习题4.3

第 1 题

计算下列第一类曲面积分.

(1) \(\iint\limits_{S}(x+y+z)\text{d}{S}\), 其中 \(S\) 是上半球面 \(x^2 + y^2 + z^2 = a^2(z \ge 0)\);

(3) \(\iint\limits_{S}\dfrac{\text{d}{S}}{(1 + x + y)^2}\), 其中 \(S\) 是四面体 \(x + y + z \le 1, x \ge 0, y \ge 0, z \ge 0\) 的边界面;

解

(1) 做坐标变换 \(\begin{cases}x = a\sin\varphi\cos\theta, \\ y = a\sin\varphi\sin\theta, \\ z = a\cos\varphi\end{cases}\), 则积分区域变为 \(D = \{(a\sin\varphi\cos\theta, a\sin\varphi\sin\theta, a\cos\varphi) | 0 \le \varphi \le \dfrac{\pi}{2}, 0 \le \theta \le 2\pi\}\). \(\text{d}{S} = a^2\sin\varphi\text{d}{\varphi}\text{d}{\theta}\). 由对称性知 \(\iint\limits_{S}(x+y)\text{d}{S} = 0\). 因此

\[ \begin{aligned} \iint\limits_{S}(x+y+z)\text{d}{S} & = \iint\limits_{S}z\text{d}{S} \\ & = \iint\limits_{S}a\cos\varphi\cdot a^2\sin\varphi\text{d}{\varphi}\text{d}{\theta} \\ & = a^3\int_{0}^{2\pi}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\sin\varphi\cos\varphi\text{d}{\varphi} \\ & = 2\pi a^3\int_{0}^{\frac{\pi}{2}}\sin\varphi\cos\varphi\text{d}{\varphi} \\ & = -\dfrac{\pi a^3}{2}\cos2\varphi \bigg\vert_{0}^{\frac{\pi}{2}} \\ & = \pi a^3 \end{aligned} \]

(3) 将 \(S\) 分解为

\[ \begin{aligned} S_1 & = \{(x, y, z) | x + y + z = 1 , x \ge 0, y \ge 0, z \ge 0\}, \\ S_2 & = \{(x, y, z) | x + y + z \le 1, x = 0, y \ge 0, z \ge 0\}, \\ S_3 & = \{(x, y, z) | x + y + z \le 1, x \ge 0, y = 0, z \ge 0\}, \\ S_4 & = \{(x, y, z) | x + y + z \le 1, x \ge 0, y \ge 0, z = 0\}. \end{aligned} \]

对于 \(S_1\), 由于 \(x + y + z = 1\), 所以 \(\text{d}{S} = \sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2}\text{d}{x}\text{d}{y} = \sqrt{3}\text{d}{x}\text{d}{y}\).

\[ \begin{aligned} \iint\limits_{S_1}\dfrac{\text{d}{S}}{(1 + x + y)^2} & = \iint\limits_{S_1}\dfrac{\sqrt{3}\text{d}{x}\text{d}{y}}{(1 + x + y)^2} \\ & = \sqrt{3}\int_{0}^{1}\text{d}{x}\int_{0}^{1 - x}\dfrac{\text{d}{y}}{(1 + x + y)^2} \\ & = \sqrt{3}\int_{0}^{1}\left(\dfrac{1}{1 + x} - \dfrac{1}{2}\right)\text{d}{x} \\ & = \sqrt{3}\left(\ln(x + 1) - \dfrac{x}{2}\right)\bigg\vert_{0}^{1} \\ & = \sqrt{3}\left(\ln 2 - \dfrac{1}{2}\right) \end{aligned} \]

同理 \(\iint\limits_{S_4}\dfrac{\text{d}{S}}{(1 + x + y)^2} = \ln2 - \dfrac{1}{2}\), 再考虑 \(S_2\):

\[ \begin{aligned} \iint\limits_{S_2}\dfrac{\text{d}{S}}{(1 + x + y)^2} & = \iint\limits_{S_2}\dfrac{\text{d}{y}\text{d}{z}}{(1 + y)^2} \\ & = \int_{0}^{1}\text{d}{y}\int_{0}^{1 - y}\dfrac{\text{d}{z}}{(1 + y)^2} \\ & = \int_{0}^{1}\dfrac{1 - x}{(1 + x)^2}\text{d}{x} \\ & = -\left(\ln(1 +x) + \dfrac{2}{1 + x}\right)\bigg\vert_{0}^{1} \\ & = 1 - \ln 2 \end{aligned} \]

同理 \(\iint\limits_{S_3}\dfrac{\text{d}{S}}{(1 + x + y)^2} = 1 - \ln 2\). 因此总和为

\[ \begin{aligned} S & = S_1 + S_2 + S_3 + S_4 \\ & = \sqrt{3}\left(\ln 2 - \dfrac{1}{2}\right) + (1 - \ln 2) + (1 - \ln 2) + \ln2 - \dfrac{1}{2} \\ & = \dfrac{3 - \sqrt{3}}{2} + (\sqrt{3} - 1)\ln 2 \end{aligned} \]

第 2 题

计算圆柱面 \(x^2 + y^2 = ax\) 被球面 \(x^2 + y^2 + z^2 = a^2\) 所截部分的面积 \((a > 0)\).

解

进行坐标变换 \(\begin{cases}x = \dfrac{a}{2} + \dfrac{a}{2}\cos\theta, \\ y = \dfrac{a}{2}\sin\theta,\end{cases}\) 则积分区域变为 \(L = \{(x, y) | x = \dfrac{a}{2} + \dfrac{a}{2}\cos\theta, y = \dfrac{a}{2}\sin\theta, 0 \le \theta \le 2\pi\}\). 同时 \(\text{d}{l} = \dfrac{a}{2}\text{d}{\theta},z^2 = a^2 - ax = \dfrac{a^2}{2}(1-\cos\theta)\). 不妨只考虑 \(z > 0\) 的部分.

\[ \begin{aligned} \int_{L}z\text{d}{l} & = \int_{0}^{2\pi}\sqrt{\dfrac{a^2}{2}(1 - \cos\theta)}\cdot\dfrac{a}{2}\text{d}{\theta} \\ & = \dfrac{a^2}{2}\int_{0}^{2\pi}\sin\dfrac{\theta}{2}\text{d}{\theta} \\ & = -a^2\cos\dfrac{\theta}{2}\bigg\vert_{0}^{2\pi} \\ & = 2a^2 \end{aligned} \]

由对称性, \(z < 0\) 的区域所截面积也为 \(2a^2\), 因此全面积为 \(4a^2\).

第 3 题

求抛物面 \(2z = x^2 + y^2\) 在 \(z \in [0, 1]\) 部分的质量, 其中质量面密度为 \(\sigma = z\).

解

抛物面的面积元素为

\[ \begin{aligned} \text{d}{S} & = \sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2}\text{d}{x}\text{d}{y} \\ & = \sqrt{1 + x^2 + y^2}\text{d}{x}\text{d}{y} \\ \end{aligned} \]

进行坐标变换 \(\begin{cases}x = r\cos\theta, \\ y = r\sin\theta,\end{cases}\) 则有

\[ \begin{aligned} \iint\limits_{0 \le x^2 + y^2 \le 2}\sigma\text{d}{S} & = \iint\limits_{0 \le x^2 + y^2 \le 2}\dfrac{x^2 + y^2}{2}\cdot\sqrt{1 + x^2 + y^2}\text{d}{x}\text{d}{y} \\ & = \int_{0}^{\sqrt{2}}\text{d}{r}\int_{0}^{2\pi}\dfrac{r^3}{2}\sqrt{1 + r^2}\text{d}{r} \\ & = \pi\int_{0}^{\sqrt{2}}r^3\sqrt{1 + r^2}\text{d}{r} \\ & = \dfrac{\pi}{2}\int_{0}^{2}u\sqrt{1 + u}\text{d}{u} \\ & = \dfrac{\pi}{15}(3u-2)(u+1)^{\frac{3}{2}}\bigg\vert_{0}^{2} \\ & = \dfrac{12\sqrt{3} + 2}{15}\pi \end{aligned} \]

第 8 题

求锥面 \(z = \sqrt{x^2 + y^2}\) 在柱面 \(z^2 = 2x\) 内的面积.

解

锥面的面积元素为

\[ \begin{aligned} \text{d}{S} & = \sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2}\text{d}{x}\text{d}{y} \\ & = \sqrt{2}\text{d}{x}\text{d}{y} \\ \end{aligned} \]

锥面被柱面所截部分记作 \(S\) (其面积也记为 \(S\)), 它在 \(xy\) 平面上的投影区域是圆域 \(D_{xy} = \{(x, y) | x^2 + y^2 \le 2x\}\), 所以

\[ S = \iint\limits_{S}1\text{d}{S} = \iint_{D_{xy}}\sqrt{2}\text{d}{x}\text{d}{y} = \sqrt{2}\pi \]