习题4.2
第 1 题
计算下列曲线积分.
(1) \(\int_L(x+y)\text{d}{l}\), 其中 \(L\) 为 \(O(0, 0), A(1, 0), B(0, 1)\) 为顶点的三角形的三条边;
(2) \(\int_L\sqrt{x^2+y^2}\text{d}{l}\), 其中 \(L\) 为圆周 \(x^2+y^2=2x\);
(3) \(\int_Ly^2\text{d}{l}\), 其中 \(L\) 为摆线 \(\begin{cases}x = a(t - \sin t), \\ y = a(1 - \cos t),\end{cases}0 \le t \le 2\pi\);
(4) \(\int_L(x^{\frac{4}{3}}+y^{\frac{4}{3}})\text{d}{l}\), 其中 \(L\) 为星形线 \(\begin{cases}x = a\cos^3t, \\ y = a\sin^3t,\end{cases}0 \le t \le 2\pi\).
解
(1) \(L = L_1 + L_2 + L_3\), 其中 \(L_1 = \begin{cases}x = x, \\ y = 0, \end{cases}L_2 = \begin{cases}x = x, \\ y = 1-x, \end{cases}L_3 = \begin{cases}x = 0, \\ y = y, \end{cases}(0 \le x \le 1, 0 \le y \le 1)\). 因此
\[
\begin{aligned}
\int_L (x+y)\text{d}{l} & = \int_{L_1} (x+y)\text{d}{l} + \int_{L_2} (x+y)\text{d}{l} + \int_{L_3} (x+y)\text{d}{l} \\
& = \int_{0}^{1}x\text{d}{x} + \int_{0}^{1}\sqrt{2}\text{d}{x} + \int_{0}^{1}y\text{d}{y} \\
& = \dfrac{1}{2} + \sqrt{2} + \dfrac{1}{2} \\
& = \sqrt{2} + 1
\end{aligned}
\]
(2) 设 \(\begin{cases}x=1+\cos\theta, \\ y = \sin\theta, \end{cases}\) 则 \(\text{d}{l} = \sqrt{(\text{d}{x})^2 + (\text{d}{y})^2} = \sqrt{(-\sin\theta)^2+(\cos\theta)^2}\text{d}{\theta} = \text{d}{\theta}\).
\[
\begin{aligned}
\int_L\sqrt{x^2+y^2}\text{d}{l} & = \int_{-\pi}^{\pi}\sqrt{(1+\cos\theta)^2 + \sin^2\theta}\text{d}{\theta} \\
& = \int_{-\pi}^{\pi}2\cos\dfrac{\theta}{2}\text{d}{\theta} \\
& = 4\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos u\text{d}{u} \\
& = 8
\end{aligned}
\]
(3) \(\text{d}{l} = \sqrt{(\text{d}{x})^2 + (\text{d}{y})^2} = \sqrt{a^2(a-\cos t)^2 + (a\sin t)^2} = \sqrt{2}a\sin\dfrac{t}{2}\). 因此
\[
\begin{aligned}
\int_Ly^2\text{d}{l} & = \int_{0}^{2\pi}a^2(1-\cos t)^2\cdot\sqrt{2}a\sin\dfrac{t}{2} \text{d}{t} \\
& = 4\sqrt{2}a^3\int_{0}^{2\pi}\sin^3\dfrac{t}{2}\text{d}{t} \\
& = 8\sqrt{2}a^3\int_{0}^{2\pi}\sin^3u\text{d}{u} \\
& = 32\sqrt{2}a^3\int_{0}^{\frac{\pi}{2}}\sin^5u\text{d}{u} \\
& = 32\sqrt{2}a^3\cdot\dfrac{4 \cdot 2}{5\cdot 3} \\
& = \dfrac{256}{15}a^3
\end{aligned}
\]
(4) \(\text{d}{l} = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2} = \sqrt{(3a\cos^2t\sin t)^2 + (3a\sin^2t\cos t)^2}\text{d}{t} = 3a\left|\sin t\cos t\right|\text{d}{t}\). 只考虑 \(0 \le t \le \dfrac{\pi}{2}\) 的情况.
\[
\begin{aligned}
\int_{L'}(x^{\frac{4}{3}}+y^{\frac{4}{3}})\text{d}{l} & = \int_{0}^{\frac{\pi}{2}}a^{\frac{4}{3}}(\cos^4t+\sin^4t)\cdot 3a(\sin t\cos t)\text{d}{t} \\
& = 3a^{\frac{7}{3}}\int_{0}^{\frac{\pi}{2}}(\cos^4t+\sin^4t)\sin t\cos t\text{d}{t} \\
& = 3a^{\frac{7}{3}}\left(\dfrac{1}{6}\sin^6t - \dfrac{1}{6}\cos^6t\right)\bigg\vert_{0}^{\frac{\pi}{2}} \\
& = a^{\frac{7}{3}}
\end{aligned}
\]
由对称性知 \(\int_{L}(x^{\frac{4}{3}}+y^{\frac{4}{3}})\text{d}{l} = 4\int_{L'}(x^{\frac{4}{3}}+y^{\frac{4}{3}})\text{d}{l} = 4a^{\frac{7}{3}}\).
第 2 题
计算下列曲线积分.
(1) \(\int_L x\sqrt{x^2-y^2}\text{d}{l}\), 其中 \(L\) 为双纽线右半支 \(r^2 = a^2\cos2\theta(-\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{4}, a > 0)\);
(2) \(\int_L(x^2+y^2+z^2)\text{d}{l}\), 其中 \(L\) 为螺线 \(x=2\cos t, y=2\sin t, z=3t(0\le t \le 2\pi)\);
(3) \(\int_L xyz \text{d}{l}\), 其中 \(L\) 的参数方程为 \(x=t, y=\dfrac{2}{3}\sqrt{2}t^{\frac{3}{2}}, z=\dfrac{1}{2}t^2(0 \le t \le 1)\);
(4) \(\int_L x\text{d}{l}\), 其中 \(L\) 为球面 \(x^2+y^2+z^2=4\) 在第一象限部分的边界.
解
(1) 由题知 \(\sqrt{x^2-y^2}=\sqrt{(r\cos\theta)^2-(r\sin\theta)^2} = \sqrt{r^2\cos2\theta} = \dfrac{r^2}{a}\). 而
\[
\begin{aligned}
\text{d}{l} & = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2} \\
& = \sqrt{(-r\sin\theta)^2+(r\cos\theta)^2} \\
& = \sqrt{(\cos\theta\text{d}{r} - r\sin\theta\text{d}{\theta})^2 + (\sin\theta\text{d}{r} + r\cos\theta\text{d}{\theta})^2} \\
& = \sqrt{(\text{d}{r})^2 + r^2(\text{d}{\theta})^2}
\end{aligned}
\]
对 \(r^2 = a^2\cos2\theta\) 微分得到 \(\text{d}{r} = -\dfrac{a^2\sin2\theta}{r}\text{d}{\theta}\), 代入有 \(\text{d}{l} = \sqrt{\left(-\dfrac{a^2\sin2\theta}{r}\text{d}{\theta}\right)^2+r^2(\text{d}{\theta})^2} = \dfrac{1}{r}\sqrt{r^4 - (a^2\sin2\theta)^2}\text{d}{\theta}\). 结合 \(r^2 = a^2\cos2\theta\) 知 \(\text{d}{l} = \dfrac{a^2}{r}\text{d}{\theta}\). 因此
\[
\begin{aligned}
\int_L x\sqrt{x^2-y^2}\text{d}{l} & = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}r\cos\theta\cdot \dfrac{r^2}{a}\cdot \dfrac{a^2}{r}\text{d}{\theta} \\
& = a\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}r^2\cos\theta\text{d}{\theta} \\
& = a\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}a^2\cos2\theta\cos\theta\text{d}{\theta} \\
& = a^3\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}(1-2\sin^2\theta)\cos\theta\text{d}{\theta} \\
& = a^3\left(\sin\theta - \dfrac{2}{3}\sin^3\theta\right)\bigg\vert_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \\
& = \dfrac{2\sqrt{2}a^3}{3}
\end{aligned}
\]
(2) \(\text{d}{l} = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2+(\text{d}{z})^2} = \sqrt{(-2\sin t)^2+(2\cos t)^2 + 3^2} = \sqrt{13}\). 因此
\[
\begin{aligned}
\int_L(x^2+y^2+z^2)\text{d}{l} & = \int_{0}^{2\pi}(4 + 9t^2)\cdot\sqrt{13}\text{d}{t} \\
& = \sqrt{13}(4t + 3t^3)\bigg\vert_{0}^{2\pi} \\
& = 24\sqrt{13}\pi^3+8\sqrt{13}\pi
\end{aligned}
\]
(3) \(\text{d}{l} = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2+(\text{d}{z})^2} = \sqrt{1^2+(\sqrt{2t})^2 + t^2} = (t+1)\text{d}{t}\). 因此
\[
\begin{aligned}
\int_Lxyz\text{d}{l} & = \int_{0}^{1}\dfrac{\sqrt{2}}{3}t^{\frac{9}{2}}\cdot(t+1)\text{d}{t} \\
& = \dfrac{2\sqrt{2}}{3}t^{\frac{11}{2}}\left(\dfrac{1}{13}t + \dfrac{1}{11}\right)\bigg\vert_{0}^{1} \\
& = \dfrac{16\sqrt{2}}{143}
\end{aligned}
\]
(4) \(L = L_1 + L_2 + L_3\), 其中 \(L_1 = \begin{cases}x = 0, \\ y = 2\cos\theta, \\ z = 2\sin\theta, \end{cases}L_2 = \begin{cases}x = 2\cos\theta, \\ y = 0, \\ z = 2\sin\theta, \end{cases} L_3 = \begin{cases}x = 2\cos\theta, \\ y = 2\sin\theta, \\ z = 0, \end{cases}(0 \le \theta \le \dfrac{\pi}{2})\). 因此
\[
\begin{aligned}
\int_L x\text{d}{l} & = \int_{L_1} x\text{d}{l} + \int_{L_2} x\text{d}{l} + \int_{L_3} x\text{d}{l} \\
& = 0 + 2\int_{0}^{\frac{\pi}{2}}2\cos\theta\cdot4\text{d}{\theta} \\
& = 8
\end{aligned}
\]
第 4 题
曲线 \(y=\ln x\) 的线密度 \(\rho(x, y) = x^2\), 试求曲线在 \(x = \sqrt{3}\) 与 \(x = \sqrt{15}\) 之间的质量.
解
\(M = \int_L\rho(x, y)\text{d}{l}\), 其中 \(L\) 是 \(y=\ln x(\sqrt{3} \le x \le \sqrt{15})\). 而 \(\text{d}{l} = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2} = \sqrt{1 + \dfrac{1}{x^2}}\text{d}{x}\), 因此
\[
\begin{aligned}
M & = \int_L\rho(x, y)\text{d}{l} \\
& = \int_{\sqrt{3}}^{\sqrt{15}}x^2\cdot \sqrt{1 + \dfrac{1}{x^2}}\text{d}{x} \\
& = \int_{\sqrt{3}}^{\sqrt{15}}x\sqrt{x^2+1}\text{d}{x} \\
& = \dfrac{1}{3}(x^2+1)^{\frac{3}{2}}\bigg\vert_{\sqrt{3}}^{\sqrt{15}} \\
& = \dfrac{56}{3}
\end{aligned}
\]
第 5 题
求圆柱面 \(x^2+y^2=a^2\) 介于曲面 \(z = a+\dfrac{x^2}{a}\) 与 \(z=0\) 之间的面积 (\(a > 0\)).
解
积分区域为 \(L = \{(x, y) | x^2+y^2=a^2\} = \{(x, y) | x = a\cos\theta, b=a\sin\theta, 0 \le \theta \le 2\pi\}\). 而 \(\text{d}{l} = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2} = \sqrt{(-a\sin\theta)^2 + (a\cos\theta)^2}\text{d}{\theta} = a\text{d}{\theta}\), 因此
\[
\begin{aligned}
S & = \int_{L}z\text{d}{l} \\
& = \int_{0}^{2\pi}(a+a\cos^2\theta)a\text{d}{\theta} \\
& = \dfrac{1}{2}a^2\int_{0}^{2\pi}(3+\cos2\theta)\text{d}{\theta} \\
& = 3\pi a^2
\end{aligned}
\]
第 6 题
求摆线 \(\begin{cases}x = a(t - \sin t), \\ y = a(1 - \cos t),\end{cases}0 \le t \le \pi\) 的形心.
解
\(\text{d}{l} = \sqrt{(\text{d}{x})^2 + (\text{d}{y})^2} = \sqrt{a^2(a-\cos t)^2 + (a\sin t)^2} = \sqrt{2}a\sin\dfrac{t}{2}\).
总质量
\[
\begin{aligned}
M & = \int_{L}\text{d}{l} \\
& = \int_{0}^{\pi}\sqrt{2}a\sin\dfrac{t}{2}\text{d}{t} \\
& = \sqrt{2}a\int_{0}^{\pi}\sin\dfrac{t}{2}\text{d}{t} \\
& = 2\sqrt{2}a\int_{0}^{\frac{\pi}{2}}\sin u\text{d}{u} \\
& = 2\sqrt{2}a
\end{aligned}
\]
对 \(x\) 轴的静力矩
\[
\begin{aligned}
M_{x} & = \int_{L}y\text{d}{l} \\
& = \int_{0}^{\pi}a(1-\cos t)\cdot\sqrt{2}a\sin\dfrac{t}{2}\text{d}{t} \\
& = 2\sqrt{2}a^2\int_{0}^{\pi}\sin^2\dfrac{t}{2}\cdot\sin\dfrac{t}{2}\text{d}{t} \\
& = 4\sqrt{2}a^2\int_{0}^{\frac{\pi}{2}}\sin^3u\text{d}{u} \\
& = 4\sqrt{2}a^2\cdot\dfrac{2}{3} \\
& = \dfrac{8\sqrt{2}a^2}{3}
\end{aligned}
\]
对 \(y\) 轴的静力矩
\[
\begin{aligned}
M_y & = \int_{L}x\text{d}{l} \\
& = \int_{0}^{\pi}a(t-\sin t)\cdot\sqrt{2}a\sin\dfrac{t}{2}\text{d}{t} \\
& = \sqrt{2}a^2\int_{0}^{\pi}t\sin\dfrac{t}{2}\text{d}{t} -\sqrt{2}a^2\int_{0}^{\pi}\sin t\sin\dfrac{t}{2}\text{d}{t} \\
& = 4\sqrt{2}a^2\int_{0}^{\frac{\pi}{2}}u\sin u\text{d}{u} - 4\sqrt{2}a^2\int_{0}^{\frac{\pi}{2}}sin^2u\cos u\text{d}{u} \\
& = 4\sqrt{2}a^2(-u\cos u + \sin u)\bigg\vert_{0}^{\frac{\pi}{2}} - \dfrac{4\sqrt{2}a^2}{3}\sin^3u\bigg\vert_{0}^{\frac{\pi}{2}} \\
& = 4\sqrt{2}a^2 - \dfrac{4\sqrt{2}a^2}{3} \\
& = \dfrac{8\sqrt{2}a^2}{3}
\end{aligned}
\]
因此形心在 \(\left(\dfrac{M_x}{M}, \dfrac{M_y}{M}\right)\) 即 \((\dfrac{4a}{3}, \dfrac{4a}{3})\) 处.
第 7 题
求螺线 \(x = a\cos t, y = a\sin t, z = \dfrac{b}{2\pi}t(0 \le t \le 2\pi)\) 绕 \(x\) 轴旋转的转动惯量 (线密度为 \(1\)).
解
\(\text{d}{l} = \sqrt{(\text{d}{x})^2+(\text{d}{y})^2+(\text{d}{z})^2} = \sqrt{(a\sin t)^2+(a\cos t)^2 + \left(\dfrac{b}{2\pi}\right)^2}\text{d}{t} = \dfrac{\sqrt{b^2+4\pi^2a^2}}{2\pi}\text{d}{t}\). 因此
\[
\begin{aligned}
J_{x} & = \int_{L}(y^2 + z^2)\text{d}{l} \\
& = \int_{0}^{2\pi}\left(a^2\sin^2t + \left(\dfrac{b}{2\pi}t\right)^2\right)\dfrac{\sqrt{b^2+4\pi^2a^2}}{2\pi}\text{d}{t} \\
& = \dfrac{\sqrt{b^2+4\pi^2a^2}}{2\pi}\int_{0}^{2\pi}\left(a^2\cdot\dfrac{1-2\cos2t}{2} + \dfrac{b^2}{4\pi^2}\cdot t^2\right)\text{d}{t} \\
& = \dfrac{\sqrt{b^2+4\pi^2a^2}}{2\pi}\cdot\left(\pi a^2 + \dfrac{2\pi b^2}{3}\right) \\
& = \left(\dfrac{a^2}{2} + \dfrac{b^2}{3}\right)\sqrt{b^2+4\pi^2a^2}
\end{aligned}
\]
第 8 题
圆周 \(L:x^2+y^2=-2y\) 上每点的质量线密度等于 \(\sqrt{x^2+y^2}\), 求曲线 \(L\) 的质量与曲线 \(L\) 对 \(x\) 轴的静力矩.
解
设 \(\begin{cases}x = \cos\theta, \\ y = -1 + \sin\theta,\end{cases}(-\dfrac{3\pi}{2} \le \theta \le \dfrac{\pi}{2})\), 此时恒有 \(\cos\dfrac{\theta}{2} \ge \sin\dfrac{\theta}{2}\).
曲线 \(L\) 的质量为
\[
\begin{aligned}
M & = \int_{L}\sqrt{x^2+y^2}\text{d}{l} \\
& = \int_{-\frac{3\pi}{2}}^{\frac{\pi}{2}}\sqrt{2 - 2\sin\theta}\text{d}{\theta} \\
& = \sqrt{2}\int_{-\frac{3\pi}{2}}^{\frac{\pi}{2}}(\cos\dfrac{\theta}{2} - \sin\dfrac{\theta}{2})\text{d}{\theta} \\
& = 2\sqrt{2}\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}(\cos\varphi - \sin\varphi)\text{d}{\varphi} \\
& = 2\sqrt{2}(\sin\varphi + \cos\varphi)\bigg\vert_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \\
& = 8
\end{aligned}
\]
对 \(x\) 轴的静力矩为
\[
\begin{aligned}
M_{x} & = \int_{L}y\sqrt{x^2+y^2}\text{d}{l} \\
& = \sqrt{2}\int_{-\frac{3\pi}{2}}^{\frac{\pi}{2}}(\sin\theta - 1)(\cos\dfrac{\theta}{2} - \sin\dfrac{\theta}{2})\text{d}{\theta} \\
& = \sqrt{2}\int_{-\frac{3\pi}{2}}^{\frac{\pi}{2}}(2\cos^2\dfrac{\theta}{2}\sin\dfrac{\theta}{2} - 2\sin^2\dfrac{\theta}{2}\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2} - \cos\dfrac{\theta}{2})\text{d}{\theta} \\
& = 2\sqrt{2}\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}(2\cos^2\varphi\sin\varphi - 2\sin^2\varphi\cos\varphi + \sin\varphi - \cos\varphi)\text{d}{\varphi} \\
& = 2\sqrt{2}\left(-\dfrac{2}{3}\cos^2\varphi - \dfrac{2}{3}\sin^3\varphi - \cos\varphi - \sin\varphi\right)\bigg\vert_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \\
& = -\dfrac{32}{3}
\end{aligned}
\]