 $y$ 的函数。
用于计算弧长的概念可推广到求旋转面的表面积。
由弧长公式与表面积公式产生的积分往往难以计算。有时必须借助计算机或计算器来近似这些积分的值。

2.5 Physical Applications 2.5 物理应用

Several physical applications of the definite integral are common in engineering and physics.
Definite integrals can be used to determine the mass of an object if its density function is known.
Work can also be calculated from integrating a force function, or when counteracting the force of gravity, as in a pumping problem.
Definite integrals can also be used to calculate the force exerted on an object submerged in a liquid.

定积分在工程与物理中有若干常见的应用。
若已知物体的密度函数，定积分可用来确定该物体的质量。
功也可通过积分一个力函数求得，或在抵抗重力（如抽水问题）时求得。
定积分也可用来计算作用在浸没于液体中的物体上的力。

2.6 Moments and Centers of Mass 2.6 力矩与质心

Mathematically, the center of mass of a system is the point at which the total mass of the system could be concentrated without changing the moment. Loosely speaking, the center of mass can be thought of as the balancing point of the system.
For point masses distributed along a number line, the moment of the system with respect to the origin is $M = \sum\limits_{i = 1}^{n}m_{i}x_{i}.$ For point masses distributed in a plane, the moments of the system with respect to the *x*- and *y*-axes, respectively, are $M_{x} = \sum\limits_{i = 1}^{n}m_{i}y_{i}$ and $M_{y} = \sum\limits_{i = 1}^{n}m_{i}x_{i},$ respectively.
For a lamina bounded above by a function $f(x),$ the moments of the system with respect to the *x*- and *y*-axes, respectively, are $M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}$ and $M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}.$
The *x*- and *y*-coordinates of the center of mass can be found by dividing the moments around the *y*-axis and around the *x*-axis, respectively, by the total mass. The symmetry principle says that if a region is symmetric with respect to a line, then the centroid of the region lies on the line.
The theorem of Pappus for volume says that if a region is revolved around an external axis, the volume of the resulting solid is equal to the area of the region multiplied by the distance traveled by the centroid of the region.

从数学上讲，一个系统的质心是该系统的全部质量可以集中于此而不改变其力矩的那一点。通俗地说，质心可看作系统的平衡点。
对于分布在一条数轴上的点质量，系统关于原点的力矩为 $M = \sum\limits_{i = 1}^{n}m_{i}x_{i}.$ 对于分布在平面上的点质量，系统分别关于 *x* 轴与 *y* 轴的力矩为 $M_{x} = \sum\limits_{i = 1}^{n}m_{i}y_{i}$ 与 $M_{y} = \sum\limits_{i = 1}^{n}m_{i}x_{i},$ 。（分别）
对于上边界由函数 $f(x)$ 所界的薄片，系统分别关于 *x* 轴与 *y* 轴的力矩为 $M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}$ 与 $M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}.$
质心的 *x* 坐标与 *y* 坐标可分别通过将绕 *y* 轴的力矩与绕 *x* 轴的力矩除以总质量得到。对称原理指出：若一个区域关于一条直线对称，则该区域的形心就在这条直线上。
帕普斯体积定理说：若一个区域绕一条外部轴旋转，则所得立体的体积等于该区域的面积乘以该区域形心所经过的路程。

2.7 Integrals, Exponential Functions, and Logarithms 2.7 积分、指数函数与对数函数

The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way.
The cornerstone of the development is the definition of the natural logarithm in terms of an integral.
The function $e^{x}$ is then defined as the inverse of the natural logarithm.
General exponential functions are defined in terms of $e^{x},$ and the corresponding inverse functions are general logarithms.
Familiar properties of logarithms and exponents still hold in this more rigorous context.

此前对对数函数与指数函数的处理并未以精确、严格的方式定义这些函数。本节以一种数学上严谨的方式发展这些概念。
这一发展的基石是用积分来定义自然对数。
函数 $e^{x}$ 随后被定义为自然对数的反函数。
一般指数函数借助 $e^{x}$ 定义，其相应的反函数即为一般对数函数。
在对数与指数方面我们所熟悉的性质，在这一更严格的框架下仍然成立。

2.8 Exponential Growth and Decay 2.8 指数增长与衰减

Exponential growth and exponential decay are two of the most common applications of exponential functions.
Systems that exhibit exponential growth follow a model of the form $y = y_{0}e^{kt}.$
In exponential growth, the rate of growth is proportional to the quantity present. In other words, $y^{\prime} = ky.$
Systems that exhibit exponential growth have a constant doubling time, which is given by ${\left( {\text{ln}\ 2} \right)\text{/}k}.$
Systems that exhibit exponential decay follow a model of the form $y = y_{0}e^{\text{−}kt}.$
Systems that exhibit exponential decay have a constant half-life, which is given by ${\left( {\text{ln}\ 2} \right)\text{/}k}.$

指数增长与指数衰减是指数函数最常见应用中的两种。
呈现指数增长的系统遵循形如 $y = y_{0}e^{kt}.$ 的模型。
在指数增长中，增长率与当前数量成正比。换言之，$y^{\prime} = ky.$
呈现指数增长的系统具有恒定的倍增时间，由 ${\left( {\text{ln}\ 2} \right)\text{/}k}.$ 给出。
呈现指数衰减的系统遵循形如 $y = y_{0}e^{\text{−}kt}.$ 的模型。
呈现指数衰减的系统具有恒定的半衰期，由 ${\left( {\text{ln}\ 2} \right)\text{/}k}.$ 给出。

2.9 Calculus of the Hyperbolic Functions 2.9 双曲函数的微积分

Hyperbolic functions are defined in terms of exponential functions.
Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiation formulas give rise, in turn, to integration formulas.
With appropriate range restrictions, the hyperbolic functions all have inverses.
Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give rise to integration formulas.
The most common physical applications of hyperbolic functions are calculations involving catenaries.

双曲函数是借助指数函数定义的。
逐项求导得到双曲函数的求导公式。这些求导公式又进而给出积分公式。
在适当的范围限制下，双曲函数都有反函数。
隐函数求导得到反双曲函数的求导公式，而这些公式又进而给出积分公式。
双曲函数最常见的物理应用是涉及悬链线的计算。

Review Exercises 复习题

*True or False?* Justify your answer with a proof or a counterexample.

*True or False?* 用证明或反例说明你的答案。

435. The amount of work to pump the water out of a half-full cylinder is half the amount of work to pump the water out of the full cylinder.

435. 将半满圆柱体中水抽出所做的功，等于将满圆柱体水抽出所做功的一半。

436\. If the force is constant, the amount of work to move an object from $x = a$ to $x = b$ is $F\left( {b - a} \right).$

436\. 若力为常量，则将物体从 $x = a$ 移动到 $x = b$ 所做的功为 $F\left( {b - a} \right).$

437. The disk method can be used in any situation in which the washer method is successful at finding the volume of a solid of revolution.

437. 圆盘法可用于垫圈法成功求得旋转体体积的任何情形。

438\. If the half-life of $\text{seaborgium-}266$ is $360$ ms, then $k = {\left( {\text{ln}(2)} \right)\text{/}360}.$

438\. 若 $\text{seaborgium-}266$ 的半衰期为 $360$ ms，则 $k = {\left( {\text{ln}(2)} \right)\text{/}360}.$

For the following exercises, use the requested method to determine the volume of the solid.

对于下列习题，使用所要求的方法确定该立体的体积。

439. The volume that has a base of the ellipse ${x^{2}\text{/}4} + {y^{2}\text{/}9} = 1$ and cross-sections of an equilateral triangle perpendicular to the $y\text{-axis}\text{.}$ Use the method of slicing.

439. 该立体以椭圆 ${x^{2}\text{/}4} + {y^{2}\text{/}9} = 1$ 为底面，且垂直于 $y\text{-axis}\text{.}$ 的横截面为等边三角形。使用切片法。

440\. The region bounded by the curve $y~ = ~x^{2}~–~x$ and the *x*-axis from $x = 1\ \text{to}\ x = 4,$ rotated around the *y*-axis using the washer method

440\. 由曲线 $y~ = ~x^{2}~–~x$ 与 *x* 轴在区间 $x = 1\ \text{to}\ x = 4$ 上围成的区域，绕 *y* 轴旋转，使用垫圈法。

441. $x = y^{2}$ and $x = 3y$ rotated around the *y*-axis using the washer method

441. $x = y^{2}$ 与 $x = 3y$ 绕 *y* 轴旋转，使用垫圈法。

442\. $x = 2y^{2} - y^{3},x = 0,\ \text{and}\ y = 0$ rotated around the *x*-axis using cylindrical shells

442\. $x = 2y^{2} - y^{3},x = 0,\ \text{and}\ y = 0$ 绕 *x* 轴旋转，使用圆柱壳法。

For the following exercises, find

对于下列习题，求

1. the area of the region,

1. 该区域的面积，

2. the volume of the solid when rotated around the *x*-axis, and

2. 绕 *x* 轴旋转所得立体的体积，以及

3. the volume of the solid when rotated around the *y*-axis. Use whichever method seems most appropriate to you.

3. 绕 *y* 轴旋转所得立体的体积。使用你认为最合适的方法。

443. $y = x^{3},x = 0,y = 0,\ \text{and}\ x = 2$

444\. $y = x^{2} - x\ \text{and}\ x = 0$

445. \[T\] $y = \text{ln}(x) + 2\ \text{and}\ y = x$

446\. $y = x^{2}$ and $y = \sqrt{x}$

446\. $y = x^{2}$ 与 $y = \sqrt{x}$

447. $y = 5 + x,$ $y = x^{2},$ $x = 0,$ and $x = 1$

447. $y = 5 + x,$ $y = x^{2},$ $x = 0,$ 与 $x = 1$

448\. Below $x^{2} + y^{2} = 1$ and above $y = 1 - x$

448\. 在 $x^{2} + y^{2} = 1$ 下方、在 $y = 1 - x$ 上方。

449. Find the mass of $\rho = \frac{1}{x^{2} + 1}$ on a disk centered at the origin with radius $4.$

449. 求密度函数为 $\rho = \frac{1}{x^{2} + 1}$、以原点为中心、半径为 $4.$ 的圆盘上的质量。

450\. Find the center of mass for $\rho = \text{tan}^{2}x$ on $x \in \left( {- \frac{\pi}{4},\frac{\pi}{4}} \right).$

450\. 求密度函数 $\rho = \text{tan}^{2}x$ 在 $x \in \left( {- \frac{\pi}{4},\frac{\pi}{4}} \right)$ 上的质心。

451. Find the mass and the center of mass of $\rho = 1$ on the region bounded by $y = x^{5}$ and $y = \sqrt{x}.$

451. 求密度函数 $\rho = 1$ 在由 $y = x^{5}$ 与 $y = \sqrt{x}.$ 所围区域上的质量与质心。

For the following exercises, find the requested arc lengths.

对于下列习题，求所要求的弧长。

452\. The length of $x$ for $y = \text{cosh}(x)$ from $x = 0\ \text{to}\ x = 2.$

452\. 求曲线 $y = \text{cosh}(x)$ 上从 $x = 0\ \text{to}\ x = 2.$ 的弧长。

453. The length of $y$ for $x = 3 - \sqrt{y}$ from $y = 0$ to $y = 4$

453. 求曲线 $x = 3 - \sqrt{y}$ 上从 $y = 0$ 到 $y = 4$ 的弧长。

For the following exercises, find the surface area and volume when the given curves are revolved around the specified axis.

对于下列习题，求给定曲线绕指定轴旋转时的表面积与体积。

454\. The shape created by revolving the region between $y = 4 + x,$ $y = 3 - x,$ $x = 0,$ and $x = 2$ rotated around the *y*-axis.

454\. 将由 $y = 4 + x,$ $y = 3 - x,$ $x = 0,$ 与 $x = 2$ 所围区域绕 *y* 轴旋转所形成的立体。

455. The loudspeaker created by revolving $y = {1\text{/}x}$ from $x = 1$ to $x = 4$ around the *x*-axis.

455. 将由 $y = {1\text{/}x}$ 从 $x = 1$ 到 $x = 4$ 绕 *x* 轴旋转所形成的扬声器形状。

456\. For this exercise, consider the Karun-3 dam in Iran. Its shape can be approximated as an inverted isosceles triangle spanning across the river, with height 205 m and width (across the top of the dam) 388 m. Assume the current depth of the water is 180 m. The density of water is 1000 kg/m³. Find the total force on the wall of the dam.

456\. 本习题中，考虑伊朗的 Karun-3 大坝。其形状可近似为一个横跨河流的倒置等腰三角形，高 205 m，顶宽（坝顶跨度）388 m。假设当前水深 180 m。水的密度为 1000 kg/m³。求作用在坝壁上的总力。

457. You are a crime scene investigator attempting to determine the time of death of a victim. It is noon and $45\text{°}\text{F}$ outside and the temperature of the body is $78\text{°}\text{F}.$ You know the cooling constant is $k = 0.00824\text{°}\text{F/min}\text{.}$ When did the victim die, assuming that a human’s temperature is $98\text{°}\text{F}$ ?

457. 你是一名犯罪现场调查员，试图确定一名受害者的死亡时间。当前为正午，室外温度为 $45\text{°}\text{F}$，尸体温度为 $78\text{°}\text{F}.$ 已知冷却常数为 $k = 0.00824\text{°}\text{F/min}\text{.}$ 若假设人体体温为 $98\text{°}\text{F}$，受害者是何时死亡的？

For the following exercises, consider the stock market crash in $1929$ in the United States. The table lists the Dow Jones industrial average per year leading up to the crash.

对于下列习题，考虑美国 $1929$ 年的股市崩盘。下表列出了崩盘前历年的道琼斯工业平均指数。

Years after 1920	Value (\$)
$1$	$63.90$
$3$	$100$
$5$	$110$
$7$	$160$
$9$	$381.17$

1920 年之后的年数	价值（\$）
$1$	$63.90$
$3$	$100$
$5$	$110$
$7$	$160$
$9$	$381.17$

458. \[T\] The best-fit exponential curve to these data is given by $y = 40.71 + 1.224^{x}.$ Why do you think the gains of the market were unsustainable? Use first and second derivatives to help justify your answer. What would this model predict the Dow Jones industrial average to be in $2014$ ?

458. \[T\] 拟合这些数据的最优指数曲线为 $y = 40.71 + 1.224^{x}.$ 你为何认为市场的上涨是不可持续的？用一阶与二阶导数来帮助论证你的答案。该模型预测 2014 年的道琼斯工业平均指数为多少？

For the following exercises, consider the catenoid, the only solid of revolution that has a minimal surface, or zero mean curvature. A catenoid in nature can be found when stretching soap between two rings.

对于下列习题，考虑悬链面（catenoid）——唯一具有极小曲面（即零平均曲率）的旋转体。在自然界中，将肥皂膜拉伸于两个圆环之间时即可观察到悬链面。

459. Find the volume of the catenoid $y = \text{cosh}(x)$ from $x = -1\ \text{to}\ x = 1$ that is created by rotating this curve around the $x\text{-axis},$ as shown here.

459. 求由曲线 $y = \text{cosh}(x)$ 从 $x = -1\ \text{to}\ x = 1$ 绕 $x\text{-axis}$ 旋转所成悬链面的体积，如图所示。

460\. Find surface area of the catenoid $y = \text{cosh}(x)$ from $x = -1$ to $x = 1$ that is created by rotating this curve around the $x\ \text{-axis.}$

460\. 求由曲线 $y = \text{cosh}(x)$ 从 $x = -1$ 到 $x = 1$ 绕 $x\ \text{-axis.}$ 旋转所成悬链面的表面积。

2 Applications of Integration 积分的应用

2.1 Areas between Curves 2.1 曲线之间的面积

Area of a Region between Two Curves 两曲线之间区域的面积

Finding the Area between Two Curves 求两曲线之间的面积

Finding the Area of a Region between Two Curves 1 求两曲线之间区域的面积 1

Solution 解答

Finding the Area of a Region between Two Curves 2 求两曲线之间区域的面积 2

Solution 解答

Areas of Compound Regions 复合区域的面积

Finding the Area of a Region between Curves That Cross 求相交曲线之间区域的面积

Finding the Area of a Region Bounded by Functions That Cross 求由相交函数所围区域的面积

Solution 解答

Finding the Area of a Complex Region 求复杂区域的面积

Solution 解答

Regions Defined with Respect to *y* 关于 *y* 定义的区域

Finding the Area between Two Curves, Integrating along the *y*-axis 求两曲线之间的面积（沿 *y* 轴积分）

Integrating with Respect to *y* 关于 *y* 积分

Solution 解答

Section 2.1 Exercises 2.1 节习题

2.2 Determining Volumes by Slicing 2.2 用切片法确定体积

Volume and the Slicing Method 体积与切片法

Finding Volumes by the Slicing Method 用切片法求体积

Deriving the Formula for the Volume of a Pyramid 推导棱锥体积公式

Solution 解答

Solids of Revolution 旋转体

Using the Slicing Method to find the Volume of a Solid of Revolution 用切片法求旋转体的体积

Solution 解答

The Disk Method 圆盘法

Using the Disk Method to Find the Volume of a Solid of Revolution 1 用圆盘法求旋转体的体积 1

Solution 解答

Using the Disk Method to Find the Volume of a Solid of Revolution 2 用圆盘法求旋转体的体积 2

Solution 解答

The Washer Method 垫圈法

Using the Washer Method 用垫圈法

Solution 解答

The Washer Method with a Different Axis of Revolution 旋转轴不同的垫圈法

Solution 解答

Section 2.2 Exercises 2.2 节习题

2.3 Volumes of Revolution: Cylindrical Shells 2.3 旋转体的体积：圆柱壳法

The Method of Cylindrical Shells 圆柱壳法

The Method of Cylindrical Shells 1 圆柱壳法 1

Solution 解答

The Method of Cylindrical Shells 2 圆柱壳法 2

Solution 解答

The Method of Cylindrical Shells for a Solid Revolved around the *x*-axis 绕 *x* 轴旋转的立体的圆柱壳法

Solution 解答

A Region of Revolution Revolved around a Line 绕一条直线旋转的旋转区域

Solution 解答

A Region of Revolution Bounded by the Graphs of Two Functions 由两个函数图形所界定的旋转区域

Solution 解答

Which Method Should We Use? 该用哪种方法？

Selecting the Best Method 选择最佳方法

Solution 解答

Section 2.3 Exercises 2.3 节习题

2.4 Arc Length of a Curve and Surface Area 2.4 曲线的弧长与表面积

Arc Length of the Curve *y* = *f*(*x*) 曲线 *y* = *f*(*x*) 的弧长

Arc Length for *y* = *f*(*x*) 曲线 *y* = *f*(*x*) 的弧长

Calculating the Arc Length of a Function of *x* 计算关于 *x* 的函数的弧长

Solution 解答

Using a Computer or Calculator to Determine the Arc Length of a Function of *x* 使用计算机或计算器确定关于 *x* 的函数的弧长

Solution 解答

Arc Length of the Curve *x* = *g*(*y*) 曲线 *x* = *g*(*y*) 的弧长

Arc Length for *x* = *g*(*y*) 曲线 *x* = *g*(*y*) 的弧长

Calculating the Arc Length of a Function of *y* 计算关于 *y* 的函数的弧长

Solution 解答

Area of a Surface of Revolution 旋转面的面积

Surface Area of a Surface of Revolution 旋转面的表面积

Calculating the Surface Area of a Surface of Revolution 1 计算旋转面的面积 1

Solution 解答

Calculating the Surface Area of a Surface of Revolution 2 计算旋转面的面积 2

Solution 解答

Section 2.4 Exercises 2.4 节习题

2.5 Physical Applications 2.5 物理应用

Mass and Density 质量与密度

Mass–Density Formula of a One-Dimensional Object 一维物体的质—密度公式

Calculating Mass from Linear Density 由线密度计算质量

Solution 解答

Mass–Density Formula of a Circular Object 圆形物体的质—密度公式

Calculating Mass from Radial Density 由径向密度计算质量

Solution 解答

Regions Defined with Respect to y 关于 y 定义的区域

Finding the Area between Two Curves, Integrating along the y-axis 求两曲线之间的面积（沿 y 轴积分）

Integrating with Respect to y 关于 y 积分

The Method of Cylindrical Shells for a Solid Revolved around the x-axis 绕 x 轴旋转的立体的圆柱壳法

Arc Length of the Curve y = f(x) 曲线 y = f(x) 的弧长

Arc Length for y = f(x) 曲线 y = f(x) 的弧长

Calculating the Arc Length of a Function of x 计算关于 x 的函数的弧长

Using a Computer or Calculator to Determine the Arc Length of a Function of x 使用计算机或计算器确定关于 x 的函数的弧长

Arc Length of the Curve x = g(y) 曲线 x = g(y) 的弧长

Arc Length for x = g(y) 曲线 x = g(y) 的弧长

Calculating the Arc Length of a Function of y 计算关于 y 的函数的弧长

Center of Mass of a Thin Plate in the xy-Plane xy 平面上薄板的质心

Integral of (1/u) du （1/u）du 的积分

Defining the Number e 定义数 e