6 Applications of Integration 积分的应用

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，我们拓展这一思想，以计算更复杂区域的面积。我们先从求两条关于 $x$ 的函数的曲线之间的面积开始，先考虑其中一条函数值始终大于另一条的简单情形。接着考察两条函数图形相交的情形。最后，我们考虑如何求两条关于 $y$ 的函数的曲线之间的面积。

设 $f(x)$ 与 $g(x)$ 为区间 $\left\lbrack {a,b} \right\rbrack$ 上的连续函数，且满足 $f(x) \geq g(x)$（于 $\left\lbrack {a,b} \right\rbrack$ 上）。我们想求这两函数图形之间的面积，如下图所示。

与之前一样，我们将对 $x\text{-axis}$ 上的区间作分割，并用矩形逼近两函数图形之间的面积。于是，对 $i = 0,1,2\text{,…},n,$ 设 $P = \left\{ x_{i} \right\}$ 为 $\left\lbrack {a,b} \right\rbrack$ 的一个正则分割。接着，对 $i = 1,2\text{,…},n,$ 选取一点 $x_{i}^{*} \in \left\lbrack {x_{i - 1},x_{i}} \right\rbrack,$ 并在每个区间 $\left\lbrack {x_{i - 1},x_{i}} \right\rbrack$ 上构造一个从 $g(x_{i}^{*})$ 竖直延伸到 $f(x_{i}^{*})$ 的矩形。图 6.3(a) 显示了当 $x_{i}^{*}$ 取为区间左端点且 $n = 10$ 时的矩形。图 6.3(b) 详细显示了一个代表矩形。

使用此计算器了解更多关于两条曲线之间面积的内容。

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

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

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

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

这些结果总结于下列定理中。

设 $f(x)$ 与 $g(x)$ 为连续函数，在 $\left\lbrack {a,b} \right\rbrack$ 上满足 $f(x) \geq g(x).$ 记 $R$ 为上方由 $f(x)$ 的图形、下方由 $g(x)$ 的图形、左右分别由直线 $x = a$ 与 $x = b$ 所围成的区域。则 $R$ 的面积由下式给出

$$A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx.$$ (6.1)

我们在下列示例中应用此定理。

若 $R$ 为在区间 $\left\lbrack {1,4} \right\rbrack$ 上、上方由函数 $f(x) = x + 4$ 的图形、下方由函数 $g(x) = 3 - \frac{x}{2}$ 的图形所围成的区域，求区域 $R$ 的面积。

该区域如下图所示。

我们有

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

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

若 $R$ 为在区间 $\left\lbrack {1,5} \right\rbrack$ 上、由函数 $f(x) = \frac{x}{2} + 5$ 与 $g(x) = x + \frac{1}{2}$ 的图形所围成的区域，求区域 $R$ 的面积。

在示例 6.1 中，我们将所关注的区间作为问题陈述的一部分给出。但通常，我们希望根据两函数图形的交点来界定所关注的区间。下面的示例说明了这一点。

若 $R$ 为上方由函数 $f(x) = 9 - \left( {x\text{/}2} \right)^{2}$ 的图形、下方由函数 $g(x) = 6 - x$ 的图形所围成的区域，求区域 $R$ 的面积。

该区域如下图所示。

我们首先要计算两函数图形的交点。令 $f(x) = g(x)$，得到

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

两函数的图形在 $x = 6$ 或 $x = -2$ 处相交，因此我们要从 $-2$ 积到 $6.$ 由于在 $-2 \leq x \leq 6$ 上 $f(x) \geq g(x)$，我们得到

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

该区域的面积为 $64\text{/}3$ units².

若 $R$ 为上方由函数 $f(x) = x$ 的图形、下方由函数 $g(x) = x^{4}$ 的图形所围成的区域，求区域 $R$ 的面积。

迄今我们一直要求 $f(x) \geq g(x)$ 在所关注的整个区间上成立，但若我们要考察由互相交叉的函数图形所围成的区域，又当如何？此时，我们通过引入绝对值函数来修正刚才建立的步骤。

设 $f(x)$ 与 $g(x)$ 为区间 $\left\lbrack {a,b} \right\rbrack$ 上的连续函数。记 $R$ 为 $f(x)$ 与 $g(x)$ 的图形之间、左右分别由直线 $x = a$ 与 $x = b$ 所界的区域。则 $R$ 的面积由下式给出

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

在实际中，应用此定理要求我们将区间 $\left\lbrack {a,b} \right\rbrack$ 分段，并视在给定部分上哪个函数值更大而计算若干个积分。我们在下面的示例中研究这一过程。

若 $R$ 为在区间 $\left\lbrack {0,\pi} \right\rbrack$ 上、由函数 $f(x) = \text{sin}\ x$ 与 $g(x) = \text{cos}\ x$ 的图形所围成的区域，求区域 $R$ 的面积。

该区域如下图所示。

两函数的图形在 $x = {\pi\text{/}4}$ 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

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

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

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

于是

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

该区域的面积为 $2\sqrt{2}$ units².

若 $R$ 为在区间 $\left\lbrack {{\pi\text{/}2},2\pi} \right\rbrack$ 上、由函数 $f(x) = \text{sin}\ x$ 与 $g(x) = \text{cos}\ x$ 的图形所围成的区域，求区域 $R$ 的面积。

考虑图 6.7 所示的区域。求 $R$ 的面积。

与示例 6.3 类似，我们需要将区间分成两段。两函数的图形在 $x = 1$ 处相交（令 $f(x) = g(x)$ 并解 *x*），因此我们分别计算两个积分：一个在 $\left\lbrack {0,1} \right\rbrack$ 上，另一个在 $\left\lbrack {1,2} \right\rbrack$ 上。

在区间 $\left\lbrack {0,1} \right\rbrack$ 上，该区域上方由 $f(x) = x^{2}$ 界定、下方由 *x* 轴界定，因此我们有

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

在区间 $\left\lbrack {1,2} \right\rbrack$ 上，该区域上方由 $g(x) = 2 - x$ 界定、下方由 $x\text{-axis}$ 界定，因此我们有

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

将这些面积相加，我们得到

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

该区域的面积为 $5\text{/}6$ units².

考虑下图所示的区域。求 $R$ 的面积。

在示例 6.4 中，我们需要计算两个独立的积分才能求出该区域的面积。然而，还有另一种只需一个积分的方法。如果我们把曲线当作 $y$ 的函数，而非 $x$ 的函数，会怎样？回顾图 6.7。注意左侧（红色）的图形由函数 $y = f(x) = x^{2}$ 表示。我们同样可以对其解 $x$，将该曲线表示为函数 $x = v(y) = \sqrt{y}.$（注意 $x = \text{−}\sqrt{y}$ 也是函数 $y = f(x) = x^{2}$ 作为 $y$ 的函数的一种有效表示。不过，根据图形可以清楚看出，我们感兴趣的是正的平方根。）类似地，右侧图形由函数 $y = g(x) = 2 - x$ 表示，但同样可以很容易地由函数 $x = u(y) = 2 - y$ 表示。当图形表示为 $y$ 的函数时，我们看到该区域左侧由其中一个函数的图形所界、右侧由另一个函数的图形所界。因此，若我们关于 $y$ 积分，只需计算一个积分即可。我们来为这类积分建立一个公式。

设 $u(y)$ 与 $v(y)$ 为区间 $\left\lbrack {c,d} \right\rbrack$ 上的连续函数，且对一切 $y \in \left\lbrack {c,d} \right\rbrack$ 满足 $u(y) \geq v(y).$ 我们想求这两函数图形之间的面积，如下图所示。

这一次，我们将对 $y\text{-axis}$ 上的区间作分割，并用水平矩形逼近函数之间的面积。于是，对 $i = 0,1,2\text{,…},n,$ 设 $Q = \left\{ y_{i} \right\}$ 为 $\left\lbrack {c,d} \right\rbrack$ 的一个正则分割。接着，对 $i = 1,2\text{,…},n,$ 选取一点 $y_{i}^{*} \in \left\lbrack {y_{i - 1},y_{i}} \right\rbrack,$ 然后在每个区间 $\left\lbrack {y_{i - 1},y_{i}} \right\rbrack$ 上构造一个从 $v\left( y_{i}^{*} \right)$ 水平延伸到 $u\left( y_{i}^{*} \right)$ 的矩形。图 6.9(a) 显示了当 $y_{i}^{*}$ 取为区间下端点且 $n = 10$ 时的矩形。图 6.9(b) 详细显示了一个代表矩形。

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，曲线之间的面积近似为

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

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

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

这些结果总结于下列定理中。

设 $u(y)$ 与 $v(y)$ 为连续函数，对一切 $y \in \left\lbrack {c,d} \right\rbrack$ 满足 $u(y) \geq v(y).$ 记 $R$ 为右侧由 $u(y)$ 的图形、左侧由 $v(y)$ 的图形、上下分别由直线 $y = d$ 与 $y = c$ 所围成的区域。则 $R$ 的面积由下式给出

$$A = \int_{c}^{d}\left\lbrack {u(y) - v(y)} \right\rbrack dy.$$ (6.2)

我们重新来看示例 6.4，只是这一次我们关于 $y$ 积分。设 $R$ 为图 6.10 所示的区域。通过对 $y$ 积分求 $R$ 的面积。

我们必须先将两图形表示为 $y$ 的函数。正如本节开头所看到的，左侧曲线可由函数 $x = v(y) = \sqrt{y}$ 表示，右侧曲线可由函数 $x = u(y) = 2 - y$ 表示。

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 $y = 0.$ 积分上限由两图形相交的点决定，该点为 $\left( {1,1} \right)$，因此积分上限为 $y = 1.$ 于是我们有 $\left\lbrack {c,d} \right\rbrack = \left\lbrack {0,1} \right\rbrack.$

计算该区域的面积，我们得到

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

该区域的面积为 $5\text{/}6$ units².

我们重新来看与示例 6.4 相关的检查点，只是这一次我们关于 $y$ 积分。设 $R$ 为下图所示的区域。通过对 $y$ 积分求 $R$ 的面积。

在以下习题中，通过对 $x\text{-axis}\text{.}$ 积分，确定所给图形中两条曲线之间区域的面积。

1.

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

2\.

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

在以下习题中，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。注意你将需要求解两个积分。

3.

$y = x^{3}$ 和 $y = x^{2} + x$

4\.

$y = \text{cos}\ \theta$ 和 $y = 0.5,$ 其中 $0 \leq \theta \leq \pi$

在以下习题中，通过对 $y\text{-axis}.$ 积分确定两条曲线之间区域的面积。

5.

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

6\.

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

在以下习题中，画出方程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。

7.

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

8\.

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

9.

$y = \text{cos}\ x$ 和 $y = \text{cos}^{2}x$ 在 $x = \left\lbrack {\text{−}\pi,\pi} \right\rbrack$ 上

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.

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

在以下习题中，画出方程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\.

$y = \text{sin}\ x$ 和 $y = \text{cos}\ x$ 在 $x = \left\lbrack {\text{−}\pi,\pi} \right\rbrack$ 上

17.

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

18\.

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

19.

$y = x^{3} + 3x$ 和 $y = 4x$

在以下习题中，画出方程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.

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

在以下习题中，画出方程图像并阴影标出曲线之间区域的面积。通过对 *x*-axis 或 *y*-axis 积分来确定其面积，以更方便者为准。

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.

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

在以下习题中，若可能，求由所给方程所围区域的精确面积。若无法用解析方法确定交点，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\.

以 $x\text{-axis}$ 为底、能放入单位圆 $y^{2} + x^{2} = 1$ 上半部分的最大三角形由 $y = 1 + x$ 和 $y = 1 - x$ 给出。见下图。半圆内部但在三角形外部的面积是多少？

49.

一家销售手机的工厂的边际成本函数为 $C(x) = 0.01x^{2} - 3x + 229$，其中 $x$ 表示手机数量，$C$ 是销售 $x$ 部手机的边际成本（单位：美元）；其边际收益函数为 $R(x) = 429 - 2x$，其中 $R$ 是销售 $x$ 部手机所赚得的收益（单位：美元）。求这些曲线图像与 $x = 0$ 之间的面积。这一面积代表什么？

50\.

一家游乐园的边际成本函数为 $C(x) = 1000e^{\text{−}x} + 5$，其中 $x$ 表示售出的门票数量，其边际收益函数为 $R(x) = 60 - 0.1x$。求售出 $550$ 张门票时所产生的总利润。如有必要，使用计算器将交点确定到两位小数。

51.

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/小时计。求从 $t = 0$ 到一小时后乌龟与兔子首次速度相同那一刻之间曲线所围的面积。这代表什么？如有必要，使用计算器将交点确定到三位小数。

52\.

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/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，使用计算器将交点确定到三位小数。

在以下习题中，先关于 $x$ 积分、再关于 $y.$ 积分求曲线之间的面积。两种方法哪一种更简便？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.

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

在以下习题中，用微积分求解，再用几何方法验证你的答案。

56\.

确定如下列图所示、与单位圆在四条边上都相切的正方形的边长方程。求该正方形周长与单位圆之间的面积。是否有不使用微积分的其它解法？

57.

求单位圆周长与由 $y = 2x + 1,y = 1 - 2x$ 和 $y = - \frac{3}{5}$ 所构成的三角形之间的面积，如下列图所示。是否有不使用微积分的解法？

在前一节中，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立体的体积。根据立体的不同特征，我们考虑三种求体积的方法——切片法、圆盘法与垫圈法。

正如面积是二维区域的数值度量，体积也是三维立体的数值度量。我们大多数人都曾用基本几何公式计算过立体的体积。例如，长方体的体积可通过长、宽、高相乘得到：$V = lwh.$ 球体 $\left( {V = \frac{4}{3}\pi r^{3}} \right)$、圆锥 $\left( {V = \frac{1}{3}\pi r^{2}h} \right)$ 以及棱锥 $\left( {V = \frac{1}{3}Ah} \right)$ 的体积公式也都已介绍过。尽管其中一些公式仅用几何方法即可推导，但这些公式全部都可以通过积分得到。

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联想到圆形底面，比如罐头或金属棒，但在数学中，*cylinder*（柱体）一词含义更为宽泛。要在这种更一般的语境下讨论柱体，我们首先需要定义一些术语。

我们定义立体的横截面为平面与该立体的交集。*cylinder*（柱体）定义为：通过将一个平面区域沿一条垂直于该区域的直线（称为柱体的 *axis*（轴））平移而生成的任何立体。因此，垂直于柱体轴的所有横截面都完全相同。图 6.11 所示的立体是一个具有非圆形底面的柱体例子。要计算柱体的体积，我们只需将横截面积乘以柱体的高：$V = A \cdot h.$ 对于正圆柱（如罐头），这变为 $V = \pi r^{2}h.$

如果一个立体没有恒定的横截面（且它也不是其它基本立体之一），我们可能就没有现成公式求它的体积。此时，我们可以用定积分来计算该立体的体积。做法是将立体切成若干片，估计每一片的体积，再将这些估计的体积相加。各切片应彼此平行，当我们把所有切片拼在一起时，应得到整个立体。例如，考虑图 6.12 所示的、沿 $x\text{-axis}\text{.}$ 延伸的立体 *S*。

我们想把 $S$ 切成垂直于 $x\text{-axis}\text{.}$ 的薄片。正如我们在本章后面会看到的，有时我们可能想沿其它方向切片——例如，切成垂直于 *y*-轴 的薄片。选择沿哪个方向切片非常重要。如果选错方向，计算会变得相当繁琐。在本章后面，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 $x\text{-axis}\text{.}$ 的切片。

由于横截面积不是恒定的，我们令 $A(x)$ 表示点 $x$ 处横截面的面积。现在令 $P = \left\{ {x_{0},x_{1}\text{…},X_{n}} \right\}$ 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 $x_{i - 1}\text{to}\ x_{i}$ 延伸出的那一片。下图展示了 $n = 3$ 时的切片立体。

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。那么切片 $S_{i}$ 的体积可由 $V\left( S_{i} \right) \approx A\left( x_{i}^{*} \right)\text{Δ}x$ 估计。把这些近似值相加，我们看到整个立体 $S$ 的体积可以近似为

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

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。于是我们得到

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

我们刚刚描述的方法称为切片法。应用它时，我们使用以下策略。

1. 考察该立体并确定其横截面的形状。如果没有提供图形，画一张往往很有帮助。

2. 确定横截面面积的计算公式。

3. 在适当的区间上对面积公式积分，得到体积。

回想一下，在本节中我们假设切片垂直于 $x\text{-axis}\text{.}$。因此，面积公式是关于 *x* 的，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选择沿哪个方向切片都有效。

由几何学我们知道，棱锥的体积公式为 $V = \frac{1}{3}Ah$。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。

我们想对底面为正方形的棱锥应用切片法。为建立积分，考虑图 6.14 所示的、沿 $x\text{-axis}\text{.}$ 放置的棱锥。

我们首先要确定棱锥横截面的形状。我们知道底面是正方形，因此横截面也是正方形（步骤 1）。现在我们要找出其中一个横截面正方形的面积公式。观察图 6.14(b)，利用比例关系（因为它们是相似三角形），我们有

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

因此，其中一个横截面正方形的面积是

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

然后我们从 $0\ \text{to}\ h$ 积分来求棱锥的体积（步骤 $3)\text{:}$

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

这正是我们所要寻找的公式。

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

若一个平面内的区域绕该平面内的一条直线旋转，所得到的立体称为旋转体，如下图所示。

旋转体在机械应用中很常见，例如车床加工出的机器零件。本节余下部分将研究这类立体。下一个示例用切片法计算一个旋转体的体积。

使用一个在线积分计算器来进一步了解。

用切片法求由 $f(x) = x^{2} - 4x + 5,x = 1,\ \text{and}\ x = 4,$ 所围成、并绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

运用解题策略，我们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。

接着，将该区域绕 *x* 轴旋转，如下图所示。

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

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

于是体积（步骤 3）为

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

体积为 $78\pi\text{/}5$。

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、并绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体的体积。参见下图。

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，用来逼近其体积的薄片是圆盘。为看清这一点，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、并绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体。函数的图像与一个代表性圆盘分别见图 6.18(a) 和 (b)。旋转区域及所得立体见图 6.18(c) 和 (d)。

我们在推导切片法时已经使用了体积公式的黎曼和严格推导。我们知道

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

圆盘法唯一的区别在于我们提前知道横截面积的公式，即圆的面积。由此得到下列法则。

设 $f(x)$ 连续且非负。定义区域 $R$ 的上边界为 $f(x)$ 的图像，下边界为 $x\text{-axis,}$ 左边界为直线 $x = a,$ 右边界为直线 $x = b.$ 则 $R$ 绕 $x\text{-axis}$ 旋转而成的旋转体体积由下式给出

$$V = {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}.$$ (6.3)

我们一直在研究的立体（图 6.18）的体积由下式给出

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

我们来看几个示例。

用圆盘法求由函数 $f(x) = \sqrt{x}$ 的图像与 $x\text{-axis}$ 在区间 $\left\lbrack {1,4} \right\rbrack$ 上所围成、并绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

函数图像与旋转体如下图所示。

我们有

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

体积为 $\left( {15\pi} \right)\text{/}2$ units³。

用圆盘法求由函数 $f(x) = \sqrt{4 - x}$ 的图像与 $x\text{-axis}$ 在区间 $\left\lbrack {0,\ 4} \right\rbrack$ 上所围成、并绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

到目前为止，我们的示例都涉及绕 $x\text{-axis,}$ 旋转的区域，但我们也可以通过将一个平面区域绕任意水平或竖直直线旋转来生成旋转体。在下一个示例中，我们来看一个由区域绕 $y\text{-axis}\text{.}$ 旋转而生成的旋转体。圆盘法的操作过程与以 $x\text{-axis}$ 为旋转轴时几乎相同，但我们要把函数表示为 $y$ 的函数，并且也要关于 *y* 积分。这一点在下列法则中作了总结。

设 $g(y)$ 连续且非负。定义区域 $Q$ 的右边界为 $g(y)$ 的图像，左边界为 $y\text{-axis,}$ 下边界为直线 $y = c,$ 上边界为直线 $y = d.$ 则 $Q$ 绕 $y\text{-axis}$ 旋转而成的旋转体体积由下式给出

$$V = {\int_{c}^{d}{\pi\left\lbrack {g(y)} \right\rbrack^{2}dy}}.$$ (6.4)

下一个示例说明这一法则在实际中如何运用。

设 $R$ 为由函数 $g(y) = \sqrt{4 - y}$ 的图像与 $y\text{-axis}$ 在 $y\text{-axis}$ 上的区间 $\left\lbrack {0,4} \right\rbrack$ 内所围成的区域。用圆盘法求 $R$ 绕 $y\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

图 6.20 显示了该函数以及一个可用来估计体积的代表性圆盘。注意，由于我们是将函数绕 $y\text{-axis,}$ 旋转，这些圆盘是水平的，而非竖直的。

待旋转的区域与完整的旋转体如下图所示。

为求体积，我们关于 $y$ 积分。得到

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

体积为 $8\pi$ units³。

用圆盘法求由函数 $g(y) = y$ 的图像与 $y\text{-axis}$ 在区间 $\left\lbrack {1,4} \right\rbrack$ 上所围成、并绕 $y\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

有些旋转体中间有空腔，它们并非一直延伸到旋转轴都是实心的。有时，这仅仅是旋转区域相对于旋转轴的形状所导致的结果。在另一些情况下，当旋转区域被定义为两个函数图像之间的区域时，就会出现空腔。第三种情况是选择了 $x\text{-axis}$ 或 $y\text{-axis}$ 之外的旋转轴。

当旋转体中间有空腔时，用来逼近体积的薄片不是圆盘，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、上边界为函数 $f(x) = \sqrt{x}$ 的图像、下边界为函数 $g(x) = 1$ 的图像的区域。当该区域绕 $x\text{-axis,}$ 旋转时，得到的是一个中间有空腔的立体，其薄片为垫圈。函数图像与一个代表性垫圈分别见图 6.22(a) 和 (b)。旋转区域及所得立体见图 6.22(c) 和 (d)。

因此横截面积为外圆面积减去内圆面积。在本例中，

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

于是该立体的体积为

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

将这一过程推广，便得到垫圈法。

设 $f(x)$ 与 $g(x)$ 为连续的非负函数，且在区间 $\left\lbrack {a,b} \right\rbrack$ 上满足 $f(x) \geq g(x)$。令 $R$ 表示上边界为 $f(x)$ 的图像、下边界为 $g(x)$ 的图像、左边界为直线 $x = a,$ 右边界为直线 $x = b$ 的区域。则 $R$ 绕 $x\text{-axis}$ 旋转而成的旋转体体积由下式给出

$$V = {\int_{a}^{b}{\pi\left\lbrack {\left( {f(x)} \right)^{2} - \left( {g(x)} \right)^{2}} \right\rbrack}}dx.$$ (6.5)

求由区间 $\left\lbrack {1,4} \right\rbrack$ 上、上边界为函数 $f(x) = x$ 的图像、下边界为函数 $g(x) = 1\text{/}x$ 的图像的区域绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

函数图像与旋转体如下图所示。

我们有

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

求由区间 $\left\lbrack {1,3} \right\rbrack$ 上、由函数 $f(x) = \sqrt{x}$ 与 $g(x) = 1\text{/}x$ 的图像所围成的区域绕 $x\text{-axis}\text{.}$ 旋转而成的旋转体的体积。

与圆盘法一样，我们也可以把垫圈法应用于由区域绕 *y* 轴旋转而得到的旋转体。此时适用下列法则。

设 $u(y)$ 与 $v(y)$ 为连续的非负函数，且对 $y \in \left\lbrack {c,d} \right\rbrack$ 满足 $v(y) \leq u(y)$。令 $Q$ 表示右边界为 $u(y)$ 的图像、左边界为 $v(y)$ 的图像、下边界为直线 $y = c,$ 上边界为直线 $y = d$ 的区域。则 $Q$ 绕 $y\text{-axis}$ 旋转而成的旋转体体积由下式给出

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

我们不再考察以 $y\text{-axis}$ 为旋转轴的垫圈法示例，而是考虑一个旋转轴为两条坐标轴之外的某条直线的例子。同样的一般方法仍然适用，但你可能需要想象如何描述该立体的横截面积。

求由区间 $\left\lbrack {0,4} \right\rbrack$ 上、上边界为 $f(x) = 4 - x$、下边界为 $x\text{-axis}$ 的区域绕直线 $y = -2$ 旋转而成的旋转体的体积。

区域与旋转体的图像如下图所示。

我们不能直接将体积公式用于本问题，因为旋转轴并非坐标轴之一。不过，我们仍然知道横截面积是外圆面积减去内圆面积。观察函数图像可知，外圆的半径为 $f(x) + 2,$ 化简为

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

内圆的半径为 $g(x) = 2.$ 因此，我们有

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

求由区间 $\left\lbrack {0,3} \right\rbrack$ 上、上边界为函数 $f(x) = x + 2$ 的图像、下边界为 $x\text{-axis}$ 的区域绕直线 $y = -1$ 旋转而成的旋转体的体积。

58\.

用切片法推导球体的体积公式。

59\.

用切片法推导圆锥体的体积公式。

60\.

用切片法推导边长为 $a$ 的四面体体积公式。

61\.

用圆盘法推导梯形柱体的体积公式。

62\.

解释你会在何时使用圆盘法而非垫圈法。它们在什么情况下可以互换？

对下列习题，画出典型切片，并用切片法求出给定立体的体积。

63.

一个高为 6 单位、底面为边长 2 单位的正方形的棱锥，如图所示。

64\.

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

65.

一个底面边长为 4 单位的四面体，如图所示。

66\.

一个高为 5 单位的棱锥，其底面为等腰三角形，边长分别为 6 单位和 8 单位，如图所示。

67.

一个半径为 $r$、高为 $h$ 的圆锥，从其顶部去掉一个半径为 $r\text{/}2$、高为 $h\text{/}2$ 的小圆锥，如图所示。所得立体称为*圆台*。

对下列习题，画出立体的轮廓，并用切片法求出体积。

68\.

底面是一个半径为 $a$ 的圆。垂直于底面的切片为正方形。

69.

底面是一个顶点为 $\left( {0,0} \right),\left( {1,0} \right),$ 与 $\left( {0,1} \right)$ 的三角形。垂直于 *x* 轴的切片为半圆。

70\.

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 轴的切片为正方形。

71.

底面是抛物线 $y = 1 - x^{2}$ 下方、且在 $x\text{-axis}$ 上方的区域。垂直于 $y\text{-axis}$ 的切片为正方形。

72\.

底面是由 $y = x^{2}$ 与 $y = 9$ 围成的区域。垂直于 *x* 轴的切片为等腰直角三角形。其中某一切片与底面的交线是该三角形的直角边。

73.

底面是 $y = x$ 与 $y = x^{2}$ 之间的区域。垂直于 *x* 轴的切片为半圆。

对下列习题，画出由曲线围成的区域。然后，用圆盘法求当该区域绕 *x* 轴旋转时的体积。

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.

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

对下列习题，画出由曲线围成的区域。然后，求该区域绕 *y* 轴旋转时的体积。

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.

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

对下列习题，画出由曲线围成的区域。然后，求该区域绕 *x* 轴旋转时的体积。

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.

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

对下列习题，画出由曲线围成的区域。然后，用垫圈法求当该区域绕 *y* 轴旋转时的体积。

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.

酸奶容器可做成圆台的形状。将直线 $y = \frac{1}{m}x$ 绕 *y* 轴旋转，以求出 $y = a\ \text{and}\ y = b.$ 之间的体积。

104\.

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

105.

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

106\.

橄榄球体积的一个更好的近似，由将 $y = \text{sin}\ x$ 从 $x = 0$ 到 $x = \pi$ 绕 *x* 轴旋转所得的立体给出。如图所示，这个橄榄球近似的体积是多少？

107.

将 $y = \text{sin}\ x$ 从 $x = 0$ 到 $x = \pi$ 绕 *y* 轴旋转得到的「Bundt 蛋糕」的体积是多少？

对下列习题，求所描述立体的体积。

108\.

底面是 $y = x$ 与 $y = x^{2}$ 之间的区域。垂直于 *x* 轴的切片为半圆。

109.

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* 轴的切片为半圆。

110\.

在一个高为 $b$、半径为 $b$ 的正圆锥的轴线上，从其底面沿轴向钻一个半径为 $a$ 的孔，如图所示。

111.

求两个半径为 $r$、球心相距 $2h$ 的球体的公共体积，如图所示。

112\.

求高为 $h$、半径为 $r$（其中 $h < r$）的球缺的体积，如图所示。

113.

求一个半径为 $R$ 的球体去掉顶部高为 $h$ 的球缺后的体积，如图所示。

6.3 Volumes of Revolution: Cylindrical Shells 6.3 旋转体体积：圆柱壳法

• 6.3.1 Calculate the volume of a solid of revolution by using the method of cylindrical shells.
• 6.3.2 Compare the different methods for calculating a volume of revolution.

• 6.3.1 使用圆柱壳法计算旋转体的体积。
• 6.3.2 比较计算旋转体体积的各种方法。

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.

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 6.25(a). We then revolve this region around the *y*-axis, as shown in Figure 6.25(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.

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 6.26(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.

To calculate the volume of this shell, consider Figure 6.27.

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

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

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

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

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

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

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 6.28).

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 6.28). 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

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

which is the same formula we had before.

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

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

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

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

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

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

$$V = {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx.$$ (6.6)

Now let’s consider an example.

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 $y\text{-axis}.$

Solution 解答

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

Then the volume of the solid is given by

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

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 $y\text{-axis}.$

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 $y\text{-axis}.$

Solution 解答

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

Then the volume of the solid is given by

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

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 $y\text{-axis}.$

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.

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

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

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.

Solution 解答

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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 $x = -1.$

Solution 解答

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

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

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

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 $x = -2.$

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.

A Region of Revolution Bounded by the Graphs of Two Functions 由两函数图像围成的旋转区域

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 $y\text{-axis}.$

Solution 解答

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

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

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

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 $y\text{-axis}.$

Which Method Should We Use? 我们该选用哪种方法？

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 6.34 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 $y\text{-axis}.$

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

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).

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

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

Solution 解答

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

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

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

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

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

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.

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

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

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

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 $y = x^{2}.$

Section 6.3 Exercises 6.3 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.

114\.

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

115.

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

116\.

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

117.

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

118\.

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

119.

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

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 $y\text{-axis}.$

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 $x\text{-axis}.$

130\.

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

131.

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

132\.

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

133.

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

134\.

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

135.

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

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

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.

140\.

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

141.

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

142\.

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

143.

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

144\.

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

145.

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

146\.

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

147.

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

148\.

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

149.

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

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.

150\.

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

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.

152\.

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

153.

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

154\.

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

155.

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

156\.

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

157.

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

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

158\.

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

159.

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

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 $x\text{-axis}.$

161.

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

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 $x = 4.$

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 $\lbrack 0,a\rbrack.$

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 $f^{-1}(y).)$

6.4 Arc Length of a Curve and Surface Area 6.4 曲线的弧长与表面积

• 6.4.1 Determine the length of a curve, $y = f(x),$ between two points.
• 6.4.2 Determine the length of a curve, $x = g(y),$ between two points.
• 6.4.3 Find the surface area of a solid of revolution.

• 6.4.1 确定曲线 $y = f(x)$ 在两点之间的弧长。
• 6.4.2 确定曲线 $x = g(y)$ 在两点之间的弧长。
• 6.4.3 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.

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.

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.)

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 6.37 depicts this construct for $n = 5.$

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 6.38. Note that some (or all) $\text{Δ}y_{i}$ may be negative.

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

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

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

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

We summarize these findings in the following theorem.

Arc Length for *y* = *f*(*x*) 适用于 *y* = *f*(*x*) 的弧长公式

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

$$\text{Arc Length} = {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx.$$ (6.7)

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.

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.

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

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

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

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

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.

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.

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 $\left\lbrack {1,3} \right\rbrack.$

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

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

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

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

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.

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 6.39 shows a representative line segment.

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 $x = g(y).$

Arc Length for *x* = *g*(*y*) 适用于 *x* = *g*(*y*) 的弧长公式

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

$$\text{Arc Length} = {\int_{c}^{d}\sqrt{1 + \left\lbrack {g^{\prime}(y)} \right\rbrack^{2}}}\ dy.$$ (6.8)

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 $\left\lbrack {1,2} \right\rbrack.$

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

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

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

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

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.

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

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.

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.

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.

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.

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

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

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

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).

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

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

Solving for $s,$ we get

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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

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

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.

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

$$\text{Surface Area} = {\int_{a}^{b}{\left( {2\pi f(x)\sqrt{1 + \left( {f^{\prime}(x)} \right)^{2}}} \right)dx}}.$$ (6.9)

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*-axis is given by

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

Calculating the Surface Area of a Surface of Revolution 1 计算旋转面的表面积 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.

Solution 解答

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

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,

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

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

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

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.

Calculating the Surface Area of a Surface of Revolution 2 计算旋转面的表面积 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*-axis.

Solution 解答

Notice that we are revolving the curve around the *y*-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.

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

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

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

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

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*-axis.

Section 6.4 Exercises 6.4 节习题

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

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.

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.

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 $(3,9).$

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.

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 $x = 1$

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 $x = 0\ \text{to}\ x = \pi$

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.

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 $y = e$

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.

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.

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 $x = \frac{\pi}{2}$

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.

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).

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.

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.

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.

For the following exercises, find the exact arc length for the following problems over the given interval.

212\.

$y = \text{ln}(\text{sin}\ x)$ from $x = {\pi\text{/}4}$ to $x = {\left( {3\pi} \right)\text{/}4}.$ (*Hint*: Recall trigonometric identities.)

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.

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?

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.

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 $x + 2y = 3?$

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.

6.5 Physical Applications 6.5 物理应用

• 6.5.1 Determine the mass of a one-dimensional object from its linear density function.
• 6.5.2 Determine the mass of a two-dimensional circular object from its radial density function.
• 6.5.3 Calculate the work done by a variable force acting along a line.
• 6.5.4 Calculate the work done in pumping a liquid from one height to another.
• 6.5.5 Find the hydrostatic force against a submerged vertical plate.

• 6.5.1 由线密度函数确定一维物体的质量。
• 6.5.2 由径向密度函数确定二维圆形物体的质量。
• 6.5.3 计算变力沿直线做功的大小。
• 6.5.4 计算把液体从某一高度抽到另一高度所做的功。
• 6.5.5 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.

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 6.48). 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.

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 6.49 shows a representative segment of the rod.

The mass $m_{i}$ of the segment of the rod from $x_{i - 1}$ to $x_{i}$ is approximated by

$$m_{i} \approx \rho(x_{i}^{*})\left( {x_{i} - x_{i - 1}} \right) = \rho(x_{i}^{*})\text{Δ}x.$$

Adding the masses of all the segments gives us an approximation for the mass of the entire rod:

$$m = {\sum\limits_{i = 1}^{n}m_{i}} \approx {\sum\limits_{i = 1}^{n}\rho}(x_{i}^{*})\text{Δ}x.$$

This is a Riemann sum. Taking the limit as $n\rightarrow\infty,$ we get an expression for the exact mass of the rod:

$$m = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\rho(x_{i}^{*})\text{Δ}x = {\int_{a}^{b}\rho}(x)dx.$$

We state this result in the following theorem.

Mass–Density Formula of a One-Dimensional Object 一维物体的质量—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

$$m = {\int_{a}^{b}\rho}(x)dx.$$ (6.10)

We apply this theorem in the next example.

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?

Solution 解答

Applying Equation 6.10 directly, we have

$$m = {\int_{a}^{b}\rho}(x)dx = \int_{\pi\text{/}2}^{\pi}\text{sin}\ x\ dx = \left. {\text{−}\text{cos}\ x} \right|_{\pi\text{/}2}^{\pi} = 1.$$

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?

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.

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

$$\begin{array}{cl} A_{i} & {= \pi{(x_{i})}^{2} - \pi{(x_{i - 1})}^{2}} \\ & {= \pi\left\lbrack {x_{i}^{2} - x_{i - 1}^{2}} \right\rbrack} \\ & {= \pi\left( {x_{i} + x_{i - 1}} \right)\left( {x_{i} - x_{i - 1}} \right)} \\ & {= \pi(x_{i} + x_{i - 1})\text{Δ}x.} \end{array}$$

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

$$A_{i} = \pi(x_{i} + x_{i - 1})\text{Δ}x \approx 2\pi x_{i}^{*}\text{Δ}x.$$

Using $\rho(x_{i}^{*})$ to approximate the density of the washer, we approximate the mass of the washer by

$$m_{i} \approx 2\pi x_{i}^{*}\rho(x_{i}^{*})\text{Δ}x.$$

Adding up the masses of the washers, we see the mass $m$ of the entire disk is approximated by

$$m = {\sum\limits_{i = 1}^{n}m_{i}} \approx {\sum\limits_{i = 1}^{n}2}\pi x_{i}^{*}\rho(x_{i}^{*})\text{Δ}x.$$

We again recognize this as a Riemann sum, and take the limit as $n\rightarrow\infty.$ This gives us

$$m = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}2\pi x_{i}^{*}\rho(x_{i}^{*})\text{Δ}x = {\int_{0}^{r}2}\pi x\rho(x)dx.$$

We summarize these findings in the following theorem.

Mass–Density Formula of a Circular Object 圆形物体的质量—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

$$m = {\int_{0}^{r}2}\pi x\rho(x)dx.$$ (6.11)

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.

Solution 解答

Applying the formula, we find

$$\begin{array}{cl} m & {= {\int_{0}^{r}2}\pi x\rho(x)dx} \\ & {= {\int_{0}^{4}2}\pi x\sqrt{x}dx = 2\pi{\int_{0}^{4}x^{3\text{/}2}}dx} \\ & {= 2\pi\left. {\frac{2}{5}x^{5\text{/}2}} \right|_{0}^{4} = \frac{4\pi}{5}\lbrack 32\rbrack = \frac{128\pi}{5}.} \end{array}$$

Let $\rho(x) = 3x + 2$ represent the radial density of a disk. Calculate the mass of a disk of radius 2.

Work Done by a Force 力所做的功

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.

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 $\left( {{\text{kg} \cdot \text{m}^{2}}\text{/}\text{s}^{2}} \right).$

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.

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

$$W_{i} \approx F(x_{i}^{*})\left( {x_{i} - x_{i - 1}} \right) = F(x_{i}^{*})\text{Δ}x.$$

Therefore, the work done over the interval $\left\lbrack {a,b} \right\rbrack$ is approximately

$$W = {\sum\limits_{i = 1}^{n}W_{i}} \approx {\sum\limits_{i = 1}^{n}F}(x_{i}^{*})\text{Δ}x.$$

Taking the limit of this expression as $n\rightarrow\infty$ gives us the exact value for work:

$$W = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}F}(x_{i}^{*})\text{Δ}x = {\int_{a}^{b}F}(x)dx.$$

Thus, we can define work as follows.

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

$$W = {\int_{a}^{b}F}(x)dx.$$ (6.12)

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.

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).

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.

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?

Solution 解答

First find the spring constant, $k.$ When $x = -0.2,$ we know $F(x) = -10,$ so

$$\begin{array}{rll} {F(x)} & = & {kx} \\ {- 10} & = & {k(-0.2)} \\ k & = & 50 \end{array}$$

and $F(x) = 50x.$ Then, to calculate work, we integrate the force function, obtaining

$$W = {\int_{a}^{b}F}(x)dx = {\int_{0}^{0.5}5}0x\ dx = \left. {25x^{2}} \right|_{0}^{0.5} = 6.25.$$

The work done to stretch the spring is $6.25$ 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?

Work Done in Pumping 抽水所做的功

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.

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?

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).

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 6.53 shows a representative layer.

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

$$V = \pi{(4)}^{2}\text{Δ}x = 16\pi\text{Δ}x.$$

Then, the force needed to lift each layer is

$$F = 9800 \cdot 16\pi\text{Δ}x = 156,800\pi\text{Δ}x.$$

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.

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

$$W_{i} \approx 156,800\pi x_{i}^{*}\text{Δ}x.$$

Adding the work for each layer, we see the approximate work to empty the tank is given by

$$W = \sum\limits_{i = 1}^{n}W_{i} \approx \sum\limits_{i = 1}^{n}156,800\pi x_{i}^{*}\text{Δ}x.$$

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

$$\begin{array}{cl} W & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}156,800\pi x_{i}^{*}\text{Δ}x} \\ & {= 156,800\pi{\int_{2}^{10}x}dx} \\ & {= 156,800\pi\left. \left\lbrack \frac{x^{2}}{2} \right\rbrack\ \right|_{2}^{10} = 7,526,400\pi \approx 23,644,883.} \end{array}$$

The work required to empty the tank is approximately 23,650,000 J.

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.

Solving Pumping Problems 求解抽水问题

1. Sketch a picture of the tank and select an appropriate frame of reference.

2. Calculate the volume of a representative layer of water.

3. Multiply the volume by the weight-density of water to get the force.

4. Calculate the distance the layer of water must be lifted.

5. Multiply the force and distance to get an estimate of the work needed to lift the layer of water.

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.

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.

We now apply this problem-solving strategy in an example with a noncylindrical tank.

A Pumping Problem with a Noncylindrical Tank 非圆柱形容器的抽水问题

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?

Solution 解答

The tank is depicted in Figure 6.54. 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).

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).

From properties of similar triangles, we have

$$\begin{array}{cll} \frac{r_{i}}{12 - x_{i}^{*}} & = & {\frac{4}{12} = \frac{1}{3}} \\ {3r_{i}} & = & {12 - x_{i}^{*}} \\ r_{i} & = & \frac{12 - x_{i}^{*}}{3} \\ & = & {4 - \frac{x_{i}^{*}}{3}.} \end{array}$$

Then the volume of the disk is

$$V_{i} = \pi\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 2).}$$

The weight-density of water is $62.4$ lb/ft³, so the force needed to lift each layer is approximately

$$F_{i} \approx 62.4\pi\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 3).}$$

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

$$W_{i} \approx 62.4\pi x_{i}^{*}\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 5).}$$

Summing the work required to lift all the layers, we get an approximate value of the total work:

$$W = \sum\limits_{i = 1}^{n}W_{i} \approx \sum\limits_{i = 1}^{n}62.4\pi x_{i}^{*}\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x\ \text{(step 6).}$$

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

$$\begin{array}{cl} W & {= \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}62.4\pi x_{i}^{*}\left( {4 - \frac{x_{i}^{*}}{3}} \right)^{2}\text{Δ}x} \\ & {= {\int_{0}^{8}{62.4\pi x\left( {4 - \frac{x}{3}} \right)^{2}dx}}} \\ & {= 62.4\pi{\int_{0}^{8}{x\left( {16 - \frac{8x}{3} + \frac{x^{2}}{9}} \right)dx}} = 62.4\pi{\int_{0}^{8}{\left( {16x - \frac{8x^{2}}{3} + \frac{x^{3}}{9}} \right)dx}}} \\ & {= 62.4\pi\left. \left\lbrack {8x^{2} - \frac{8x^{3}}{9} + \frac{x^{4}}{36}} \right\rbrack\ \right|_{0}^{8} = 10,649.6\pi \approx 33,456.7.} \end{array}$$

It takes approximately $33,450$ ft-lb of work to empty the tank to the desired level.

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?

Hydrostatic Force and Pressure 静水压力与静水压强

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*.

Let’s begin with the simple case of a plate of area $A$ submerged horizontally in water at a depth *s* (Figure 6.56). 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 $p = {F\text{/}A} = \rho s.$

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.

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 $x.$

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).

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

$$F_{i} = \rho As = \rho\left\lbrack {w(x_{i}^{*})\text{Δ}x} \right\rbrack s(x_{i}^{*}).$$

Adding the forces, we get an estimate for the force on the plate:

$$F \approx {\sum\limits_{i = 1}^{n}F_{i}} = \sum\limits_{i = 1}^{n}\rho\left\lbrack {w(x_{i}^{*})\text{Δ}x} \right\rbrack s(x_{i}^{*}).$$

This is a Riemann sum, so taking the limit gives us the exact force. We obtain

$$F = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\rho\left\lbrack {w(x_{i}^{*})\text{Δ}x} \right\rbrack s(x_{i}^{*}) = {\int_{a}^{b}\rho}w(x)s(x)dx.$$ (6.13)

Evaluating this integral gives us the force on the plate. We summarize this in the following problem-solving strategy.

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 6.13 accordingly.)

2. Determine the depth and width functions, $s(x)$ and $w(x).$

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.

4. Use the equation to calculate the total force.

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.

Solution 解答

Figure 6.58 shows the trough and a more detailed view of one end.

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 6.13, we obtain

$$\begin{array}{cl} F & {= {\int_{a}^{b}\rho}w(x)s(x)dx} \\ & {= {\int_{0}^{3}{62.4\left( {8 - \frac{8}{3}x} \right)x\ dx}} = 62.4{\int_{0}^{3}{\left( {8x - \frac{8}{3}x^{2}} \right)dx}}} \\ & {= 62.4\left. \left\lbrack {4x^{2} - \frac{8}{9}x^{3}} \right\rbrack\ \right|_{0}^{3} = 748.8.} \end{array}$$

The water exerts a force of 748.8 lb on the end of the trough (step 4).

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.

Chapter Opener: Finding Hydrostatic Force 章首题：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).

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).

1. Find the force on the face of the dam when the reservoir is full.

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?

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).

To find the width function, we again turn to similar triangles as shown in the figure below.

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,

$$w(x) = 1250 - \frac{2}{3}x\ \text{(step 2).}$$

Using a weight-density of $62.4$ lb/ft³ (step 3) and applying Equation 6.13, we get

$$\begin{array}{cl} F & {= {\int_{a}^{b}\rho}w(x)s(x)dx} \\ & {= {\int_{10}^{540}{62.4\left( {1250 - \frac{2}{3}x} \right)\left( {x - 10} \right)dx}} = 62.4{\int_{10}^{540}{- \frac{2}{3}\left\lbrack {x^{2} - 1885x + 18750} \right\rbrack}}dx} \\ & {= -62.4\left( \frac{2}{3} \right)\left. \left\lbrack {\frac{x^{3}}{3} - \frac{1885x^{2}}{2} + 18750x} \right\rbrack\ \right|_{10}^{540} \approx 8,832,245,000\ \text{lb} = 4,416,122.5\ \text{t}\text{.}} \end{array}$$

Note the change from pounds to tons $(2000$ lb = $1$ ton) (step 4).

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

$$\begin{array}{cl} F & {= {\int_{a}^{b}\rho}w(x)s(x)dx} \\ & {= {\int_{135}^{540}{62.4\left( {1250 - \frac{2}{3}x} \right)\left( {x - 135} \right)dx}}} \\ & {= -62.4\left( \frac{2}{3} \right){\int_{135}^{540}{\left( {x - 1875} \right)\left( {x - 135} \right)dx}} = -62.4\left( \frac{2}{3} \right){\int_{135}^{540}{\left( {x^{2} - 2010x + 253125} \right)dx}}} \\ & {= -62.4\left( \frac{2}{3} \right)\left. \left\lbrack {\frac{x^{3}}{3} - 1005x^{2} + 253125x} \right\rbrack\ \right|_{135}^{540} \approx 5,015,230,000\ \text{lb} = \ 2,507,615\ \text{t}\text{.}} \end{array}$$

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?

To learn more about Hoover Dam, see this article published by the History Channel.

Section 6.5 Exercises 6.5 节习题

For the following exercises, find the work done.

218\.

Find the work done when a constant force $F = 12$ lb moves a chair from $x = 0.9$ to $x = 1.1$ ft.

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?

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.)

221.

Find the work done when you push a box along the floor $2$ m, when you apply a constant force of $F = 100\ \text{N}.$

222\.

Compute the work done for a force $F = {12\text{/}x^{2}}$ N from $x = 1$ to $x = 2$ m.

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?

For the following exercises, find the mass of the one-dimensional object.

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

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

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.

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.

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.

For the following exercises, find the mass of the two-dimensional object that is centered at the origin.

229.

An oversized hockey puck of radius $2$ in. with density function $\rho(x) = x^{3} - 2x + 5$

230\.

A frisbee of radius $6$ in. with density function $\rho(x) = e^{\text{−}x}$

231.

A plate of radius $10$ in. with density function $\rho(x) = 1 + \text{cos}\left( {\pi x} \right)$

232\.

A jar lid of radius $3$ in. with density function $\rho(x) = \text{ln}(x + 1)$

233.

A disk of radius $5$ cm with density function $\rho(x) = \sqrt{3x}$

234\.

A $12$-in. spring is stretched to $15$ in. by a force of $75$ lb. What is the spring constant?

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?

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?

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?

238\.

A shock absorber is compressed 1 in. by a weight of 1 t. What is the spring constant?

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?

240\.

Find the work done by winding up a hanging cable of length $100$ ft and weight-density $5$ lb/ft.

241.

For the cable in the preceding exercise, how much work is done to lift the cable $50$ ft?

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?

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?

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?

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 $M = 6\ \times \ 10^{24}\ \text{kg}\text{.})$

246\.

\[T\] For the rocket in the preceding exercise, find the work to lift the rocket from $x = 6400$ to $x = \infty.$

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

1. the surface of the water is at the top of the dam and

2. the surface of the water is halfway down the dam.

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.

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.

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.

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.

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.

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.

6.6 Moments and Centers of Mass 6.6 力矩与质心

• 6.6.1 Find the center of mass of objects distributed along a line.
• 6.6.2 Locate the center of mass of a thin plate.
• 6.6.3 Use symmetry to help locate the centroid of a thin plate.
• 6.6.4 Apply the theorem of Pappus for volume.

• 6.6.1 求沿直线分布物体的质心。
• 6.6.2 确定薄板的质心。
• 6.6.3 利用对称性帮助确定薄板的形心。
• 6.6.4 应用帕普斯体积定理。

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*.

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.

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 6.62(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 6.62(b).

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 $m_{1}d_{1} = m_{2}d_{2}.$

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 6.63). The center of mass, $\overset{–}{x},$ is the point where the fulcrum should be placed to make the system balance.

Thus, we have

$$\begin{matrix} {m_{1}\left| \overset{–}{x} - x_{1} \right|} & = & {m_{2}\left| x_{2} - \overset{–}{x} \right|} \\ {m_{1}\left( \overset{–}{x} - x_{1} \right)} & = & {m_{2}\left( x_{2} - \overset{–}{x} \right)} \\ {m_{1}\overset{–}{x} - m_{1}x_{1}} & = & {m_{2}x_{2} - m_{2}\overset{–}{x}} \\ {\overset{–}{x}\left( m_{1} + m_{2} \right)} & = & {m_{1}x_{1} + m_{2}x_{2}} \\ \overset{–}{x} & = & {\frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}.} \end{matrix}$$

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.

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

$$\overset{–}{x} = \frac{{\sum\limits_{i = 1}^{n}m_{i}}x_{i}}{\sum\limits_{i = 1}^{n}m_{i}}.$$

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

$$M = \sum\limits_{i = 1}^{n}m_{i}x_{i}$$ (6.14)

and the center of mass of the system is given by

$$\overset{–}{x} = \frac{M}{m}.$$ (6.15)

We apply this theorem in the following example.

Finding the Center of Mass of Objects along a Line 求沿直线的物体的质心

Suppose four point masses are placed on a number line as follows:

$$\begin{array}{lccl} {m_{1} = 30\ \text{kg,}\ \text{placed at}\ x_{1} = -2\ \text{m}} & & & {m_{2} = 5\ \text{kg,}\ \text{placed at}\ x_{2} = 3\ \text{m}} \\ {m_{3} = 10\ \text{kg,}\ \text{placed at}\ x_{3} = 6\ \text{m}} & & & {m_{4} = 15\ \text{kg,}\ \text{placed at}\ x_{4} = -3\ \text{m}.} \end{array}$$

Find the moment of the system with respect to the origin and find the center of mass of the system.

Solution 解答

First, we need to calculate the moment of the system:

$$\begin{array}{cl} M & {= {\sum\limits_{i = 1}^{4}m_{i}}x_{i}} \\ & {= -60 + 15 + 60 - 45 = -30.} \end{array}$$

Now, to find the center of mass, we need the total mass of the system:

$$\begin{array}{cl} m & {= {\sum\limits_{i = 1}^{4}m_{i}}} \\ & {= 30 + 5 + 10 + 15 = 60\ \text{kg}\text{.}} \end{array}$$

Then we have

$$\overset{–}{x} = \frac{M}{m} = \frac{-30}{60} = - \frac{1}{2}.$$

The center of mass is located 1/2 m to the left of the origin.

Suppose four point masses are placed on a number line as follows:

$$\begin{array}{lccl} {m_{1} = 12\ \text{kg,}\ \text{placed at}\ x_{1} = -4\ \text{m}} & & & {m_{2} = 12\ \text{kg,}\ \text{placed at}\ x_{2} = 4\ \text{m}} \\ {m_{3} = 30\ \text{kg,}\ \text{placed at}\ x_{3} = 2\ \text{m}} & & & {m_{4} = 6\ \text{kg,}\ \text{placed at}\ x_{4} = -6\ \text{m}.} \end{array}$$

Find the moment of the system with respect to the origin and find the center of mass of the system.

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).

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.

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

$$M_{x} = \sum\limits_{i = 1}^{n}m_{i}y_{i}\quad\text{and}\quad M_{y} = \sum\limits_{i = 1}^{n}m_{i}x_{i}.$$ (6.16)

Also, the coordinates of the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ of the system are

$$\overset{–}{x} = \frac{M_{y}}{m}\quad\text{and}\quad\overset{–}{y} = \frac{M_{x}}{m}.$$ (6.17)

The next example demonstrates how to apply this theorem.

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):

$$\begin{array}{l} {m_{1} = 2\ \text{kg, placed at}\ {\left( {-1,3} \right),}} \\ {m_{2} = 6\ \text{kg, placed at}\ {\left( {1,1} \right),}} \\ {m_{3} = 4\ \text{kg, placed at}\ \left( {2,-2} \right).} \end{array}$$

Find the center of mass of the system.

Solution 解答

First we calculate the total mass of the system:

$$m = {\sum\limits_{i = 1}^{3}m_{i}} = 2 + 6 + 4 = 12\ \text{kg}\text{.}$$

Next we find the moments with respect to the *x*- and *y*-axes:

$$\begin{array}{l} \\ \\ {M_{y} = {\sum\limits_{i = 1}^{3}m_{i}}x_{i} = -2 + 6 + 8 = 12,} \\ {M_{x} = {\sum\limits_{i = 1}^{3}m_{i}}y_{i} = 6 + 6 - 8 = 4.} \end{array}$$

Then we have

$$\overset{–}{x} = \frac{M_{y}}{m} = \frac{12}{12} = 1\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m} = \frac{4}{12} = \frac{1}{3}.$$

The center of mass of the system is $\left( {1,{1\text{/}3}} \right),$ in meters.

Suppose three point masses are placed on a number line as follows (assume coordinates are given in meters):

$$\begin{array}{l} {m_{1} = 5\ \text{kg, placed at}\ {\left( {-2,-3} \right),}} \\ {m_{2} = 3\ \text{kg, placed at}\ {\left( {2,3} \right),}} \\ {m_{3} = 2\ \text{kg, placed at}\ \left( {-3,-2} \right).} \end{array}$$

Find the center of mass of the system.

Center of Mass of Thin Plates 薄片的质心

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.

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.

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.

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*.

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.

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.

If $i = 0,1,2\text{,…},n,$ 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.

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 $\rho f(x_{i}^{*})\text{Δ}x.$

To get the approximate mass of the lamina, we add the masses of all the rectangles to get

$$m \approx {\sum\limits_{i = 1}^{n}\rho}f(x_{i}^{*})\text{Δ}x.$$

This is a Riemann sum. Taking the limit as $n\rightarrow\infty$ gives the exact mass of the lamina:

$$m = \underset{n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{n}\rho}f(x_{i}^{*})\text{Δ}x = \rho\int_{a}^{b}f(x)dx.$$

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

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

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

$$M_{y} = \underset{n\rightarrow\infty}{\text{lim}}\sum\limits_{i = 1}^{n}\rho x_{i}^{*}f(x_{i}^{*})\text{Δ}x = \rho{\int_{a}^{b}{xf(x)dx}}.$$

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.

We summarize these findings in the following theorem.

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:

1. The mass of the lamina is

$$m = \rho\int_{a}^{b}f(x)dx.$$ (6.18)

2. The moments $M_{x}$ and $M_{y}$ of the lamina with respect to the *x*- and *y*-axes, respectively, are

$$M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}.$$ (6.19)

3. The coordinates of the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ are

$$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}.$$ (6.20)

In the next example, we use this theorem to find the center of mass of a lamina.

Finding the Center of Mass of a Lamina 求薄片的质心

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.

Solution 解答

The region is depicted in the following figure.

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 $\rho = 1.$

First, we need to calculate the total mass:

$$\begin{array}{cl} m & {= \rho\int_{a}^{b}f(x)dx = {\int_{0}^{4}\sqrt{x}}\ dx} \\ & {= \left. {\frac{2}{3}x^{3\text{/}2}} \right|_{0}^{4} = \frac{2}{3}\left\lbrack {8 - 0} \right\rbrack = \frac{16}{3}.} \end{array}$$

Next, we compute the moments:

$$\begin{array}{cl} M_{x} & {= \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}} \\ & {= {\int_{0}^{4}{\frac{x}{2}dx}} = \left. {\frac{1}{4}x^{2}} \right|_{0}^{4} = 4} \end{array}$$

and

$$\begin{array}{cl} M_{y} & {= \rho{\int_{a}^{b}{xf(x)dx}}} \\ & {= {\int_{0}^{4}{x\sqrt{x}dx}} = {\int_{0}^{4}x^{3\text{/}2}}dx} \\ & {= \left. {\frac{2}{5}x^{5\text{/}2}} \right|_{0}^{4} = \frac{2}{5}\left\lbrack {32 - 0} \right\rbrack = \frac{64}{5}.} \end{array}$$

Thus, we have

$$\overset{–}{x} = \frac{M_{y}}{m} = \frac{64\text{/}5}{16\text{/}3} = \frac{64}{5} \cdot \frac{3}{16} = \frac{12}{5}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m} = \frac{4}{16\text{/}3} = 4 \cdot \frac{3}{16} = \frac{3}{4}.$$

The centroid of the region is $\left( {{12\text{/}5},{3\text{/}4}} \right).$

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.

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.

Again, we partition the interval $\left\lbrack {a,b} \right\rbrack$ and construct rectangles. A representative rectangle is shown in the following figure.

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.

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.

Center of Mass of a Lamina Bounded by Two Functions 由两函数界定的薄片的质心

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:

1. The mass of the lamina is

$$m = \rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx.$$ (6.21)

2. The moments $M_{x}$ and $M_{y}$ of the lamina with respect to the *x*- and *y*-axes, respectively, are

$$M_{x} = \rho{\int_{a}^{b}{\frac{1}{2}\left( {\left\lbrack {f(x)} \right\rbrack^{2} - \left\lbrack {g(x)} \right\rbrack^{2}} \right)}}dx\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}.$$ (6.22)

3. The coordinates of the center of mass $\left( {\overset{–}{x},\overset{–}{y}} \right)$ are

$$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}.$$ (6.23)

We illustrate this theorem in the following example.

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.

Solution 解答

The region is depicted in the following figure.

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 $\rho = 1.$

First, we need to calculate the total mass:

$$\begin{array}{cl} m & {= \rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx} \\ & {= {\int_{-2}^{1}\left\lbrack {1 - x^{2} - (x - 1)} \right\rbrack}dx = \int_{-2}^{1}(2 - x^{2} - x)dx} \\ & {= \left. \left\lbrack {2x - \frac{1}{3}x^{3} - \frac{1}{2}x^{2}} \right\rbrack\ \right|_{-2}^{1} = \left\lbrack {2 - \frac{1}{3} - \frac{1}{2}} \right\rbrack - \left\lbrack {-4 + \frac{8}{3} - 2} \right\rbrack = \frac{9}{2}.} \end{array}$$

Next, we compute the moments:

$$\begin{array}{cl} M_{x} & {= \rho{\int_{a}^{b}{\frac{1}{2}\left( {\left\lbrack {f(x)} \right\rbrack^{2} - \left\lbrack {g(x)} \right\rbrack^{2}} \right)}}dx} \\ & {= \frac{1}{2}{\int_{-2}^{1}{\left( {\left( {1 - x^{2}} \right)^{2} - \left( {x - 1} \right)^{2}} \right)dx}} = \frac{1}{2}\int_{-2}^{1}\left( {x^{4} - 3x^{2} + 2x} \right)dx} \\ & {= \frac{1}{2}\left. \left\lbrack {\frac{x^{5}}{5} - x^{3} + x^{2}} \right\rbrack\ \right|_{-2}^{1} = - \frac{27}{10}} \end{array}$$

and

$$\begin{matrix} M_{y} & {= \rho\int_{a}^{b}x\left\lbrack {f(x) - g(x)} \right\rbrack dx} \\ & {= \int_{-2}^{1}x\left\lbrack {\left( 1 - x^{2} \right) - (x - 1)} \right\rbrack dx = \int_{-2}^{1}x\left\lbrack 2 - x^{2} - x \right\rbrack dx} \\ {= \int_{-2}^{1}\left( 2x - x^{3} - x^{2} \right)dx} & \\ & {= \left. \left\lbrack x^{2} - \frac{x^{4}}{4} - \frac{x^{3}}{3} \right\rbrack\ \right|_{-2}^{1} = - \frac{9}{4}.} \end{matrix}$$

Therefore, we have

$$\overset{–}{x} = \frac{M_{y}}{m} = - \frac{9}{4} \cdot \frac{2}{9} = - \frac{1}{2}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m} = - \frac{27}{10} \cdot \frac{2}{9} = - \frac{3}{5}.$$

The centroid of the region is $\left( {\text{−}\left( {1\text{/}2} \right),\text{−}\left( {3\text{/}5} \right)} \right).$

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.

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.

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.

Solution 解答

The region is depicted in the following figure.

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 $\rho = 1.$

First, we calculate the total mass:

$$\begin{array}{cl} m & {= \rho\int_{a}^{b}f(x)dx} \\ & {= {\int_{-2}^{2}\left( {4 - x^{2}} \right)}dx} \\ & {= \left. \left\lbrack {4x - \frac{x^{3}}{3}} \right\rbrack\ \right|_{-2}^{2} = \frac{32}{3}.} \end{array}$$

Next, we calculate the moments. We only need $M_{x}\text{:}$

$$\begin{array}{cl} M_{x} & {= \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}} \\ & {= \frac{1}{2}\int_{-2}^{2}\left\lbrack {4 - x^{2}} \right\rbrack^{2}dx = \frac{1}{2}\int_{-2}^{2}\left( {16 - 8x^{2} + x^{4}} \right)dx} \\ & {= \frac{1}{2}\left. \left\lbrack {\frac{x^{5}}{5} - \frac{8x^{3}}{3} + 16x} \right\rbrack\ \right|_{-2}^{2} = \frac{256}{15}.} \end{array}$$

Then we have

$$\overset{–}{y} = \frac{M_{x}}{m} = \frac{256}{15} \cdot \frac{3}{32} = \frac{8}{5}.$$

The centroid of the region is $\left( {0,{8\text{/}5}} \right).$

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.

The Grand Canyon Skywalk 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).

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.

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.

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.

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).

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 $g = 32\ \text{ft/sec}^{2}.$

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.

2. Determine the mass associated with each of the three sub-regions.

3. Calculate the center of mass of each of the three sub-regions.

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.

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?

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.)

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.)

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.*

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

$$\overset{–}{x} = \frac{M_{y}}{m},$$

where

$$m = \rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx\ \text{and}\ M_{y} = \rho{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}.$$

Then,

$$d = 2\pi\frac{\rho{\int_{a}^{b}{x\left\lbrack {f(x) - g(x)} \right\rbrack dx}}}{\rho\int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx}$$

and thus

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

However, using the method of cylindrical shells, we have

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

So,

$$V = d \cdot A$$

and the proof is complete.

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Using the Theorem of Pappus for Volume 应用帕普斯体积定理

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.

Solution 解答

The region and torus are depicted in the following figure.

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.

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.

Section 6.6 Exercises 6.6 节习题

For the following exercises, calculate the center of mass for the collection of masses given.

254\.

$m_{1} = 2$ at $x_{1} = 1$ and $m_{2} = 4$ at $x_{2} = 2$

255.

$m_{1} = 1$ at $x_{1} = -1$ and $m_{2} = 3$ at $x_{2} = 2$

256\.

$m = 3$ at $x = 0,1,2,6$

257.

Unit masses at $(x,y) = (1,0),(0,1),(1,1)$

258\.

$m_{1} = 1$ at $(1,0)$ and $m_{2} = 4$ at $(0,1)$

259.

$m_{1} = 1$ at $(1,0)$ and $m_{2} = 3$ at $(2,2)$

For the following exercises, compute the center of mass $\overset{–}{x}.$

260\.

$\rho = 1$ for $x \in (-1,3)$

261.

$\rho = x^{2}$ for $x \in (0,L)$

262\.

$\rho = 1$ for $x \in (0,1)$ and $\rho = 2$ for $x \in (1,2)$

263.

$\rho = \text{sin}\ x$ for $x \in (0,\pi)$

264\.

$\rho = \text{cos}\ x$ for $x \in \left( {0,\frac{\pi}{2}} \right)$

265.

$\rho = e^{x}$ for $x \in \left( {0,2} \right)$

266\.

$\rho = x^{3} + xe^{\text{−}x}$ for $x \in (0,1)$

267.

$\rho = x\ \text{sin}\ x$ for $x \in (0,\pi)$

268\.

$\rho = \sqrt{x}$ for $x \in \left( {1,4} \right)$

269.

$\rho = \text{ln}\ x$ for $x \in \left( {1,e} \right)$

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.

270\.

$\rho = 7$ in the square $0 \leq x \leq 1,$ $0 \leq y \leq 1$

271.

$\rho = 3$ in the triangle with vertices $(0,0),$ $(a,0),$ and $(0,b)$

272\.

$\rho = 2$ for the region bounded by $y = \text{cos}(x),$ $y = \text{−}\text{cos}(x),$ $x = - \frac{\pi}{2},$ and $x = \frac{\pi}{2}$

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.

273.

\[T\] The region bounded by $y = \text{cos}(2x),$ $x = - \frac{\pi}{4},$ and $x = \frac{\pi}{4}$

274\.

\[T\] The region between $y = 2x^{2},$ $y = 0,$ $x = 0,$ and $x = 1$

275.

\[T\] The region between $y = \frac{5}{4}x^{2}$ and $y = 5$

276\.

\[T\] Region between $y = \sqrt{x},$ $y = \text{ln}(x),$ $x = 1,$ and $x = 4$

277.

\[T\] The region bounded by $y = 0,$ $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

278\.

\[T\] The region bounded by $y = 0,$ $x = 0,$ and $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

279.

\[T\] The region bounded by $y = x^{2}$ and $y = x^{4}$ in the first quadrant

For the following exercises, use the theorem of Pappus to determine the volume of the shape.

280\.

Rotating $y = mx$ around the $x$-axis between $x = 0$ and $x = 1$

281.

Rotating $y = mx$ around the $y$-axis between $x = 0$ and $x = 1$

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?

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?

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?

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.

285.

\[T\] Quarter-circle: $y = \sqrt{1 - x^{2}},$ $y = 0,$ and $x = 0$

286\.

\[T\] Triangle: $y = x,$ $y = 2 - x,$ and $y = 0$

287.

\[T\] Lens: $y = x^{2}$ and $y = x$

288\.

\[T\] Ring: $y^{2} + x^{2} = 1$ and $y^{2} + x^{2} = 4$

289.

\[T\] Half-ring: $y^{2} + x^{2} = 1,$ $y^{2} + x^{2} = 4,$ and $y = 0$

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.

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.

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.

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.

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.)

6.7 Integrals, Exponential Functions, and Logarithms 6.7 积分、指数函数与对数

• 6.7.1 Write the definition of the natural logarithm as an integral.
• 6.7.2 Recognize the derivative of the natural logarithm.
• 6.7.3 Integrate functions involving the natural logarithmic function.
• 6.7.4 Define the number $e$ through an integral.
• 6.7.5 Recognize the derivative and integral of the exponential function.
• 6.7.6 Prove properties of logarithms and exponential functions using integrals.
• 6.7.7 Express general logarithmic and exponential functions in terms of natural logarithms and exponentials.

• 6.7.1 用积分写出自然对数的定义。
• 6.7.2 识别自然对数的导数。
• 6.7.3 对涉及自然对数函数的函数求积分。
• 6.7.4 通过积分定义数 $e$。
• 6.7.5 识别指数函数的导数与积分。
• 6.7.6 用积分证明对数函数与指数函数的性质。
• 6.7.7 用自然对数和自然指数表示一般的对数函数和指数函数。

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.

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).

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.

The Natural Logarithm as an Integral 用积分定义的自然对数

Recall the power rule for integrals:

$${\int{x^{n}dx = \frac{x^{n + 1}}{n + 1}}} + C,\ n \neq \text{−}1.$$

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.

For $x > 0,$ define the natural logarithm function by

$$\text{ln}\ x = {\int_{1}^{x}{\frac{1}{t}dt}}.$$ (6.24)

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).

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 $x > 0.$

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 to the Fundamental Theorem of Calculus.

Derivative of the Natural Logarithm 自然对数的导数

For $x > 0,$ the derivative of the natural logarithm is given by

$$\frac{d}{dx}\text{ln}\ x = \frac{1}{x}.$$

Corollary to the Derivative of the Natural Logarithm 自然对数导数的推论

The function $\text{ln}\ x$ is differentiable; therefore, it is continuous.

A graph of $\text{ln}\ x$ is shown in Figure 6.76. Notice that it is continuous throughout its domain of $\left( {0,\infty} \right).$

Calculating Derivatives of Natural Logarithms 计算自然对数的导数

Calculate the following derivatives:

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.

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:

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.

Integral of (1/*u*) *du* (1/*u*) *du* 的积分

The natural logarithm is the antiderivative of the function $f(u) = 1\text{/}u\text{:}$

$${\int\frac{1}{u}}du = \text{ln}\ |u| + C.$$

Calculating Integrals Involving Natural Logarithms 计算含自然对数的积分

Calculate the integral ${\int\frac{x}{x^{2} + 4}}dx.$

Solution 解答

Using $u$-substitution, let $u = x^{2} + 4.$ Then $du = 2x\ dx$ and we have

$${\int\frac{x}{x^{2} + 4}}dx = \frac{1}{2}{\int{\frac{1}{u}du = \frac{1}{2}\text{ln}\ |u| + C =}}\frac{1}{2}\text{ln}\ \left| {x^{2} + 4} \right| + C = \frac{1}{2}\text{ln}\left( {x^{2} + 4} \right) + C.$$

Calculate the integral ${\int\frac{x^{2}}{x^{3} + 6}}dx.$

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.

If $a,b > 0$ and $r$ is a rational number, then

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, $\text{ln}\ 1 = {\int_{1}^{1}\frac{1}{t}}dt = 0.$

ii\. We have

$$\text{ln}\left( {ab} \right) = {\int_{1}^{ab}\frac{1}{t}}dt = {\int_{1}^{a}\frac{1}{t}}dt + {\int_{a}^{ab}\frac{1}{t}}dt.$$

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

$$\text{ln}\left( {ab} \right) = {\int_{1}^{a}\frac{1}{t}}dt + {\int_{a}^{ab}\frac{1}{t}}dt = {\int_{1}^{a}\frac{1}{t}}dt + {\int_{a}^{ab}{\frac{a}{t} \cdot \frac{1}{a}}}dt = {\int_{1}^{a}{\frac{1}{t}dt +}}{\int_{1}^{b}{\frac{1}{u}du = \text{ln}\ a + \text{ln}\ b.}}$$

iv\. Note that

$$\frac{d}{dx}\text{ln}\left( x^{r} \right) = \frac{rx^{r - 1}}{x^{r}} = \frac{r}{x}.$$

Furthermore,

$$\frac{d}{dx}\left( {r\ \text{ln}\ x} \right) = \frac{r}{x}.$$

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

$$\text{ln}\left( x^{r} \right) = r\ \text{ln}\ x + C$$

for some constant $C.$ Taking $x = 1,$ we get

$$\begin{array}{rll} {\text{ln}\left( 1^{r} \right)} & = & {r\ \text{ln}(1) + C} \\ 0 & = & {r(0) + C} \\ C & = & {0.} \end{array}$$

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.

Part iii. follows from parts ii. and iv. and the proof is left to you.

□

Using Properties of Logarithms 利用对数的性质

Use properties of logarithms to simplify the following expression into a single logarithm:

$$\text{ln}\ 9 - 2\ \text{ln}\ 3 + \text{ln}\left( \frac{1}{3} \right).$$

Solution 解答

We have

$$\text{ln}\ 9 - 2\ \text{ln}\ 3 + \text{ln}\left( \frac{1}{3} \right) = \text{ln}\left( 3^{2} \right) - 2\ \text{ln}\ 3 + \text{ln}\left( 3^{-1} \right) = 2\ \text{ln}\ 3 - 2\ \text{ln}\ 3 - \text{ln}\ 3 = \text{−}\text{ln}\ 3.$$

Use properties of logarithms to simplify the following expression into a single logarithm:

$$\text{ln}\ 8 - \text{ln}\ 2 - \text{ln}\left( \frac{1}{4} \right).$$

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 $e.$

The number $e$ is defined to be the real number such that

$$\text{ln}\ e = 1.$$

To put it another way, the area under the curve $y = 1\text{/}t$ between $t = 1$ and $t = e$ is $1$ (Figure 6.77). 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.)

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

$$e \approx 2.71828182846.$$

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,

$$\text{exp}\left( {\text{ln}\ x} \right) = x\ \text{for}\ x > 0\ \text{and}\ \text{ln}\left( {\text{exp}\ x} \right) = x\ \text{for all}\ x.$$

The following figure shows the graphs of $\text{exp}\ x$ and $\text{ln}\ x.$

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 $\text{ln}\ x.$

For any real number $x,$ define $y = e^{x}$ to be the number for which

$$\text{ln}\ y = \text{ln}\left( e^{x} \right) = x.$$ (6.25)

Then we have $e^{x} = \text{exp}(x)$ for all $x,$ and thus

$$e^{\text{ln}\ x} = x\ \text{for}\ x > 0\ \text{and}\ \text{ln}\left( e^{x} \right) = x$$ (6.26)

for all $x.$

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 $e^{x}.$

If $p$ and $q$ are any real numbers and $r$ is a rational number, then

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

$$\text{ln}\left( {e^{p}e^{q}} \right) = \text{ln}\left( e^{p} \right) + \text{ln}\left( e^{q} \right) = p + q = \text{ln}\left( e^{p + q} \right).$$

Since $\text{ln}\ x$ is one-to-one, then

$$e^{p}e^{q} = e^{p + q}.$$

□

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.

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

$$\begin{array}{rll} {\text{ln}\ y} & = & x \\ {\frac{d}{dx}\text{ln}\ y} & = & {\frac{d}{dx}x} \\ {\frac{1}{y}\mspace{2mu}\frac{dy}{dx}} & = & 1 \\ \frac{dy}{dx} & = & {y.} \end{array}$$

Thus, we see

$$\frac{d}{dx}e^{x} = e^{x}$$

as desired, which leads immediately to the integration formula

$${\int{e^{x}dx}} = e^{x} + C.$$

We apply these formulas in the following examples.

Using Properties of Exponential Functions 使用指数函数的性质

Evaluate the following derivatives:

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.

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:

1. $\frac{d}{dx}\left( \frac{e^{x^{2}}}{e^{5x}} \right)$

2. $\frac{d}{dt}\left( e^{2t} \right)^{3}$

Evaluate the following integral: ${\int{2xe^{\text{−}x^{2}}dx}}.$

Solution 解答

Using $u$-substitution, let $u = \text{−}x^{2}.$ Then $du = -2x\ dx,$ and we have

$${\int{2xe^{\text{−}x^{2}}dx}} = \text{−}{\int{e^{u}du}} = \text{−}e^{u} + C = \text{−}e^{\text{−}x^{2}} + C.$$

Evaluate the following integral: ${\int\frac{4}{e^{3x}}}dx.$

General Logarithmic and Exponential Functions 一般对数函数与指数函数

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.

For any $a > 0,$ and for any real number $x,$ define $y = a^{x}$ as follows:

$$y = a^{x} = e^{x\ \text{ln}\ a}.$$

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.

Let's now apply this definition to calculate a differentiation formula for $a^{x}.$ We have

$$\frac{d}{dx}a^{x} = \frac{d}{dx}e^{x\ \text{ln}\ a} = e^{x\ \text{ln}\ a}\text{ln}\ a = a^{x}\text{ln}\ a.$$

The corresponding integration formula follows immediately.

Derivatives and Integrals Involving General Exponential Functions 含一般指数函数的导数与积分

Let $a > 0.$ Then,

$$\frac{d}{dx}a^{x} = a^{x}\text{ln}\ a$$

and

$${\int{a^{x}dx}} = \frac{1}{\text{ln}\ a}a^{x} + C.$$

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,

$$y = \text{log}_{a}x\ \text{if and only if}\ x = a^{y}.$$

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

$$\begin{array}{rll} {\text{ln}\ x} & = & {\text{ln}\left( a^{y} \right)} \\ {\text{ln}\ x} & = & {y\ \text{ln}\ a} \\ y & = & \frac{\text{ln}\ x}{\text{ln}\ a} \\ {\text{log}_{a}x} & = & {\frac{\text{ln}\ x}{\text{ln}\ a}.} \end{array}$$

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,

$$\begin{array}{cl} \frac{dy}{dx} & {= \frac{d}{dx}\left( {\text{log}_{a}x} \right)} \\ & {= \frac{d}{dx}\left( \frac{\text{ln}\ x}{\text{ln}\ a} \right)} \\ & {= \left( \frac{1}{\text{ln}\ a} \right)\frac{d}{dx}\left( {\text{ln}\ x} \right)} \\ & {= \frac{1}{\text{ln}\ a} \cdot \frac{1}{x}} \\ & {= \frac{1}{x\ \text{ln}\ a}.} \end{array}$$

Derivatives of General Logarithm Functions 一般对数函数的导数

Let $a > 0.$ Then,

$$\frac{d}{dx}\text{log}_{a}x = \frac{1}{x\ \text{ln}\ a}.$$

Calculating Derivatives of General Exponential and Logarithm Functions 计算一般指数函数与对数函数的导数

Evaluate the following derivatives:

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.

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:

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: ${\int\frac{3}{2^{3x}}}dx.$

Solution 解答

Use $u\text{-substitution}$ and let $u = -3x.$ Then $du = -3dx$ and we have

$${\int\frac{3}{2^{3x}}}dx = {\int{3 \cdot 2^{-3x}}}dx = \text{−}{\int{2^{u}du}} = - \frac{1}{\text{ln}\ 2}2^{u} + C = - \frac{1}{\text{ln}\ 2}2^{-3x} + C.$$

Evaluate the following integral: ${\int{x^{2}2^{x^{3}}dx}}.$

Section 6.7 Exercises 6.7 节习题

For the following exercises, find the derivative $\frac{dy}{dx}.$

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.

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.)

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.

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 $\text{ln}\ y.$

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.

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.

335.

Find the area of the region enclosed by $x = 1$ and $y = 5$ above $y = \text{ln}\ x.$

336\.

\[T\] Find the arc length of $\text{ln}\ x$ from $x = 1$ to $x = 2.$

337.

Find the area between $\text{ln}\ x$ and the *x*-axis from $x = 1\ \text{to}\ x = 2.$

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.

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.

If you are unable to find intersection points analytically in the following exercises, use a calculator.

340\.

Find the area of the hyperbolic quarter-circle enclosed by $x = 2\ \text{and}\ y = 2$ above $y = 1\text{/}x.$

341.

\[T\] Find the arc length of $y = 1\text{/}x$ from $x = 1\ \text{to}\ x = 4.$

342\.

Find the area under $y = 1\text{/}x$ and above the *x*-axis from $x = 1\ \text{to}\ x = 4.$

For the following exercises, verify the derivatives and antiderivatives.

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\.

${\int\frac{dx}{x\ \text{ln}(x)\text{ln}\left( {\text{ln}\ x} \right)}} = \text{ln}\left( {\text{ln}\left( {\text{ln}\ x} \right)} \right) + C$

6.8 Exponential Growth and Decay 6.8 指数增长与衰减

• 6.8.1 Use the exponential growth model in applications, including population growth and compound interest.
• 6.8.2 Explain the concept of doubling time.
• 6.8.3 Use the exponential decay model in applications, including radioactive decay and Newton’s law of cooling.
• 6.8.4 Explain the concept of half-life.

• 6.8.1 在各类应用中使用指数增长模型，包括种群增长与复利。
• 6.8.2 解释倍增时间的概念。
• 6.8.3 在各类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。
• 6.8.4 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.

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

$$y^{\prime} = ky_{0}e^{kt} = ky.$$ (6.27)

That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth. Equation 6.27 involves derivatives and is called a *differential equation.* We learn more about differential equations in Introduction to Differential Equations.

Systems that exhibit exponential growth increase according to the mathematical model

$$y = y_{0}e^{kt},$$

where $y_{0}$ represents the initial state of the system and $k > 0$ is a constant, called the *growth constant*.

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 6.79 and Table 6.1 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!

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）	种群大小（细菌数）
$10$	$244$
$20$	$298$
$30$	$364$
$40$	$445$
$50$	$544$
$60$	$664$
$70$	$811$
$80$	$991$
$90$	$1210$
$100$	$1478$
$110$	$1805$
$120$	$2205$

Table 6.1 Exponential Growth of a Bacterial Population

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.

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?

Solution 解答

We have $f(t) = 200e^{0.02t}.$ Then

$$f(300) = 200e^{0.02{(300)}} \approx 80,686.$$

There are $80,686$ bacteria in the population after $5$ hours.

To find when the population reaches $100,000$ bacteria, we solve the equation

$$\begin{array}{rll} 100,000 & = & {200e^{0.02t}} \\ 500 & = & e^{0.02t} \\ {\text{ln}\ 500} & = & {0.02t} \\ t & = & {\frac{\text{ln}\ 500}{0.02} \approx 310.73.} \end{array}$$

The population reaches $100,000$ bacteria after $310.73$ minutes.

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?

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

$$1000\left( {1 + 0.02} \right) = \text{\$}1020.$$

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

$$1000\left( {1 + \frac{0.02}{2}} \right)^{2} = \text{\$}1020.10.$$

Similarly, if the interest is compounded every $4$ months, we have

$$1000\left( {1 + \frac{0.02}{3}} \right)^{3} = \text{\$}1020.13,$$

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

$$1000\underset{n\rightarrow\infty}{\text{lim}}\left( {1 + \frac{0.02}{n}} \right)^{nt}.$$

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:

$$e = \underset{m\rightarrow\infty}{\text{lim}}\left( {1 + \frac{1}{m}} \right)^{m}.$$

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

$$1000\underset{n\rightarrow\infty}{\text{lim}}\left( {1 + \frac{0.02}{n}} \right)^{nt} = 1000\underset{m\rightarrow\infty}{\text{lim}}\left( {1 + \frac{0.02}{0.02m}} \right)^{0.02mt} = 1000\left\lbrack {\underset{m\rightarrow\infty}{\text{lim}}\left( {1 + \frac{1}{m}} \right)^{m}} \right\rbrack^{0.02t}.$$

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

$$\text{Balance} = Pe^{rt}.$$

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?

Solution 解答

We have

$$\begin{array}{rll} 1,000,000 & = & {Pe^{0.05{(40)}}} \\ P & = & 135,335.28. \end{array}$$

She must invest $\text{\$}135,335.28$ at $5\text{\%}$ interest.

If, instead, she is able to earn $6\text{\%},$ then the equation becomes

$$\begin{array}{rll} 1,000,000 & = & {Pe^{0.06{(40)}}} \\ P & = & 90,717.95. \end{array}$$

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.

Suppose instead of investing at age $25$, the student waits until age $35.$ How much would she have to invest at $5\text{\%}?$ At $6\text{\%}?$

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

$$\begin{array}{rll} {2y_{0}} & = & {y_{0}e^{kt}} \\ 2 & = & e^{kt} \\ {\text{ln}\ 2} & = & {kt} \\ t & = & {\frac{\text{ln}\ 2}{k}.} \end{array}$$

If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by

$$\text{Doubling time}\ = \frac{\text{ln}\ 2}{k}.$$

Using the Doubling Time 使用倍增时间

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?

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:

$$\begin{array}{rll} 10,000 & = & {500e^{(\text{ln}\ 2\text{/}6)t}} \\ 20 & = & e^{(\text{ln}\ 2\text{/}6)t} \\ {\text{ln}\ 20} & = & {\left( \frac{\text{ln}\ 2}{6} \right)t} \\ t & = & {\frac{6\left( {\text{ln}\ 20} \right)}{\text{ln}\ 2} \approx 25.93.} \end{array}$$

The owner’s friends have to wait $25.93$ months (a little more than $2$ years) to fish in the pond.

Suppose it takes $9$ months for the fish population in Example 6.44 to reach $1000$ fish. Under these circumstances, how long do the owner’s friends have to wait?

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 $y = y_{0}e^{\text{−}kt}.$

As with exponential growth, there is a differential equation associated with exponential decay. We have

$$y^{\prime} = \text{−}ky_{0}e^{\text{−}kt} = \text{−}ky.$$

Systems that exhibit exponential decay behave according to the model

$$y = y_{0}e^{\text{−}kt},$$

where $y_{0}$ represents the initial state of the system and $k > 0$ is a constant, called the *decay constant*.

The following figure shows a graph of a representative exponential decay function.

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

$$T^{\prime} = \text{−}k\left( {T - T_{a}} \right).$$

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

$$y^{\prime} = \text{−}ky.$$

From our previous work, we know this relationship between *y* and its derivative leads to exponential decay. Thus,

$$y = y_{0}e^{\text{−}kt},$$

and we see that

$$\begin{array}{rll} {T - T_{a}} & = & {\left( {T_{0} - T_{a}} \right)e^{\text{−}kt}} \\ T & = & {\left( {T_{0} - T_{a}} \right)e^{\text{−}kt} + T_{a}} \end{array}$$

where $T_{0}$ represents the initial temperature. Let’s apply this formula in the following example.

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.

Solution 解答

We have

$$\begin{array}{rll} T & = & {\left( {T_{0} - T_{a}} \right)e^{\text{−}kt} + T_{a}} \\ 180 & = & {(200 - 70)e^{\text{−}k{(2)}} + 70} \\ 110 & = & {130e^{-2k}} \\ \frac{11}{13} & = & e^{-2k} \\ {\text{ln}\ \frac{11}{13}} & = & {-2k} \\ {\text{ln}\ 11 - \text{ln}\ 13} & = & {-2k} \\ k & = & {\frac{\text{ln}\ 13 - \text{ln}\ 11}{2}.} \end{array}$$

Then, the model is

$$T = 130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} + 70.$$

The coffee reaches $175\text{°}\text{F}$ when

$$\begin{array}{rll} 175 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} + 70} \\ 105 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t}} \\ \frac{21}{26} & = & e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} \\ {\text{ln}\ \frac{21}{26}} & = & {\frac{\text{ln}\ 11 - \text{ln}\ 13}{2}t} \\ {\text{ln}\ 21 - \text{ln}\ 26} & = & {\frac{\text{ln}\ 11 - \text{ln}\ 13}{2}t} \\ t & = & {\frac{2\left( {\text{ln}\ 21 - \text{ln}\ 26} \right)}{\text{ln}\ 11 - \text{ln}\ 13} \approx 2.56.} \end{array}$$

The coffee can be served about $2.5$ minutes after it is poured. The coffee reaches $155\text{°}\text{F}$ at

$$\begin{array}{rll} 155 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} + 70} \\ 85 & = & {130e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t}} \\ \frac{17}{26} & = & e^{{(\frac{\text{ln}\ 11 - \text{ln}\ 13}{2})}t} \\ {\text{ln}\ 17 - \text{ln}\ 26} & = & {\left( \frac{\text{ln}\ 11 - \text{ln}\ 13}{2} \right)t} \\ t & = & {\frac{2\left( {\text{ln}\ 17 - \text{ln}\ 26} \right)}{\text{ln}\ 11 - \text{ln}\ 13} \approx 5.09.} \end{array}$$

The coffee is too cold to be served about $5$ minutes after it is poured.

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.

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

$$\begin{array}{rll} \frac{y_{0}}{2} & = & {y_{0}e^{\text{−}kt}} \\ \frac{1}{2} & = & e^{\text{−}kt} \\ {- \text{ln}\ 2} & = & {\text{−}kt} \\ t & = & {\frac{\text{ln}\ 2}{k}.} \end{array}$$

*Note*: This is the same expression we came up with for doubling time.

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

$$\text{Half-life} = \frac{\text{ln}\ 2}{k}.$$

Radiocarbon Dating 放射性碳定年法

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.

Solution 解答

We have

$$\begin{array}{rll} 5730 & = & \frac{\text{ln}\ 2}{k} \\ k & = & {\frac{\text{ln}\ 2}{5730}.} \end{array}$$

So, the model says

$$y = 100e^{\text{−}{({\text{ln}\ 2\text{/}5730})}t}.$$

In $50$ years, we have

$$\begin{array}{cll} y & = & {100e^{\text{−}{({\text{ln}\ 2\text{/}5730})}{(50)}}} \\ & \approx & {99.40.} \end{array}$$

Therefore, in $50$ years, $99.40$ g of $\text{carbon-}14$ remains.

To determine the age of the artifact, we must solve

$$\begin{array}{rll} 10 & = & {100e^{\text{−}{({\text{ln}\ 2\text{/}5730})}t}} \\ \frac{1}{10} & = & e^{\text{−}{({\text{ln}\ 2\text{/}5730})}t} \\ t & \approx & 19035. \end{array}$$

The artifact is about $19,000$ years old.

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.

Section 6.8 Exercises 6.8 节习题

*True or False*? If true, prove it. If false, find the true answer.

348\.

The doubling time for $y = e^{ct}$ is $\left( {\text{ln}\ (2)} \right)\text{/}\left( {\text{ln}\ (c)} \right).$

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.

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.

351.

If given a half-life of *t* years, the constant $k$ for $y = e^{kt}$ is calculated by $k = {{\text{ln}\ \left( {1\text{/}2} \right)}\text{/}t}.$

For the following exercises, use $y = y_{0}e^{kt}.$

352\.

If a culture of bacteria doubles in $3$ hours, how many hours does it take to multiply by $10?$

353.

If bacteria increase by a factor of $10$ in $10$ hours, how many hours does it take to increase by $100?$

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.

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.

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.

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.

358\.

Suppose the value of $\text{\$}1$ in Japanese yen decreases at $2\text{\%}$ per year. Starting from $\text{\$}1 = \text{¥}250,$ when will $\text{\$}1 = \text{¥}1?$

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?

360\.

If $y = 1000$ at $t = 3$ and $y = 3000$ at $t = 4,$ what was $y_{0}$ at $t = 0?$

361.

If $y = 100$ at $t = 4$ and $y = 10$ at $t = 8,$ when does $y = 1?$

362\.

If a bank offers annual interest of $7.5\text{\%}$ or continuous interest of $7.25\text{\%},$ which has a better annual yield?

363.

What continuous interest rate has the same yield as an annual rate of $9\text{\%}?$

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?

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?

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?

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 $33\text{°}\text{F}\text{.}$

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?

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?

For the next set of exercises, use the following table, which features the world population by decade.

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 年以来的年数	人口（百万）
$0$	$2,556$
$10$	$3,039$
$20$	$3,706$
$30$	$4,453$
$40$	$5,279$
$50$	$6,083$
$60$	$6,849$

*Source*: 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.

371.

\[T\] Find and graph the derivative $y^{\prime}$ of your equation. Where is it increasing and what is the meaning of this increase?

372\.

\[T\] Find and graph the second derivative of your equation. Where is it increasing and what is the meaning of this increase?

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.

For the next set of exercises, use the following table, which shows the population of San Francisco during the 19th century.

Years since 1850	Population (thousands)
$0$	$21.00$
$10$	$56.80$
$20$	$149.5$
$30$	$234.0$

自 1850 年以来的年数	人口（千）
$0$	$21.00$
$10$	$56.80$
$20$	$149.5$
$30$	$234.0$

*Source*: 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.

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?

376\.

\[T\] Find and graph the second derivative of your equation. Where is it increasing? What is the meaning of this increase?

6.9 Calculus of the Hyperbolic Functions 6.9 双曲函数的微积分

• 6.9.1 Apply the formulas for derivatives and integrals of the hyperbolic functions.
• 6.9.2 Apply the formulas for the derivatives of the inverse hyperbolic functions and their associated integrals.
• 6.9.3 Describe the common applied conditions of a catenary curve.

• 6.9.1 应用双曲函数的导数与积分公式。
• 6.9.2 应用反双曲函数的导数公式及其相关的积分公式。
• 6.9.3 描述悬链线曲线的常见应用情形。

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.

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

$$\text{sinh}\ x = \frac{e^{x} - e^{\text{−}x}}{2}\ \text{and}\ \text{cosh}\ x = \frac{e^{x} + e^{\text{−}x}}{2}.$$

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.

It is easy to develop differentiation formulas for the hyperbolic functions. For example, looking at $\text{sinh}\ x$ we have

$$\begin{array}{cl} {\frac{d}{dx}\left( {\text{sinh}\ x} \right)} & {= \frac{d}{dx}\left( \frac{e^{x} - e^{\text{−}x}}{2} \right)} \\ & {= \frac{1}{2}\left\lbrack {\frac{d}{dx}\left( e^{x} \right) - \frac{d}{dx}\left( e^{\text{−}x} \right)} \right\rbrack} \\ & {= \frac{1}{2}\left\lbrack {e^{x} + e^{\text{−}x}} \right\rbrack = \text{cosh}\ x.} \end{array}$$

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.

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}\ x$	$\text{cosh}\ x$
$\text{cosh}\ x$	$\text{sinh}\ x$
$\text{tanh}\ x$	$\text{sech}^{2}\ x$
$\text{coth}\ x$	$\text{−}\text{csch}^{2}\ x$
$\text{sech}\ x$	$\text{−}\text{sech}\ x\ \text{tanh}\ x$
$\text{csch}\ x$	$\text{−}\text{csch}\ x\ \text{coth}\ x$

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}\ x$	$\text{cosh}\ x$
$\text{cosh}\ x$	$\text{sinh}\ x$
$\text{tanh}\ x$	$\text{sech}^{2}\ x$
$\text{coth}\ x$	$\text{−}\text{csch}^{2}\ x$
$\text{sech}\ x$	$\text{−}\text{sech}\ x\ \text{tanh}\ x$
$\text{csch}\ x$	$\text{−}\text{csch}\ x\ \text{coth}\ x$

Table 6.2 Derivatives of the Hyperbolic Functions

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.

These differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.

$$\begin{array}{rllcccrll} {{\int{\text{sinh}\ u\ d}}u} & = & {\text{cosh}\ u + C} & & & {{\int{\text{csch}^{2}\ u\ d}}u} & = & {\text{−}\text{coth}\ u + C} \\ {{\int{\text{cosh}\ u\ d}}u} & = & {\text{sinh}\ u + C} & & & {{\int{\text{sech}\ u\ \text{tanh}\ u\ d}}u} & = & {\text{−}\text{sech}\ u + C} \\ {{\int{\text{sech}^{2}u\ d}}u} & = & {\text{tanh}\ u + C} & & & {{\int{\text{csch}\ u\ \text{coth}\ u\ d}}u} & = & {\text{−}\text{csch}\ u + C} \end{array}$$

Differentiating Hyperbolic Functions 求双曲函数的导数

Evaluate the following derivatives:

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 6.2 and the chain rule, we get

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:

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 含双曲函数的积分

Evaluate the following integrals:

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.

1. Let $u = x^{2}.$ Then, $du = 2x\ dx$ and

$${\int{x\ \text{cosh}\left( x^{2} \right)d}}x = {\int\frac{1}{2}}\text{cosh}\ u\ du = \frac{1}{2}\text{sinh}\ u + C = \frac{1}{2}\text{sinh}\left( x^{2} \right) + C.$$

2. Let $u = \text{cosh}\ x.$ Then, $du = \text{sinh}\ x\ dx$ and

$${\int{\text{tanh}\ x\ d}}x = {\int{\frac{\text{sinh}\ x}{\text{cosh}\ x}d}}x = {\int{\frac{1}{u}d}}u = \text{ln}|u| + C = \text{ln}\left| {\text{cosh}\ x} \right| + C.$$

Note that $\text{cosh}\ x > 0$ for all $x,$ so we can eliminate the absolute value signs and obtain

$${\int{\text{tanh}\ x\ d}}x = \text{ln}\left( {\text{cosh}\ x} \right) + C.$$

Evaluate the following integrals:

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.

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 6.3 Domains and Ranges of the Inverse Hyperbolic Functions

The graphs of the inverse hyperbolic functions are shown in the following figure.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

$$\begin{array}{rll} y & = & {\text{sinh}^{-1}\ x} \\ {\text{sinh}\ y} & = & x \\ {\frac{d}{dx}\text{sinh}\ y} & = & {\frac{d}{dx}x} \\ {\text{cosh}\ y\frac{dy}{dx}} & = & {1.} \end{array}$$

Recall that $\text{cosh}^{2}y - \text{sinh}^{2}y = 1,$ so $\text{cosh}\ y = \sqrt{1 + \text{sinh}^{2}y}.$ Then,

$$\frac{dy}{dx} = \frac{1}{\text{cosh}\ y} = \frac{1}{\sqrt{1 + \text{sinh}^{2}y}} = \frac{1}{\sqrt{1 + x^{2}}}.$$

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}^{-1}x$	$\frac{1}{\sqrt{1 + x^{2}}}$
$\text{cosh}^{-1}x$	$\frac{1}{\sqrt{x^{2} - 1}}$
$\text{tanh}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{coth}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{sech}^{-1}x$	$\frac{-1}{x\sqrt{1 - x^{2}}}$
$\text{csch}^{-1}x$	$\frac{-1}{|x|\sqrt{1 + x^{2}}}$

$f(x)$	$\frac{d}{dx}f(x)$
$\text{sinh}^{-1}x$	$\frac{1}{\sqrt{1 + x^{2}}}$
$\text{cosh}^{-1}x$	$\frac{1}{\sqrt{x^{2} - 1}}$
$\text{tanh}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{coth}^{-1}x$	$\frac{1}{1 - x^{2}}$
$\text{sech}^{-1}x$	$\frac{-1}{x\sqrt{1 - x^{2}}}$
$\text{csch}^{-1}x$	$\frac{-1}{|x|\sqrt{1 + x^{2}}}$

Table 6.4 Derivatives of the Inverse Hyperbolic Functions

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.

$$\begin{array}{rllccccc} {{\int{\frac{1}{\sqrt{1 + u^{2}}}d}}u} & = & {\text{sinh}^{-1}u + C} & & & {{\int\frac{1}{u\sqrt{1 - u^{2}}}}du} & = & {\text{−}\text{sech}^{-1}|u| + C} \\ {{\int\frac{1}{\sqrt{u^{2} - 1}}}du} & = & {\text{cosh}^{-1}u + C} & & & {{\int\frac{1}{u\sqrt{1 + u^{2}}}}du} & = & {\text{−}\text{csch}^{-1}|u| + C} \\ {{\int\frac{1}{1 - u^{2}}}du} & = & \left\{ \begin{array}{l} {\text{tanh}^{-1}u + C\ \text{if}\ |u| < 1} \\ {\text{coth}^{-1}u + C\ \text{if}\ |u| > 1} \end{array} \right. & & & & & \end{array}$$

Differentiating Inverse Hyperbolic Functions 求反双曲函数的导数

Evaluate the following derivatives:

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 6.4 and the chain rule, we obtain the following results:

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:

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 含反双曲函数的积分

Evaluate the following integrals:

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.

1. Let $u = 2x.$ Then, $du = 2dx$ and we have

$${\int{\frac{1}{\sqrt{4x^{2} - 1}}d}}x = {\int\frac{1}{2\sqrt{u^{2} - 1}}}du = \frac{1}{2}\text{cosh}^{-1}u + C = \frac{1}{2}\text{cosh}^{-1}\left( {2x} \right) + C.$$

2. Let $u = 3x.$ Then, $du = 3dx$ and we obtain

$${\int{\frac{1}{2x\sqrt{1 - 9x^{2}}}dx = \frac{1}{2}{\int{\frac{1}{u\sqrt{1 - u^{2}}}du = - \frac{1}{2}\text{sech}^{-1}|u| + C = - \frac{1}{2}\text{sech}^{-1}\left| {3x} \right| + C}}}}.$$

Evaluate the following integrals:

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.

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 6.84 shows the graph of $y = 2\ \text{cosh}\left( {x\text{/}2} \right).$

Using a Catenary to Find the Length of a Cable 利用悬链线求缆绳的长度

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).

Solution 解答

Recall from Section $2.4$ that the formula for arc length is

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

We have $f(x) = 10\ \text{cosh}\left( {x\text{/}10} \right),$ so $f^{\prime}(x) = \text{sinh}\left( {x\text{/}10} \right).$ Then

$$\begin{array}{cl} \text{Arc Length} & {= {\int_{a}^{b}\sqrt{1 + \left\lbrack {f^{\prime}(x)} \right\rbrack^{2}}}\ dx} \\ & {= {\int_{-15}^{15}\sqrt{1 + \text{sinh}^{2}\left( \frac{x}{10} \right)}}\ dx.} \end{array}$$

Now recall that $1 + \text{sinh}^{2}x = \text{cosh}^{2}x,$ so we have

$$\begin{array}{cl} \text{Arc Length} & {= {\int_{-15}^{15}\sqrt{1 + \text{sinh}^{2}\left( \frac{x}{10} \right)}}\ dx} \\ & {= \int_{-15}^{15}\text{cosh}\left( \frac{x}{10} \right)dx} \\ & {= 10\ \text{sinh}\left. \left( \frac{x}{10} \right) \right|_{-15}^{15} = 10\left\lbrack {\text{sinh}\left( \frac{3}{2} \right) - \text{sinh}\left( {- \frac{3}{2}} \right)} \right\rbrack = 20\ \text{sinh}\left( \frac{3}{2} \right)} \\ & {\approx 42.586\ \text{ft}\text{.}} \end{array}$$

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).

Section 6.9 Exercises 6.9 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.

378\.

From the definitions of $\text{cosh}(x)$ and $\text{sinh}(x),$ find their antiderivatives.

379.

Show that $\text{cosh}(x)$ and $\text{sinh}(x)$ satisfy $y^{''} = y.$

380\.

Use the quotient rule to verify that $\text{tanh}(x)\prime = \text{sech}^{2}(x).$

381.

Derive $\text{cosh}^{2}(x) + \text{sinh}^{2}(x) = \text{cosh}\left( {2x} \right)$ from the definition.

382\.

Take the derivative of the previous expression to find an expression for $\text{sinh}\left( {2x} \right).$

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.

384\.

Take the derivative of the previous expression to find an expression for $\text{cosh}\left( {x + y} \right).$

For the following exercises, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.

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.

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.

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.

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 ${{dv}\text{/}{dt}} = g - v^{2}.$

419.

Show that $v(t) = \sqrt{g}\ \text{tanh}\left( \left( \sqrt{g} \right)t \right)$ satisfies this equation.

420\.

Derive the previous expression for $v(t)$ by integrating $\frac{dv}{g - v^{2}} = dt.$

421.

\[T\] Estimate how far a body has fallen in $12$ seconds by finding the area underneath the curve of $v(t).$

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.

422\.

Show that $S = \text{sinh}(cx)$ satisfies this equation.

423.

Integrate ${{dy}\text{/}{dx}} = \text{sinh}(cx)$ to find the cable height $y(x)$ if $y(0) = {1\text{/}c}.$

424\.

Sketch the cable and determine how far down it sags at $x = 0.$

For the following exercises, solve each problem.

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.

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).

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?

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?

429\.

Prove the formula for the derivative of $y = \text{sinh}^{-1}(x)$ by differentiating $x = \text{sinh}(y).$ (*Hint:* Use hyperbolic trigonometric identities.)

430\.

Prove the formula for the derivative of $y = \text{cosh}^{-1}(x)$ by differentiating $x = \text{cosh}(y).$

(*Hint:* Use hyperbolic trigonometric identities.)

431\.

Prove the formula for the derivative of $y = \text{sech}^{-1}(x)$ by differentiating $x = \text{sech}(y).$ (*Hint:* Use hyperbolic trigonometric identities.)

432\.

Prove that $\left( {\text{cosh}(x) + \text{sinh}(x)} \right)^{n} = \text{cosh}(nx) + \text{sinh}(nx).$

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?

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?

Key Terms 关键术语

arc length

the arc length of a curve can be thought of as the distance a person would travel along the path of the curve

catenary

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

center of mass

the point at which the total mass of the system could be concentrated without changing the moment

centroid

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

cross-section

the intersection of a plane and a solid object

density function

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

disk method

a special case of the slicing method used with solids of revolution when the slices are disks

doubling time

if a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double, and is given by $\left( {\text{ln}\ 2} \right)\text{/}k$

exponential decay

systems that exhibit exponential decay follow a model of the form $y = y_{0}e^{\text{−}kt}$

exponential growth

systems that exhibit exponential growth follow a model of the form $y = y_{0}e^{kt}$

frustum

a portion of a cone; a frustum is constructed by cutting the cone with a plane parallel to the base

half-life

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 $\left( {\text{ln}\ 2} \right)\text{/}k$

Hooke’s law

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

hydrostatic pressure

the pressure exerted by water on a submerged object

lamina

a thin sheet of material; laminas are thin enough that, for mathematical purposes, they can be treated as if they are two-dimensional

method of cylindrical shells

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

moment

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

slicing method

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

solid of revolution

a solid generated by revolving a region in a plane around a line in that plane

surface area

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

symmetry principle

the symmetry principle states that if a region *R* is symmetric about a line *l*, then the centroid of *R* lies on *l*

theorem of Pappus for volume

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

washer method

a special case of the slicing method used with solids of revolution when the slices are washers

work

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

Key Equations 关键公式

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

两条曲线之间的面积（关于 x 积分）	$A = \int_{a}^{b}\left\lbrack {f(x) - g(x)} \right\rbrack dx$
两条曲线之间的面积（关于 y 积分）	$A = \int_{c}^{d}\left\lbrack {u(y) - v(y)} \right\rbrack dy$

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

圆盘法（沿 x 轴）	$V = {\int_{a}^{b}{\pi\left\lbrack {f(x)} \right\rbrack^{2}dx}}$
圆盘法（沿 y 轴）	$V = {\int_{c}^{d}{\pi\left\lbrack {g(y)} \right\rbrack^{2}dy}}$
垫圈法	$V = {\int_{a}^{b}{\pi\left\lbrack {\left( {f(x)} \right)^{2} - \left( {g(x)} \right)^{2}} \right\rbrack}}dx$

Method of Cylindrical Shells

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

圆柱壳法	$V = {\int_{a}^{b}\left( {2\pi xf(x)} \right)}dx$

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

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

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	$F = {\int_{a}^{b}\rho}w(x)s(x)dx$

一维物体的质量	$m = {\int_{a}^{b}\rho}(x)dx$
圆环形物体的质量	$m = {\int_{0}^{r}2}\pi x\rho(x)dx$
对物体所做的功	$W = {\int_{a}^{b}F}(x)dx$
平板所受的静水压力	$F = {\int_{a}^{b}\rho}w(x)s(x)dx$

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	$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}$

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	$\overset{–}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{–}{y} = \frac{M_{x}}{m}$

Natural logarithm function	$\text{ln}\ x = {\int_{1}^{x}\frac{1}{t}}dt$
Exponential function $y = e^{x}$	$\text{ln}\ y = \text{ln}\left( e^{x} \right) = x$

自然对数函数	$\text{ln}\ x = {\int_{1}^{x}\frac{1}{t}}dt$
指数函数 $y = e^{x}$	$\text{ln}\ y = \text{ln}\left( e^{x} \right) = x$

Key Concepts 关键概念

6.1 Areas between Curves 6.1 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的面积一样，它们也可用来求两条曲线之间的面积。
• 要求由函数定义的两条曲线之间的面积，就对这两个函数之差求积分。
• 若函数的图像相交，或区域比较复杂，就使用函数之差的绝对值。在这种情况下，可能需要计算两个或更多个积分并把结果相加，才能求出该区域的面积。
• 有时关于 *y* 积分求面积会更简便。无论用哪个变量作为积分变量，其原理都是相同的。

6.2 Determining Volumes by Slicing 6.2 用切片法求体积

• 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。使用切片法，我们可以通过对横截面积积分求出体积。
• 对于旋转体，体积切片通常是圆盘，其横截面为圆。圆盘法就是在横截面为圆的特殊情形下应用切片法，并使用圆面积公式。
• 若旋转体的中心有空洞，则体积切片为垫圈。用垫圈法时，先由外圆面积减去内圆面积，再进行积分。

6.3 Volumes of Revolution: Cylindrical Shells 6.3 旋转体的体积：圆柱壳法

• 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.

• 圆柱壳法是利用定积分计算旋转体体积的另一种方法。由于这种方法是对另一个变量积分，因此有时比圆盘法或垫圈法更可取。在某些情况下，其中一个积分比另一个复杂得多。
• 函数的几何形状与积分的难易程度，是决定选用哪种积分方法的主要因素。

6.4 Arc Length of a Curve and Surface Area 6.4 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.

• 曲线的弧长可以用定积分来计算。
• 弧长首先用线段来逼近，从而产生一个黎曼和。取极限后便得到定积分公式。同样的过程可以应用于关于 $y.$ 的函数。
• 计算弧长所用的概念可以推广，用来求旋转面的表面积。
• 由弧长公式与表面积公式生成的积分往往难以求值，必要时可以用计算机或计算器来近似这些积分的值。

6.5 Physical Applications 6.5 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.

• 定积分的若干物理应用在工程与物理中十分常见。
• 若已知物体的密度函数，就可以用定积分确定其质量。
• 功也可以通过积分力函数来计算，或在与重力对抗时计算，例如在抽水问题中。
• 定积分还可用来计算浸没在液体中的物体所受到的力。

6.6 Moments and Centers of Mass 6.6 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，系统的质心是指：若将系统的总质量集中于此点，系统的力矩保持不变。粗略地说，可以把质心视为系统的平衡点。
• 对于沿数轴分布的质点，系统关于原点的力矩为 $M = \sum\limits_{i = 1}^{n}m_{i}x_{i}.$ 对于在平面内分布的质点，系统关于 *x* 轴与 *y* 轴的力矩分别为 $M_{x} = \sum\limits_{i = 1}^{n}m_{i}y_{i}$ 和 $M_{y} = \sum\limits_{i = 1}^{n}m_{i}x_{i},$。
• 对于由函数 $f(x),$ 界定其上边界的薄片，系统关于 *x* 轴与 *y* 轴的力矩分别为 $M_{x} = \rho{\int_{a}^{b}{\frac{\left\lbrack {f(x)} \right\rbrack^{2}}{2}dx}}$ 和 $M_{y} = \rho{\int_{a}^{b}{xf(x)dx}}.$。
• 质心的 *x* 坐标与 *y* 坐标，可分别用绕 *y* 轴与绕 *x* 轴的力矩除以总质量来求得。对称原理指出：若一个区域关于某条直线对称，则该区域的形心位于这条直线上。
• 帕普斯体积定理指出：若一个区域绕某条外部轴旋转，所得立体的体积等于该区域的面积乘以区域形心所经过的距离。

6.7 Integrals, Exponential Functions, and Logarithms 6.7 积分、指数函数与对数

• 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.

• 早先对对数与指数函数的处理，并没有精确而正式地定义这些函数。本节将以数学上严格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。
• 然后定义函数 $e^{x}$ 为自然对数的反函数。
• 一般指数函数用 $e^{x},$ 来定义，其对应的反函数即为一般对数函数。
• 对数与指数的那些熟悉性质，在这个更严格的框架下依然成立。

6.8 Exponential Growth and Decay 6.8 指数增长与指数衰减

• 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的两种应用。
• 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，增长率与现有的量成正比。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。

6.9 Calculus of the Hyperbolic Functions 6.9 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.

• 双曲函数用指数函数来定义。
• 逐项求导得出双曲函数的求导公式。这些求导公式反过来又导出积分公式。
• 在适当的定义域限制下，双曲函数都有反函数。
• 隐式求导得出反双曲函数的求导公式，这些公式反过来又导出积分公式。
• 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.

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.

436\.

If the force is constant, the amount of work to move an object from $x = a$ to $x = b$ is $F\left( {b - a} \right).$

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.

438\.

If the half-life of $\text{seaborgium-}266$ is $360$ ms, then $k = {\left( {\text{ln}(2)} \right)\text{/}360}.$

For the following exercises, use the requested method to determine the volume of the solid.

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.

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

441.

$x = y^{2}$ and $x = 3y$ rotated around the *y*-axis using the washer method

442\.

$x = 2y^{2} - y^{3},x = 0,\ \text{and}\ y = 0$ rotated around the *x*-axis using cylindrical shells

For the following exercises, find

1. the area of the region,

2. the volume of the solid when rotated around the *x*-axis, and

3. the volume of the solid when rotated around the *y*-axis. Use whichever method seems most appropriate to you.

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 $y = \sqrt{x}$

447.

$y = 5 + x,$ $y = x^{2},$ $x = 0,$ and $x = 1$

448\.

Below $x^{2} + y^{2} = 1$ and above $y = 1 - x$

449.

Find the mass of $\rho = \frac{1}{x^{2} + 1}$ on a disk centered at the origin with radius $4.$

450\.

Find the center of mass for $\rho = \text{tan}^{2}x$ on $x \in \left( {- \frac{\pi}{4},\frac{\pi}{4}} \right).$

451.

Find the mass and the center of mass of $\rho = 1$ on the region bounded by $y = x^{5}$ and $y = \sqrt{x}.$

For the following exercises, find the requested arc lengths.

452\.

The length of $x$ for $y = \text{cosh}(x)$ from $x = 0\ \text{to}\ x = 2.$

453.

The length of $y$ for $x = 3 - \sqrt{y}$ from $y = 0$ to $y = 4$

For the following exercises, find the surface area and volume when the given curves are revolved around the specified axis.

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.

455.

The loudspeaker created by revolving $y = {1\text{/}x}$ from $x = 1$ to $x = 4$ around the *x*-axis.

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.

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 $98\text{°}\text{F}$ ?

For the following exercise, 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.

Years after 1920	Value (\$)
$1$	$63.90$
$3$	$100$
$5$	$110$
$7$	$160$
$9$	$381.17$

1920 年之后的年数	数值（\$）
$1$	$63.90$
$3$	$100$
$5$	$110$
$7$	$160$
$9$	$381.17$

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 $2014$ ?

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.

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.

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 $x\ \text{-axis.}$

A \| Table of Integrals A 积分表

Basic Integrals 基本积分

1\. ${\int{u^{n}\ du =}}\frac{u^{n + 1}}{n + 1} + C,n \neq \text{−}1$

2\. ${\int{\frac{du}{u} =}}\text{ln}\mspace{2mu}|u| + C$

3\. ${\int{e^{u}\ du}} = e^{u} + C$

4\. ${\int{a^{u}\ du =}}\frac{a^{u}}{\text{ln}\mspace{2mu} a} + C$

5\. $\int{\text{sin}\ u\ du = \text{−cos}\ u + C}$

6\. $\int{\text{cos}\ u\ du = \text{sin}\ u + C}$

7\. $\int{\text{sec}^{2}u\ du = \text{tan}\ u + C}$

8\. $\int{\text{csc}^{2}u\ du = \text{−cot}\ u + C}$

9\. $\int{\text{sec}\ u\ \text{tan}\ u\ du = \text{sec}\ u + C}$

10\. $\int{\text{csc}\ u\ \text{cot}\ u\ du = \text{−csc}\ u + C}$

11\. $\int{\text{tan}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sec}\ u} \right| + C}$

12\. $\int{\text{cot}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sin}\ u} \right| + C}$

13\. $\int{\text{sec}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sec}\ u + \text{tan}\ u} \right| + C}$

14\. $\int{\text{csc}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{csc}\ u - \text{cot}\ u} \right| + C}$

15\. ${\int\frac{\ du}{\sqrt{a^{2} - u^{2}}}} = \text{sin}^{-1}\frac{u}{a} + C$

16\. ${\int\frac{\ du}{a^{2} + u^{2}}} = \frac{1}{a}\text{tan}^{-1}\frac{u}{a} + C$

17\. ${\int\frac{\ du}{u\sqrt{u^{2} - a^{2}}}} = \frac{1}{a}\text{sec}^{-1}\frac{u}{a} + C$

Trigonometric Integrals 三角积分

18\. ${\int{\text{sin}^{2}u\ du = \frac{1}{2}u -}}\frac{1}{4}\text{sin}\ 2u + C$

19\. ${\int{\text{cos}^{2}u\ du = \frac{1}{2}u +}}\frac{1}{4}\text{sin}\ 2u + C$

20\. ${\int{\text{tan}^{2}u\ du = \text{tan}\ u - u}} + C$

21\. ${\int{\text{cot}^{2}u\ du = \text{−}\text{cot}\ u - u}} + C$

22\. ${\int{\text{sin}^{3}u\ du = - \frac{1}{3}\left( {2 + \text{sin}^{2}u} \right)\mspace{2mu}\text{cos}\ u}} + C$

23\. ${\int{\text{cos}^{3}u\ du = \frac{1}{3}\left( {2 + \text{cos}^{2}u} \right)\mspace{2mu}\text{sin}\ u}} + C$

24\. ${\int{\text{tan}^{3}u\ du = \frac{1}{2}\text{tan}^{2}u +}}\text{ln}\mspace{2mu}\left| {\text{cos}\ u} \right| + C$

25\. ${\int{\text{cot}^{3}u\ du = - \frac{1}{2}\text{cot}^{2}u -}}\text{ln}\mspace{2mu}\left| {\text{sin}\ u} \right| + C$

26\. ${\int{\text{sec}^{3}u\ du = \frac{1}{2}\text{sec}\ u\ \text{tan}\ u + \frac{1}{2}}}\text{ln}\mspace{2mu}\left| {\text{sec}\ u + \text{tan}\ u} \right| + C$

27\. ${\int{\text{csc}^{3}u\ du = - \frac{1}{2}\text{csc}\ u\ \text{cot}\ u + \frac{1}{2}}}\text{ln}\mspace{2mu}\left| {\text{csc}\ u - \text{cot}\ u} \right| + C$

28\. ${\int{\text{sin}^{n}u\ du = - \frac{1}{n}\text{sin}^{n - 1}u\ \text{cos}\ u + \frac{n - 1}{n}}}{\int{\text{sin}^{n - 2}u}}\ du$

29\. ${\int{\text{cos}^{n}u\ du = \frac{1}{n}\text{cos}^{n - 1}u\ \text{sin}\ u + \frac{n - 1}{n}}}{\int{\text{cos}^{n - 2}u}}\ du$

30\. ${\int{\text{tan}^{n}u\ du = \frac{1}{n - 1}\text{tan}^{n - 1}u -}}{\int{\text{tan}^{n - 2}u}}\ du$

31\. ${\int{\text{cot}^{n}u\ du = \frac{-1}{n - 1}\text{cot}^{n - 1}u -}}{\int{\text{cot}^{n - 2}u}}\ du$

32\. ${\int{\text{sec}^{n}u\ du = \frac{1}{n - 1}\text{tan}\ u\ \text{sec}^{n - 2}u + \frac{n - 2}{n - 1}}}{\int{\text{sec}^{n - 2}u}}\ du$

33\. ${\int{\text{csc}^{n}u\ du = \frac{-1}{n - 1}\text{cot}\ u\ \text{csc}^{n - 2}u + \frac{n - 2}{n - 1}}}{\int{\text{csc}^{n - 2}u}}\ du$

34\. ${\int{\text{sin}\ au\ \text{sin}\ bu\ du = \frac{\left. \text{sin}\left( \left( {a - b} \right) \right.u \right)}{2\left( {a - b} \right)}}} - \frac{\left. \text{sin}\left( \left( {a + b} \right) \right.u \right)}{2\left( {a + b} \right)} + C$

35\. ${\int{\text{cos}\ au\ \text{cos}\ bu\ du = \frac{\left. \text{sin}\left( \left( {a - b} \right) \right.u \right)}{2\left( {a - b} \right)}}} + \frac{\left. \text{sin}\left( \left( {a + b} \right) \right.u \right)}{2\left( {a + b} \right)} + C$

36\. ${\int{\text{sin}\ au\ \text{cos}\ bu\ du = - \frac{\left. \text{cos}\left( \left( {a - b} \right) \right.u \right)}{2\left( {a - b} \right)}}} - \frac{\left. \text{cos}\left( \left( {a + b} \right) \right.u \right)}{2\left( {a + b} \right)} + C$

37\. $\int{u\ \text{sin}\ u\ du = \text{sin}\ u - u\ \text{cos}\ u + C}$

38\. $\int{u\ \text{cos}\ u\ du = \text{cos}\ u + u\ \text{sin}\ u + C}$

39\. $\int{u^{n}\text{sin}\ u\ du = \text{−}u^{n}\text{cos}\ u + n{\int{u^{n - 1}\text{cos}\ u}}\ du}$

40\. $\int{u^{n}\text{cos}\ u\ du = u^{n}\text{sin}\ u - n{\int{u^{n - 1}\text{sin}\ u}}\ du}$

41\. $\begin{array}{cl} {{\int\text{sin}^{n}}u\ \text{cos}^{m}\ u\ du} & {= - \frac{\text{sin}^{n - 1}\ u\ \text{cos}^{m + 1}\ u}{n + m} + \frac{n - 1}{n + m}{\int{\text{sin}^{n - 2}u\ \text{cos}^{m}u}}\ du} \\ & {= \frac{\text{sin}^{n + 1}u\ \text{cos}^{m - 1}u}{n + m} + \frac{m - 1}{n + m}{\int{\text{sin}^{n}u\ \text{cos}^{m - 2}u}}\ du} \end{array}$

Exponential and Logarithmic Integrals 指数与对数积分

42\. ${\int{ue^{au}\ du =}}\frac{1}{a^{2}}\left( {au - 1} \right)e^{au} + C$

43\. ${\int{u^{n}e^{au}\ du}} = \frac{1}{a}u^{n}e^{au} - \frac{n}{a}{\int{u^{n - 1}e^{au}\ du}}$

44\. ${\int{e^{au}\text{sin}\ bu\ du}} = \frac{e^{au}}{a^{2} + b^{2}}\left( {a\mspace{2mu}\text{sin}\ bu - b\ \text{cos}\ bu} \right) + C$

45\. ${\int{e^{au}\text{cos}\ bu\ du}} = \frac{e^{au}}{a^{2} + b^{2}}\left( {a\ \text{cos}\ bu + b\ \text{sin}\ bu} \right) + C$

46\. $\int{\text{ln}\mspace{2mu} u\ du = u\ \text{ln}\mspace{2mu} u - u + C}$

47\. ${\int u^{n}}\text{ln}\mspace{2mu} u\ du = \frac{u^{n + 1}}{\left( {n + 1} \right)^{2}}\left\lbrack {\left( {n + 1} \right)\text{ln}\mspace{2mu} u - 1} \right\rbrack + C$

48\. ${\int{\frac{1}{u\ \text{ln}\mspace{2mu} u}\ du = \text{ln}\mspace{2mu}\left| {\text{ln}\mspace{2mu} u} \right|}} + C$

Hyperbolic Integrals 双曲积分

49\. $\int{\text{sinh}\ u\ du = \text{cosh}\ u + C}$

50\. $\int{\text{cosh}\ u\ du = \text{sinh}\ u + C}$

51\. $\int{\text{tanh}\ u\ du = \text{ln}\mspace{2mu}\text{cosh}\ u + C}$

52\. $\int{\text{coth}\ u\ du = \text{ln}\mspace{2mu}\left| {\text{sinh}\ u} \right| + C}$

53\. $\int{\text{sech}\ u\ du = \text{tan}^{-1}\left| {\text{sinh}\ u} \right| + C}$

54\. ${\int{\text{csch}\ u\ du =}}\text{ln}\mspace{2mu}\left| {\text{tanh}\mspace{2mu}\frac{1}{2}u} \right| + C$

55\. $\int{\text{sech}^{2}u\ du = \text{tanh}\ u + C}$

56\. $\int{\text{csch}^{2}u\ du = \text{−}\text{coth}\ u + C}$

57\. $\int{\text{sech}\ u\ \text{tanh}\ u\ du = \text{−}\text{sech}\ u + C}$

58\. $\int{\text{csch}\ u\ \text{coth}\ u\ du = \text{−}\text{csch}\ u + C}$

Inverse Trigonometric Integrals 反三角积分

59\. ${\int{\text{sin}^{-1}u\ du =}}u\ \text{sin}^{-1}u + \sqrt{1 - u^{2}} + C$

60\. ${\int{\text{cos}^{-1}u\ du =}}u\ \text{cos}^{-1}u - \sqrt{1 - u^{2}} + C$

61\. ${\int{\text{tan}^{-1}u\ du =}}u\ \text{tan}^{-1}u - \frac{1}{2}\text{ln}\mspace{2mu}\left( {1 + u^{2}} \right) + C$

62\. ${\int{u\ \text{sin}^{-1}u\ du =}}\frac{2u^{2} - 1}{4}\text{sin}^{-1}u + \frac{u\sqrt{1 - u^{2}}}{4} + C$

63\. ${\int{u\ \text{cos}^{-1}u\ du =}}\frac{2u^{2} - 1}{4}\text{cos}^{-1}u - \frac{u\sqrt{1 - u^{2}}}{4} + C$

64\. ${\int{u\ \text{tan}^{-1}u\ du =}}\frac{u^{2} + 1}{2}\text{tan}^{-1}u - \frac{u}{2} + C$

65\. ${\int{u^{n}\text{sin}^{-1}u\ du =}}\frac{1}{n + 1}\left\lbrack {u^{n + 1}\text{sin}^{-1}u - {\int\frac{u^{n + 1}\ du}{\sqrt{1 - u^{2}}}}} \right\rbrack,n \neq \text{−}1$

66\. ${\int{u^{n}\text{cos}^{-1}u\ du =}}\frac{1}{n + 1}\left\lbrack {u^{n + 1}\text{cos}^{-1}u + {\int\frac{u^{n + 1}\ du}{\sqrt{1 - u^{2}}}}} \right\rbrack,n \neq \text{−}1$

67\. ${\int{u^{n}\text{tan}^{-1}u\ du =}}\frac{1}{n + 1}\left\lbrack {u^{n + 1}\text{tan}^{-1}u - {\int\frac{u^{n + 1}\ du}{1 + u^{2}}}} \right\rbrack,n \neq \text{−}1$

Integrals Involving *a*² + *u*², *a* \> 0 含 a² + u² 的积分，a \> 0

68\. ${\int\sqrt{a^{2} + u^{2}}}\ du = \frac{u}{2}\sqrt{a^{2} + u^{2}} + \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

69\. ${\int{u^{2}\sqrt{a^{2} + u^{2}}}}\ du = \frac{u}{8}\left( {a^{2} + 2u^{2}} \right)\sqrt{a^{2} + u^{2}} - \frac{a^{4}}{8}\text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

70\. ${\int\frac{\sqrt{a^{2} + u^{2}}}{u}}\ du = \sqrt{a^{2} + u^{2}} - a\ \text{ln}\mspace{2mu}\left| \frac{a + \sqrt{a^{2} + u^{2}}}{u} \right| + C$

71\. ${\int\frac{\sqrt{a^{2} + u^{2}}}{u^{2}}}\ du = - \frac{\sqrt{a^{2} + u^{2}}}{u} + \text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

72\. ${\int\frac{\ du}{\sqrt{a^{2} + u^{2}}}} = \text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

73\. ${\int\frac{u^{2}\ du}{\sqrt{a^{2} + u^{2}}}} = \frac{u}{2}\left( \sqrt{a^{2} + u^{2}} \right) - \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left( {u + \sqrt{a^{2} + u^{2}}} \right) + C$

74\. ${\int\frac{\ du}{u\sqrt{a^{2} + u^{2}}}} = - \frac{1}{a}\text{ln}\mspace{2mu}\left| \frac{\sqrt{a^{2} + u^{2}} + a}{u} \right| + C$

75\. ${\int\frac{\ du}{u^{2}\sqrt{a^{2} + u^{2}}}} = - \frac{\sqrt{a^{2} + u^{2}}}{a^{2}u} + C$

76\. ${\int\frac{\ du}{\left( {a^{2} + u^{2}} \right)^{3\text{/}2}}} = \frac{u}{a^{2}\sqrt{a^{2} + u^{2}}} + C$

Integrals Involving *u*² − *a*², *a* \> 0 含 u² − a² 的积分，a \> 0

77\. ${\int{\sqrt{u^{2} - a^{2}}\ du = \frac{u}{2}}}\sqrt{u^{2} - a^{2}} - \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

78\. ${\int{u^{2}\sqrt{u^{2} - a^{2}}\ du = \frac{u}{8}}}\left( {2u^{2} - a^{2}} \right)\sqrt{u^{2} - a^{2}} - \frac{a^{4}}{8}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

79\. ${\int{\frac{\sqrt{u^{2} - a^{2}}}{u}\ du =}}\sqrt{u^{2} - a^{2}} - a\text{cos}^{-1}\frac{a}{|u|} + C$

80\. ${\int{\frac{\sqrt{u^{2} - a^{2}}}{u^{2}}\ du =}} - \frac{\sqrt{u^{2} - a^{2}}}{u} + \text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

81\. ${\int{\frac{\ du}{\sqrt{u^{2} - a^{2}}} =}}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

82\. ${\int{\frac{u^{2}\ du}{\sqrt{u^{2} - a^{2}}} =}}\frac{u}{2}\sqrt{u^{2} - a^{2}} + \frac{a^{2}}{2}\text{ln}\mspace{2mu}\left| {u + \sqrt{u^{2} - a^{2}}} \right| + C$

83\. ${\int{\frac{\ du}{u^{2}\sqrt{u^{2} - a^{2}}} =}}\frac{\sqrt{u^{2} - a^{2}}}{a^{2}u} + C$

84a. ${\int{\frac{\ du}{\left( {u^{2} - a^{2}} \right)^{3\text{/}2}} = \text{−}}}\frac{u}{a^{2}\sqrt{u^{2} - a^{2}}} + C$

84b. $\int\frac{du}{u^{2} - a^{2}} = \frac{1}{2a}\ln\left| \frac{u - a}{u + a} \right| + C$

Integrals Involving *a*² − *u*², *a* \> 0 含 a² − u² 的积分，a \> 0

85\. ${\int{\sqrt{a^{2} - u^{2}}\ du = \frac{u}{2}}}\sqrt{a^{2} - u^{2}} + \frac{a^{2}}{2}\text{sin}^{-1}\frac{u}{a} + C$

86\. ${\int{u^{2}\sqrt{a^{2} - u^{2}}\ du = \frac{u}{8}}}\left( {2u^{2} - a^{2}} \right)\sqrt{a^{2} - u^{2}} + \frac{a^{4}}{8}\text{sin}^{-1}\frac{u}{a} + C$

87\. ${\int{\frac{\sqrt{a^{2} - u^{2}}}{u}\ du =}}\sqrt{a^{2} - u^{2}} - a\text{ln}\mspace{2mu}\left| \frac{a + \sqrt{a^{2} - u^{2}}}{u} \right| + C$

88\. ${\int{\frac{\sqrt{a^{2} - u^{2}}}{u^{2}}\ du = - \frac{1}{u}}}\sqrt{a^{2} - u^{2}} - \text{sin}^{-1}\frac{u}{a} + C$

89\. $\int\frac{u^{2}\ du}{\sqrt{a^{2} - u^{2}}} = - \frac{u}{2}\sqrt{a^{2} - u^{2}} + \frac{a^{2}}{2}\text{sin}^{-1}\frac{u}{a} + C$

90\. ${\int{\frac{\ du}{u\sqrt{a^{2} - u^{2}}} =}} - \frac{1}{a}\text{ln}\mspace{2mu}\left| \frac{a + \sqrt{a^{2} - u^{2}}}{u} \right| + C$

91\. ${\int{\frac{\ du}{u^{2}\sqrt{a^{2} - u^{2}}} =}} - \frac{1}{a^{2}u}\sqrt{a^{2} - u^{2}} + C$

92\. ${\int{\left( {a^{2} - u^{2}} \right)^{3\text{/}2}\ du = - \frac{u}{8}\left( {2u^{2} - 5a^{2}} \right)}}\sqrt{a^{2} - u^{2}} + \frac{3a^{4}}{8}\text{sin}^{-1}\frac{u}{a} + C$

93a. ${\int{\frac{\ du}{\left( {a^{2} - u^{2}} \right)^{3\text{/}2}} = \frac{u}{a^{2}\sqrt{a^{2} - u^{2}}} +}}C$

93b. $\int\frac{du}{a^{2} - u^{2}} = \frac{1}{2a}\ln\left| \frac{u + a}{u - a} \right| + C$

Integrals Involving 2*au* − *u*², *a* \> 0 含 2au − u² 的积分，a \> 0

94\. ${\int\sqrt{2au - u^{2}}}\ du = \frac{u - a}{2}\sqrt{2au - u^{2}} + \frac{a^{2}}{2}\text{cos}^{-1}\left( \frac{a - u}{a} \right) + C$

95\. ${\int\frac{\ du}{\sqrt{2au - u^{2}}}} = \text{cos}^{-1}\left( \frac{a - u}{a} \right) + C$

96\. ${\int{u\sqrt{2au - u^{2}}}}\ du = \frac{2u^{2} - au - 3a^{2}}{6}\sqrt{2au - u^{2}} + \frac{a^{3}}{2}\text{cos}^{-1}\left( \frac{a - u}{a} \right) + C$

97\. ${\int\frac{\ du}{u\sqrt{2au - u^{2}}}} = - \frac{\sqrt{2au - u^{2}}}{au} + C$

Integrals Involving *a* + *bu*, *a* ≠ 0 含 a + bu 的积分，a ≠ 0

98\. ${\int\frac{u\ du}{a + bu}} = \frac{1}{b^{2}}\left( {a + bu - a\text{ln}\mspace{2mu}\left| {a + bu} \right|} \right) + C$

99\. ${\int\frac{u^{2}\ du}{a + bu}} = \frac{1}{2b^{3}}\left\lbrack {\left( {a + bu} \right)^{2} - 4a\left( {a + bu} \right) + 2a^{2}\text{ln}\mspace{2mu}\left| {a + bu} \right|} \right\rbrack + C$

100\. ${\int\frac{\ du}{u\left( {a + bu} \right)}} = \frac{1}{a}\text{ln}\mspace{2mu}\left| \frac{u}{a + bu} \right| + C$

101\. ${\int\frac{\ du}{u^{2}\left( {a + bu} \right)}} = - \frac{1}{au} + \frac{b}{a^{2}}\text{ln}\mspace{2mu}\left| \frac{a + bu}{u} \right| + C$

102\. ${\int\frac{u\ du}{\left( {a + bu} \right)^{2}}} = \frac{a}{b^{2}\left( {a + bu} \right)} + \frac{1}{b^{2}}\text{ln}\mspace{2mu}\left| {a + bu} \right| + C$

103\. ${\int\frac{u\ du}{u\ \left( {a + bu} \right)^{2}}} = \frac{1}{a\left( {a + bu} \right)} - \frac{1}{a^{2}}\text{ln}\mspace{2mu}\left| \frac{a + bu}{u} \right| + C$

104\. ${\int\frac{u^{2}\ du}{\left( {a + bu} \right)^{2}}} = \frac{1}{b^{3}}\left( {a + bu - \frac{a^{2}}{a + bu} - 2a\text{ln}\mspace{2mu}\left| {a + bu} \right|} \right) + C$

105\. ${\int{u\sqrt{a + bu}}}\ du = \frac{2}{15b^{2}}\left( {3bu - 2a} \right)\left( {a + bu} \right)^{3\text{/}2} + C$

106\. ${\int\frac{u\ du}{\sqrt{a + bu}}} = \frac{2}{3b^{2}}\left( {bu - 2a} \right)\sqrt{a + bu} + C$

107\. ${\int\frac{u^{2}\ du}{\sqrt{a + bu}}} = \frac{2}{15b^{3}}\left( {8a^{2} + 3b^{2}u^{2} - 4abu} \right)\sqrt{a + bu} + C$

108\. $\begin{array}{cll} {\int\frac{\ du}{u\sqrt{a + bu}}} & {= \frac{1}{\sqrt{a}}\text{ln}\mspace{2mu}\left| \frac{\sqrt{a + bu} - \sqrt{a}}{\sqrt{a + bu} + \sqrt{a}} \right| + C,} & {\text{if}\ a > 0} \\ & {= \frac{2}{\sqrt{\text{−}a}}\text{tan} - 1\sqrt{\frac{a + bu}{\text{−}a}} + C,} & {\text{if}\ a < 0} \end{array}$

109\. ${\int\frac{\sqrt{a + bu}}{u}}\ du = 2\sqrt{a + bu} + a{\int\frac{\ du}{u\sqrt{a + bu}}}$

110\. ${\int\frac{\sqrt{a + bu}}{u^{2}}}\ du = - \frac{\sqrt{a + bu}}{u} + \frac{b}{2}{\int\frac{\ du}{u\sqrt{a + bu}}}$

111\. ${\int{u^{n}\sqrt{a + bu}}}\ du = \frac{2}{b\left( {2n + 3} \right)}\left\lbrack {u^{n}\left( {a + bu} \right)^{3\text{/}2} - na{\int{u^{n - 1}\sqrt{a + bu}\ du}}} \right\rbrack$

112\. ${\int\frac{u^{n}\ du}{\sqrt{a + bu}}} = \frac{2u^{n}\sqrt{a + bu}}{b\left( {2n + 1} \right)} - \frac{2na}{b\left( {2n + 1} \right)}{\int\frac{u^{n - 1}\ du}{\sqrt{a + bu}}}$

113\. ${\int\frac{\ du}{u^{n}\sqrt{a + bu}}} = - \frac{\sqrt{a + bu}}{a\left( {n - 1} \right)u^{n - 1}} - \frac{b\left( {2n - 3} \right)}{2a\left( {n - 1} \right)}{\int\frac{\ du}{u^{n - 1}\sqrt{a + bu}}}$

B \| Table of Derivatives B \| 导数表

General Formulas 一般公式

1\. $\frac{d}{dx}(c) = 0$

2\. $\frac{d}{dx}\left( {f(x) + g(x)} \right) = f^{\prime}(x) + g^{\prime}(x)$

3\. $\frac{d}{dx}\left( {f(x)g(x)} \right) = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$

4\. $\frac{d}{dx}\left( x^{n} \right) = nx^{n - 1},\ \text{for real numbers}\ n$

5\. $\frac{d}{dx}\left( {cf(x)} \right) = cf^{\prime}(x)$

6\. $\frac{d}{dx}\left( {f(x) - g(x)} \right) = f^{\prime}(x) - g^{\prime}(x)$

7\. $\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f^{\prime}(x) - f(x)g^{\prime}(x)}{\left( {g(x)} \right)^{2}}$

8\. $\frac{d}{dx}\left\lbrack {f\left( {g(x)} \right)} \right\rbrack = f^{\prime}\left( {g(x)} \right) \cdot g^{\prime}(x)$

Trigonometric Functions 三角函数

9\. $\frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x$

10\. $\frac{d}{dx}\left( {\text{tan}x} \right) = \text{sec}^{2}x$

11\. $\frac{d}{dx}\left( {\text{sec}x} \right) = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$

12\. $\frac{d}{dx}\left( {\text{cos}\mspace{2mu} x} \right) = \text{−}\text{sin}\mspace{2mu} x$

13\. $\frac{d}{dx}\left( {\text{cot}x} \right) = \text{−}\text{csc}^{2}x$

14\. $\frac{d}{dx}\left( {\text{csc}x} \right) = \text{−csc}\mspace{2mu} x\mspace{2mu}\text{cot}\mspace{2mu} x$

Inverse Trigonometric Functions 反三角函数

15\. $\frac{d}{dx}\left( {\text{sin}^{-1}x} \right) = \frac{1}{\sqrt{1 - x^{2}}}$

16\. $\frac{d}{dx}\left( {\text{tan}^{-1}x} \right) = \frac{1}{1 + x^{2}}$

17\. $\frac{d}{dx}\left( {\text{sec}^{-1}x} \right) = \frac{1}{|x|\sqrt{x^{2} - 1}}$

18\. $\frac{d}{dx}\left( {\text{cos}^{-1}x} \right) = - \frac{1}{\sqrt{1 - x^{2}}}$

19\. $\frac{d}{dx}\left( {\text{cot}^{-1}x} \right) = - \frac{1}{1 + x^{2}}$

20\. $\frac{d}{dx}\left( {\text{csc}^{-1}x} \right) = - \frac{1}{|x|\sqrt{x^{2} - 1}}$

Exponential and Logarithmic Functions 指数函数与对数函数

21\. $\frac{d}{dx}\left( e^{x} \right) = e^{x}$

22\. $\frac{d}{dx}\left( {\text{ln}\mspace{2mu}|x|} \right) = \frac{1}{x}$

23\. $\frac{d}{dx}\left( b^{x} \right) = b^{x}\text{ln}\mspace{2mu} b$

24\. $\frac{d}{dx}\left( {\text{log}_{b}x} \right) = \frac{1}{x\mspace{2mu}\text{ln}\mspace{2mu} b}$

Hyperbolic Functions 双曲函数

25\. $\frac{d}{dx}\left( {\text{sinh}\mspace{2mu} x} \right) = \text{cosh}\mspace{2mu} x$

26\. $\frac{d}{dx}\left( {\text{tanh}\mspace{2mu} x} \right) = \text{sech}^{2}\mspace{2mu} x$

27\. $\frac{d}{dx}\left( {\text{sech}\mspace{2mu} x} \right) = \text{−sech}\mspace{2mu} x\ \text{tanh}\mspace{2mu} x$

28\. $\frac{d}{dx}\left( {\text{cosh}\mspace{2mu} x} \right) = \text{sinh}\mspace{2mu} x$

29\. $\frac{d}{dx}\left( {\text{coth}\mspace{2mu} x} \right) = \text{−}\text{csch}^{2}\mspace{2mu} x$

30\. $\frac{d}{dx}\left( {\text{csch}\mspace{2mu} x} \right) = \text{−csch}\mspace{2mu} x\ \text{coth}\mspace{2mu} x$

Inverse Hyperbolic Functions 反双曲函数

31\. $\frac{d}{dx}\left( {\text{sinh}^{-1}x} \right) = \frac{1}{\sqrt{x^{2} + 1}}$

32\. $\frac{d}{dx}\left( {\text{tanh}^{-1}x} \right) = \frac{1}{1 - x^{2}}\left( {|x| < 1} \right)$

33\. $\frac{d}{dx}\left( {\text{sech}^{-1}x} \right) = - \frac{1}{x\sqrt{1 - x^{2}}}\quad\left( {0 < x < 1} \right)$

34\. $\frac{d}{dx}\left( {\text{cosh}^{-1}x} \right) = \frac{1}{\sqrt{x^{2} - 1}}\quad\left( {x > 1} \right)$

35\. $\frac{d}{dx}\left( {\text{coth}^{-1}x} \right) = \frac{1}{1 - x^{2}}\quad\left( {|x| > 1} \right)$

36\. $\frac{d}{dx}\left( {\text{csch}^{-1}x} \right) = - \frac{1}{|x|\sqrt{1 + x^{2}}}\ \left( {x \neq 0} \right)$

Formulas from Geometry 几何公式

$A = \text{area},$ $V = \text{Volume},\ \text{and}$ $S = \text{lateral surface area}$

Formulas from Algebra 代数公式

Laws of Exponents 指数定律

$\begin{array}{rllcccccccccc} {x^{m}x^{n}} & = & x^{m + n} & & & \frac{x^{m}}{x^{n}} & = & x^{m - n} & & & \left( x^{m} \right)^{n} & = & x^{mn} \\ x^{\text{−}n} & = & \frac{1}{x^{n}} & & & \left( {xy} \right)^{n} & = & {x^{n}y^{n}} & & & \left( \frac{x}{y} \right)^{n} & = & \frac{x^{n}}{y^{n}} \\ x^{1\text{/}n} & = & \sqrt[n]{x} & & & \sqrt[n]{xy} & = & {\sqrt[n]{x}\sqrt[n]{y}} & & & \sqrt[n]{\frac{x}{y}} & = & \frac{\sqrt[n]{x}}{\sqrt[n]{y}} \\ x^{m\text{/}n} & = & {\sqrt[n]{x^{m}} = \left( \sqrt[n]{x} \right)^{m}} & & & & & & & & & & \end{array}$

Special Factorizations 特殊因式分解

$\begin{array}{rll} {x^{2} - y^{2}} & = & {\left( {x + y} \right)\left( {x - y} \right)} \\ {x^{3} + y^{3}} & = & {\left( {x + y} \right)\left( {x^{2} - xy + y^{2}} \right)} \\ {x^{3} - y^{3}} & = & {\left( {x - y} \right)\left( {x^{2} + xy + y^{2}} \right)} \end{array}$

Quadratic Formula 二次方程求根公式

If $ax^{2} + bx + c = 0,$ then $x = \frac{\text{−}b \pm \sqrt{b^{2} - 4ac}}{2a}.$

Binomial Theorem 二项式定理

$\left( {a + b} \right)^{n} = a^{n} + \left( \begin{array}{l} n \\ 1 \end{array} \right)a^{n - 1}b + \left( \begin{array}{l} n \\ 2 \end{array} \right)a^{n - 2}b^{2} + \cdots + \begin{pmatrix} n \\ {n - 1} \end{pmatrix}ab^{n - 1} + b^{n},$

where $\left( \begin{array}{l} n \\ k \end{array} \right) = \frac{n\left( {n - 1} \right)\left( {n - 2} \right)\cdots\left( {n - k + 1} \right)}{k\left( {k - 1} \right)\left( {k - 2} \right)\cdots 3 \cdot 2 \cdot 1} = \frac{n!}{k!\left( {n - k} \right)!}$

Formulas from Trigonometry 三角公式

Right-Angle Trigonometry 直角三角形三角学

$\begin{array}{lccl} {\text{sin}\mspace{2mu}\theta = \frac{\text{opp}}{\text{hyp}}} & & & {\text{csc}\mspace{2mu}\theta = \frac{\text{hyp}}{\text{opp}}} \\ {\text{cos}\mspace{2mu}\theta = \frac{\text{adj}}{\text{hyp}}} & & & {\text{sec}\mspace{2mu}\theta = \frac{\text{hyp}}{\text{adj}}} \\ {\text{tan}\mspace{2mu}\theta = \frac{\text{opp}}{\text{adj}}} & & & {\text{cot}\mspace{2mu}\theta = \frac{\text{adj}}{\text{opp}}} \end{array}$

Trigonometric Functions of Important Angles 重要角的三角函数

$\theta$	$\text{Radians}$	$\text{sin}\mspace{2mu}\theta$	$\text{cos}\mspace{2mu}\theta$	$\text{tan}\mspace{2mu}\theta$
$0\text{°}$	$0$	$0$	$1$	$0$
$30\text{°}$	$\text{π}\text{/}\text{6}$	$1\text{/}2$	$\sqrt{3}\text{/}2$	$\sqrt{3}\text{/}3$
$45\text{°}$	$\text{π}\text{/}\text{4}$	$\sqrt{2}\text{/}2$	$\sqrt{2}\text{/}2$	$1$
$60\text{°}$	$\text{π}\text{/}\text{3}$	$\sqrt{3}\text{/}2$	$1\text{/}2$	$\sqrt{3}$
$90\text{°}$	$\text{π}\text{/}2$	$1$	$0$	—

$\theta$	$\text{Radians}$	$\text{sin}\mspace{2mu}\theta$	$\text{cos}\mspace{2mu}\theta$	$\text{tan}\mspace{2mu}\theta$
$0\text{°}$	$0$	$0$	$1$	$0$
$30\text{°}$	$\text{π}\text{/}\text{6}$	$1\text{/}2$	$\sqrt{3}\text{/}2$	$\sqrt{3}\text{/}3$
$45\text{°}$	$\text{π}\text{/}\text{4}$	$\sqrt{2}\text{/}2$	$\sqrt{2}\text{/}2$	$1$
$60\text{°}$	$\text{π}\text{/}\text{3}$	$\sqrt{3}\text{/}2$	$1\text{/}2$	$\sqrt{3}$
$90\text{°}$	$\text{π}\text{/}2$	$1$	$0$	—

Fundamental Identities 基本恒等式

$\begin{array}{rllccrll} {\text{sin}^{2}\theta + \text{cos}^{2}\theta} & = & 1 & & & {\text{sin}\left( {\text{−}\mspace{2mu}\theta} \right)} & = & {\text{−}\text{sin}\mspace{2mu}\theta} \\ {1 + \text{tan}^{2}\theta} & = & {\text{sec}^{2}\theta} & & & {\text{cos}\left( {\text{−}\mspace{2mu}\theta} \right)} & = & {\text{cos}\mspace{2mu}\theta} \\ {1 + \text{cot}^{2}\theta} & = & {\text{csc}^{2}\theta} & & & {\text{tan}\left( {\text{−}\mspace{2mu}\theta} \right)} & = & {\text{−}\text{tan}\mspace{2mu}\theta} \\ {\text{sin}\left( {\frac{\pi}{2} - \theta} \right)} & = & {\text{cos}\mspace{2mu}\theta} & & & {\text{sin}\left( {\theta + 2\pi} \right)} & = & {\text{sin}\mspace{2mu}\theta} \\ {\text{cos}\left( {\frac{\pi}{2} - \theta} \right)} & = & {\text{sin}\mspace{2mu}\theta} & & & {\text{cos}\left( {\theta + 2\pi} \right)} & = & {\text{cos}\mspace{2mu}\theta} \\ {\text{tan}\left( {\frac{\pi}{2} - \theta} \right)} & = & {\text{cot}\mspace{2mu}\theta} & & & {\text{tan}\left( {\theta + \pi} \right)} & = & {\text{tan}\mspace{2mu}\theta} \end{array}$

Law of Sines 正弦定律

$\frac{\text{sin}\mspace{2mu} A}{a} = \frac{\text{sin}\mspace{2mu} B}{b} = \frac{\text{sin}\mspace{2mu} C}{c}$

Law of Cosines 余弦定律

$\begin{array}{rll} a^{2} & = & {b^{2} + c^{2} - 2bc\ \text{cos}\ A} \\ b^{2} & = & {a^{2} + c^{2} - 2ac\ \text{cos}\ B} \\ c^{2} & = & {a^{2} + b^{2} - 2ab\ \text{cos}\ C} \end{array}$

Addition and Subtraction Formulas 和角与差角公式

$\begin{matrix} {\text{sin}\ (x + y)} & = & {\text{sin}\ x\ \text{cos}\ y + \text{cos}\ x\ \text{sin}\ y} \\ {\text{sin}\mspace{2mu}(x - y)} & = & {\text{sin}\ x\ \text{cos}\ y - \text{cos}\ x\ \text{sin}\ y} \\ {\text{cos}\mspace{2mu}(x + y)} & = & {\text{cos}\ x\ \text{cos}\ y - \text{sin}\ x\ \text{sin}\ y} \\ {\text{cos}\mspace{2mu}(x - y)} & = & {\text{cos}\ x\ \text{cos}\ y + \text{sin}\ x\ \text{sin}\ y} \\ {\text{tan}\mspace{2mu}(x + y)} & = & \frac{\text{tan}\ x + \text{tan~}y}{1 - \text{tan}\ x\ \text{tan~}y} \\ {\text{tan}(x - y)} & = & \frac{\text{tan}\ x - \text{tan~}y}{1 + \text{tan}\ x\ \text{tan~}y} \end{matrix}$

Double-Angle Formulas 倍角公式

$\begin{array}{rll} {\text{sin}\ 2x} & = & {2\mspace{2mu}\text{sin}\ x\ \text{cos}\ x} \\ {\text{cos}\ 2x} & = & {\text{cos}^{2}x - \text{sin}^{2}x = 2\mspace{2mu}\text{cos}^{2}x - 1 = 1 - 2\mspace{2mu}\text{sin}^{2}x} \\ {\text{tan}\ 2x} & = & \frac{2\mspace{2mu}\text{tan}\ x}{1 - \text{tan}^{2}x} \end{array}$

Half-Angle Formulas 半角公式

$\begin{array}{rll} {\text{sin}^{2}x} & = & \frac{1 - \text{cos}\ 2x}{2} \\ {\text{cos}^{2}x} & = & \frac{1 + \text{cos}\ 2x}{2} \end{array}$