习题4.5

第 1 题

计算下列第二类曲面积分, 其中 \(S^{+}\) 是球面 \(x^2 + y^2 + (z-R)^2 = R^2\) 的外侧.

(1) \(\oint\limits_{S^{+}} \text{d}{x}\wedge\text{d}{y}\); % 这里有问题, 应该是 \oiint

(2) \(\oint\limits_{S^{+}} z\text{d}{x}\wedge\text{d}{y}\); % 这里有问题, 应该是 \oiint

(3) \(\oint\limits_{S^{+}} z^2\text{d}{x} \wedge \text{d}{y}\). % 这里有问题, 应该是 \oiint

解

(1) 由对称性知 \(\oint\limits_{S^{+}} \text{d}{x}\wedge\text{d}{y} = 0\). % 这里有问题, 应该是 \oiint

(2) 记上半球面为 \(S_1^{+}\), 满足 \(z = R + \sqrt{R^2 - x^2 - y^2}\), \(\text{d}{x}\wedge\text{d}{y} = \text{d}{x}\text{d}{y}\), 下半球面为 \(S_2^{+}\), 满足 \(z = R - \sqrt{R^2 - x^2 - y^2}\), \(\text{d}{x} \wedge\text{d}{y} = -\text{d}{x}\text{d}{y}\). 则 \(D_{xy} = \{(x, y) | x^2 + y^2 \le R^2\}\). 换元 \(\begin{cases}x = r\cos\theta, \\ y = r\sin\theta\end{cases}\), 则 \(D_{r\theta} = \{(r, \theta) | 0 \le r \le R, 0 \le \theta \le 2\pi\}\). 因此

\[ I_1 = \int_{0}^{R}r\text{d}{r}\int_{0}^{2\pi}(R + \sqrt{R^2-r^2})\text{d}{\theta} \]

\[ I_2 = \int_{0}^{R}r\text{d}{r}\int_{0}^{2\pi}(-R + \sqrt{R^2 - r^2})\text{d}{\theta} \]

\[ \begin{aligned} I & = I_1 + I_2 \\ & = 4\pi\int_{0}^{R}r\sqrt{R^2 -r^2}\text{d}{r} \\ & = -\dfrac{4\pi}{3}(R^2 - r^2)^{\frac{3}{2}}\bigg\vert_{0}^{R} \\ & = \dfrac{4\pi R^3}{3} \end{aligned} \]

(3) 同 (2) 可知

\[ \begin{aligned} I & = I_1' + I_2' \\ & = 8\pi\int_{0}^{R}r\sqrt{R^2 -r^2}\text{d}{r} \\ & = \dfrac{8\pi R^3}{3} \end{aligned} \]

第 3 题

计算下列曲面积分.

(1) \(\iint_{S^{+}}x\text{d}{y}\wedge\text{d}{z} + y\text{d}{z}\wedge\text{d}{x} + z\text{d}{x}\wedge\text{d}{y}\), 其中 \(S^{+}\) 为平面 \(x = 0, y = 0, z = 0, x = 1, y = 1, z = 1\) 所围立方体表面的外侧;

(2) \(\iint\limits_{S^{+}} x^2\text{d}{y} \wedge \text{d}{z} + y^2\text{d}{z}\wedge\text{d}{x}+z^2\text{d}{x}\wedge\text{d}{y}\), 其中 \(S^{+}\) 是柱面 \(x^2 + y^2 = 1\) 被平面 \(z = 0, z = 3\) 所截部分的外侧;

解

(1) 设 \(S_1 = \{(x, y, z) | 0 \le x, y \le 1, z = 0\}\). 由对称性知 \(\iint_{S^{+}}x\text{d}{y}\wedge\text{d}{z} + y\text{d}{z}\wedge\text{d}{x} + z\text{d}{x}\wedge\text{d}{y} = 3\iint\limits_{S_1}\text{d}{x}\text{d}{y} = 3\).

(2) 将 \(S^{+}\) 分为三个部分: \(S_1^{+}\) 上底面, \(S_2^{+}\) 下底面, \(S_3^{+}\) 侧面.

\[ \begin{aligned} \quad & \iint\limits_{S_1^{+}}x^2\text{d}{y} \wedge \text{d}{z} + y^2\text{d}{z}\wedge\text{d}{x}+z^2\text{d}{x}\wedge\text{d}{y} \\ & = \iint\limits_{S_1^{+}}z^2\text{d}{x}\wedge\text{d}{y} \\ & = \iint\limits_{S_1^{+}}9\text{d}{x}\wedge\text{d}{y} \\ & = 9\pi \end{aligned} \]

由对称性知 \(\iint\limits_{S_2^{+}}x^2\text{d}{y} \wedge \text{d}{z} + y^2\text{d}{z}\wedge\text{d}{x}+z^2\text{d}{x}\wedge\text{d}{y} = \iint\limits_{S_3^{+}}x^2\text{d}{y} \wedge \text{d}{z} + y^2\text{d}{z}\wedge\text{d}{x}+z^2\text{d}{x}\wedge\text{d}{y} = 0\).

故 \(\iint\limits_{S^{+}} x^2\text{d}{y} \wedge \text{d}{z} + y^2\text{d}{z}\wedge\text{d}{x}+z^2\text{d}{x}\wedge\text{d}{y} = 9\pi\).

第 4 题

计算 \(\iint\limits_{S^{+}}\mathbf{A}\cdot\text{d}{\mathbf{S}}\), 其中 \(\mathbf{A} = \dfrac{x\mathbf{i} + y\mathbf{j} + z\mathbf{k}}{\sqrt{x^2 + y^2 + z^2}}\), \(S^{+}\) 是上半球面 \(z = \sqrt{R^2 - x^2 - y^2}\) 的下侧.

解

做坐标变换 \(\begin{cases}x = R\sin\varphi\cos\theta, \\ y = R\sin\varphi\sin\theta, \\ z = R\cos\varphi, \end{cases}\) 则 \(A = \dfrac{D(y, z)}{D(\varphi, \theta)} = R^2\sin^2\varphi\cos\theta\), \(B = \dfrac{D(z, x)}{D(\varphi, \theta)} = R^2\sin^2\varphi\sin\theta\), \(C = \dfrac{D(x, y)}{D(\varphi, \theta)} = R^2\sin\varphi\cos\varphi\). 积分区域为 \(D_{\varphi\theta} = \{(\varphi, \theta) | -\dfrac{\pi}{2} \le \varphi \le 0, 0 \le \theta \le 2\pi\}\). 因此

\[ \begin{aligned} \iint\limits_{S^{+}}\mathbf{A}\cdot\text{d}{\mathbf{S}} & = \iint\limits_{D_{\varphi\theta}}(\sin\varphi\cos\theta, \sin\varphi\sin\theta, \cos\varphi)\cdot(R^2\sin^2\varphi\cos\theta, R^2\sin^2\varphi\sin\theta, R^2\sin\varphi\cos\varphi)\text{d}{\varphi}\text{d}{\theta} \\ & = R^2\iint\limits_{D_{\varphi\theta}}(\sin^3\varphi\cos^2\theta + \sin^3\varphi\sin^2\theta+\sin\varphi\cos^2\varphi)\text{d}{\varphi}\text{d}{\theta} \\ & = R^2\int_{0}^{2\pi}\text{d}{\theta}\int_{-\frac{\pi}{2}}^{0}\sin\varphi\text{d}{\varphi} \\ & = -2\pi R^2 \end{aligned} \]

第 5 题

求流速场 \(\mathbf{V} = xy\mathbf{i} + yz\mathbf{j} + zx\mathbf{k}\) 由里往外穿过球面 \(x^2 + y^2 + z^2 = 1\) 在第一象限部分的流量.

解

做坐标变换 \(\begin{cases}x = \sin\varphi\cos\theta, \\ y = \sin\varphi\sin\theta, \\ z = \cos\varphi, \end{cases}\) 则 \(A = \dfrac{D(y, z)}{D(\varphi, \theta)} = \sin^2\varphi\cos\theta\), \(B = \dfrac{D(z, x)}{D(\varphi, \theta)} = \sin^2\varphi\sin\theta\), \(C = \dfrac{D(x, y)}{D(\varphi, \theta)} = \sin\varphi\cos\varphi\). 积分区域为 \(D_{\varphi\theta} = \{(\varphi, \theta) | 0 \le \varphi \le \dfrac{\pi}{2}, 0 \le \theta \le \dfrac{\pi}{2}\}\). 因此

\[ \begin{aligned} \iint\limits_{S^{+}}\mathbf{V}\cdot\text{d}{\mathbf{S}} & = \iint\limits_{D_{\varphi\theta}}(\sin^2\varphi\sin\theta\cos\theta, \sin\varphi\cos\varphi\sin\theta, \sin^2\varphi\cos^2\varphi\cos\varphi)\cdot(\sin^2\varphi\cos\theta, \sin^2\varphi\sin\theta, \sin\varphi\cos\varphi)\text{d}{\varphi}\text{d}{\theta} \\ & = \iint\limits_{D_{\varphi\theta}}(\sin^4\varphi\sin\theta\cos^2\theta + \sin^3\varphi\cos\varphi\sin^2\theta+\sin^2\varphi\cos^2\varphi\cos\theta)\text{d}{\varphi}\text{d}{\theta} \\ & = \int_{0}^{\frac{\pi}{2}}\sin^4\varphi\text{d}{\varphi}\int_{0}^{\frac{\pi}{2}}\cos^2\theta\sin\theta\text{d}{\theta} + \int_{0}^{\frac{\pi}{2}}\sin^3\varphi\cos\varphi\text{d}{\varphi}\int_{0}^{\frac{\pi}{2}}\sin^2\theta\text{d}{\theta} + \int_{0}^{\frac{\pi}{2}}\sin^2\varphi\cos^2\varphi\text{d}{\varphi}\int_{0}^{\frac{\pi}{2}}\cos\theta\text{d}{\theta} \\ & = \dfrac{3\cdot 1}{4\cdot 2}\cdot \dfrac{\pi}{2} \cdot \dfrac{1}{3} + \dfrac{1}{4} \cdot \dfrac{1}{2} \cdot \dfrac{\pi}{2} + 1\cdot\dfrac{\pi}{16} \\ & = \dfrac{3\pi}{16} \end{aligned} \]

第 7 题

\(\iint\limits_{S^{+}}(x^2 + y^2)\text{d}{x}\wedge\text{d}{y} + y^2\text{d}{y}\wedge\text{d}{z}+z^2\text{d}{z}\wedge\text{d}{x}\), 其中 \(S\) 是螺旋面 \(x = u\cos v, y = u\sin v, z = av\) 在

\[ D_{uv} = \{(u, v) | 0 \le u \le 1, 0 \le v \le 2\pi\} \]

的部分, 上侧为正.

解

\(A = \dfrac{D(y, z)}{D(u, v)} = a\sin v\), \(B = \dfrac{D(z, x)}{D(u, v)} = -a\cos v\), \(C = \dfrac{D(x, y)}{D(u, v)} = u\). 因此

\[ \begin{aligned} \quad & \iint\limits_{S^{+}}(x^2 + y^2)\text{d}{x}\wedge\text{d}{y} + y^2\text{d}{y}\wedge\text{d}{z}+z^2\text{d}{z}\wedge\text{d}{x} \\ & = \iint\limits_{D_{uv}}(u^2, u^2\sin^2v, a^2v^2)\cdot(a\sin v, -a\cos v, u)\text{d}{u}\text{d}{v} \\ & = \iint\limits_{D_{uv}}(au^2\sin v - au^2\sin^2v\cos v + a^2uv^2)\text{d}{u}\text{d}{v} \\ & = \int_{0}^{1}u^2\text{d}{u}\int_{0}^{2\pi}a\sin v\text{d}{v} + \int_{0}^{1}u^2\text{d}{u}\int_{0}^{2\pi}\sin^2v\cos v\text{d}{v} + \int_{0}^{1}u\text{d}{u}\int_{0}^{2\pi}a^2v^2\text{d}{v} \\ & = 0 + 0 + \dfrac{4\pi^3a^2}{3} \\ & = \dfrac{4\pi^3a^2}{3} \end{aligned} \]