下册 9.3 第二型曲面积分及高斯公式 第10题

\section*{解题过程：} 如图9．127所示，补面 $\displaystyle S^{\prime}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1, z=0$ ，方向向下。 $S^{\prime}+S$ 为一封闭的半球面，所围空间记为 $\Omega$ ．曲面 $S$ 在 $x y$ 平面的投影区域记为 $$ D_{x y}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1 . $$ 作广义球坐标变换 $x=\operatorname{ar} \sin \varphi \cos \theta, y=\operatorname{br} \sin \varphi \sin \theta, z=c r \cos \varphi, \Omega$ 可表为 \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-332.jpg?height=1127&width=1348&top_left_y=1091&top_left_x=4095} \captionsetup{labelformat=empty} \caption{图 9.127} \end{figure} $$ 0 \leqslant \varphi \leqslant \frac{\pi}{2}, 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r<1, J(r, \varphi, \theta)=a b c r^{2} \sin \varphi . $$ 所以 $$ \iiint_{\Omega} \frac{x}{a} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=a b c \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{1} r \cos \theta \sin \varphi r^{2} \sin \varphi \mathrm{~d} r=0 . $$ 同理 $$ \iiint_{\Omega} \frac{y}{b} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=0 . $$ $$ \iiint_{\Omega} \frac{z}{c} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=a b c \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{1} r \cos \varphi r^{2} \sin \varphi \mathrm{~d} r=\frac{\pi}{4} a b c . $$ $\displaystyle \iiint_{\Omega}\left(\frac{x}{a}\right)^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\iiint_{\Omega}\left(\frac{y}{b}\right)^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\iiint_{\Omega}\left(\frac{z}{c}\right)^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=a b c \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{1}(r \cos \theta \sin \varphi)^{2} r^{2} \sin \varphi \mathrm{~d} r=\frac{2}{15} \pi a b c$. 由高斯公式得： （1） $\displaystyle \iint_{S+S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z=3 \iiint_{\Omega} x^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=3 a^{2} \iiint_{\Omega}\left(\frac{x}{a}\right)^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{-6}{15} \pi a^{3} b c$ ． 又 $\iint_{S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z=0$ ．所以 $\displaystyle \iint_{S} x^{3} \mathrm{~d} y \mathrm{~d} z=\frac{6}{15} \pi a^{3} b c$ ． （2） $\iint_{S+S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y$ $$ \begin{aligned} & =\iiint_{\Omega}\left(3 x^{2}+3 y^{2}+3 z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=3 \iiint_{\Omega} x^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z+3 \iiint_{\Omega} y^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z+3 \iiint_{\Omega} z^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \\ & =\left(\frac{6}{15} \pi a^{3} b c+\frac{6}{15} \pi a b^{3} c+\frac{6}{15} \pi a b c^{3}\right)=\frac{2 \pi}{5} a b c\left(a^{2}+b^{2}+c^{2}\right) \end{aligned} $$ 又 $\quad \iint_{S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y=0$ ， 所以 $\displaystyle \iint_{S} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y=\frac{2 \pi}{5} a b c\left(a^{2}+b^{2}+c^{2}\right)$ ． （3） $\displaystyle \iint_{S+S^{\prime}} \frac{x z^{2}}{c^{2}} \mathrm{~d} y \mathrm{~d} z+\frac{x^{2} y-z^{2}}{a^{2}} \mathrm{~d} z \mathrm{~d} x+\frac{2 \sin \left(x^{2} y\right)+y^{2} z}{b^{2}} \mathrm{~d} x \mathrm{~d} y=\iiint_{\Omega}\left[\left(\frac{z}{c}\right)^{2}+\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}\right] \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{6}{15} \pi a b c$ ． 又 $\displaystyle \iint_{S^{\prime}} \frac{x z^{2}}{c^{2}} \mathrm{~d} y \mathrm{~d} z+\frac{x^{2} y-z^{2}}{a^{2}} \mathrm{~d} z \mathrm{~d} x+\frac{2 \sin \left(x^{2} y\right)+y^{2} z}{b^{2}} \mathrm{~d} x \mathrm{~d} y=\frac{2}{b^{2}} \iint_{S^{\prime}} \sin \left(x^{2} y\right) \mathrm{d} x \mathrm{~d} y=-\frac{2}{b^{2}} \iint_{D_{x y}} \sin \left(x^{2} y\right) \mathrm{d} x \mathrm{~d} y=0$ ． 所以 $\displaystyle \iint_{S} \frac{x z^{2}}{c^{2}} \mathrm{~d} y \mathrm{~d} z+\frac{x^{2} y-z^{2}}{a^{2}} \mathrm{~d} z \mathrm{~d} x+\frac{2 \sin \left(x^{2} y\right)+y^{2} z}{b^{2}} \mathrm{~d} x \mathrm{~d} y=\frac{6}{15} \pi a b c$ ． （4） $\displaystyle \iint_{S+S^{\prime}} y z \mathrm{~d} z \mathrm{~d} x=\iiint_{\Omega} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=c \iiint_{\Omega} \frac{z}{c} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{\pi}{4} a b c^{2}$ 。又 $\iint_{S^{\prime}} y z \mathrm{~d} z \mathrm{~d} x=0$ ，所以 $\displaystyle \iint_{S} y z \mathrm{~d} z \mathrm{~d} x=\frac{\pi}{4} a b c^{2}$ 。 （5） $\displaystyle \iint_{S+S^{\prime}}\left(\lambda \frac{x^{2}}{a^{2}}+\mu \frac{y^{2}}{b^{2}}+v \frac{z^{2}}{c^{2}}\right) \mathrm{d} S$ $$ \begin{aligned} & =\iint_{S+S^{\prime}} \frac{x^{2}}{a^{2}} \mathrm{~d} y \mathrm{~d} z+\frac{y^{2}}{b^{2}} \mathrm{~d} z \mathrm{~d} x+\frac{z^{2}}{c^{2}} \mathrm{~d} x \mathrm{~d} y=\iiint_{\Omega}\left(2 \frac{x}{a^{2}}+2 \frac{y}{b^{2}}+2 \frac{z}{c^{2}}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =\iiint_{\Omega} 2 \frac{z}{c^{2}} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{c} \iiint_{\Omega} \frac{z}{c} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{c} \cdot \frac{\pi}{4} a b c=\frac{\pi a b}{2} \end{aligned} $$ 又 $\displaystyle \quad \iint_{S^{\prime}}\left(\lambda \frac{x^{2}}{a^{2}}+\mu \frac{y^{2}}{b^{2}}+v \frac{z^{2}}{c^{2}}\right) \mathrm{d} S=\iint_{S^{\prime}} \frac{x^{2}}{a^{2}} \mathrm{~d} y \mathrm{~d} z+\frac{y^{2}}{b^{2}} \mathrm{~d} z \mathrm{~d} x+\frac{z^{2}}{c^{2}} \mathrm{~d} x \mathrm{~d} y=0$ ． 所以 $\displaystyle \iint_{S}\left(\lambda \frac{x^{2}}{a^{2}}+\mu \frac{y^{2}}{b^{2}}+v \frac{z^{2}}{c^{2}}\right) \mathrm{d} S=\frac{\pi a b}{2}$ ．