下册 9.3 第二型曲面积分及高斯公式 第3题

\section*{解题过程：} 如图9．115所示，补充圆面 $S^{\prime}: z=0, x^{2}+y^{2} \leqslant a^{2}$ ，并取下侧为正向，则它与曲面 $S$ 构成一封闭的半球，记为 $\Omega$ 。面 $S$ 在 $x y$ 平面的投影区域记为 $D_{x y}: x^{2}+y^{2} \leqslant a^{2} . \Omega$ 的体积为 $\displaystyle \iiint_{1} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{3} \pi a^{3}$ 。所以 $$ \iiint_{\Omega}\left(y^{2}+z^{2}+x^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{a} r^{2} r^{2} \sin \varphi \mathrm{~d} r=\frac{2}{5} \pi a^{5} $$ 由高斯公式得： \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-322.jpg?height=1169&width=1362&top_left_y=5359&top_left_x=4088} \captionsetup{labelformat=empty} \caption{图9．115} \end{figure} （1） $\iint_{S} x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y=\iiint_{\Omega} 3 \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z-\iint_{S^{\prime}} x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y=2 \pi a^{3}$ ． （2） $\iint_{S+S^{\prime}}\left(x+y^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y+z^{2}\right) \mathrm{d} x \mathrm{~d} z+\left(z+x^{2}\right) \mathrm{d} x \mathrm{~d} y=\iiint_{\Omega}(1+1+1) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=3 \iiint_{\Omega} \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=2 \pi a^{3}$ ． 又 $$ \begin{aligned} J & =\iint_{S^{\prime}}\left(x+y^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y+z^{2}\right) \mathrm{d} x \mathrm{~d} z+\left(z+x^{2}\right) \mathrm{d} x \mathrm{~d} y \\ & =\iint_{S^{\prime}} x^{2} \mathrm{~d} x \mathrm{~d} y=-\iint_{D_{x y}} x^{2} \mathrm{~d} x \mathrm{~d} y=-\frac{1}{2} \iint_{D_{x y}}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y=-\frac{1}{4} \pi a^{4} \end{aligned} $$ 于是 $\displaystyle \quad \iint_{S}\left(x+y^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y+z^{2}\right) \mathrm{d} x \mathrm{~d} z+\left(z+x^{2}\right) \mathrm{d} x \mathrm{~d} y=2 \pi a^{3}-J=2 \pi a^{3}+\frac{1}{4} \pi a^{4}$ ． （3） $\iint_{S} x^{2} \mathrm{~d} y \mathrm{~d} z+y^{2} \mathrm{~d} z \mathrm{~d} x+z^{2} \mathrm{~d} x \mathrm{~d} y$ $$ \begin{aligned} & =\iiint_{V}(2 x+2 y+2 z) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z-\iint_{S^{\prime}} x^{2} \mathrm{~d} y \mathrm{~d} z+y^{2} \mathrm{~d} z \mathrm{~d} x+z^{2} \mathrm{~d} x \mathrm{~d} y \\ & =2 \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{1}(r \sin \varphi \cos \theta+r \sin \varphi \sin \theta+r \cos \varphi) r^{2} \sin \varphi \mathrm{~d} r+\iint_{D_{y y}} 0 \mathrm{~d} x \mathrm{~d} y=\frac{\pi}{2} . \end{aligned} $$ （4） $\displaystyle \iint_{S} y z \mathrm{~d} x \mathrm{~d} z=\iiint_{\Omega} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z-\iint_{S^{\prime}} y z \mathrm{~d} x \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \cos \varphi \mathrm{~d} \varphi \int_{0}^{a} r^{3} \sin \varphi \mathrm{~d} r=\frac{\pi}{4} a^{4} \xlongequal{a=1} \frac{\pi}{4}$ ． （5） $\iint_{S+S^{\prime}} y z \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} x \mathrm{~d} z+\left(z^{2}+1\right) \mathrm{d} x \mathrm{~d} y$ $$ \begin{aligned} & =\iiint_{\Omega}(1+2 z) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\iiint_{\Omega} \mathrm{d} x \mathrm{~d} y \mathrm{~d} z+2 \iiint_{\Omega} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{3} \pi a^{3}+2 \iiint_{\Omega} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \\ & \iiint_{\Omega} z \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \varphi \int_{0}^{a} r \cos \varphi r^{2} \sin \varphi \mathrm{~d} r=2 \pi \cdot \frac{1}{2} \cdot \frac{1}{4} a^{4}=\frac{1}{4} \pi a^{4} \end{aligned} $$ 又 $\quad \iint_{S^{\prime}} y z \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} x \mathrm{~d} z+\left(z^{2}+1\right) \mathrm{d} x \mathrm{~d} y=\iint_{S^{\prime}} \mathrm{d} x \mathrm{~d} y=-\iint_{D_{v}} \mathrm{~d} x \mathrm{~d} y=-\pi a^{2}$ ． 所以 $\displaystyle \iint_{S} y z \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} x \mathrm{~d} z+\left(z^{2}+1\right) \mathrm{d} x \mathrm{~d} y=\frac{2}{3} \pi a^{3}+\frac{1}{2} \pi a^{4}-\left(-\pi a^{2}\right)=\frac{2}{3} \pi a^{3}+\frac{1}{2} \pi a^{4}+\pi a^{2}$ ． （6） $\iint_{S} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y$ $$ =3 \iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z-\iint_{S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y=\frac{6}{5} \pi a^{5}-0=\frac{6}{5} \pi a^{5} . $$ （7） $\displaystyle \iint_{S+S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+\left(z^{3}-a^{3}\right) \mathrm{d} x \mathrm{~d} y=3 \iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{6}{5} \pi a^{5}$ ． 又 $\quad \iint_{S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+\left(z^{3}-a^{3}\right) \mathrm{d} x \mathrm{~d} y=\iint_{S^{\prime}}-a^{3} \mathrm{~d} x \mathrm{~d} y=-\iint_{D_{\mathrm{w}}}-a^{3} \mathrm{~d} x \mathrm{~d} y=\pi a^{5}$ ， 所以 $\displaystyle \quad \iint_{S} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+\left(z^{3}-a^{3}\right) \mathrm{d} x \mathrm{~d} y=\frac{6}{5} \pi a^{5}-\pi a^{5}=\frac{1}{5} \pi a^{5}$ ． （8）记 $I=\iint_{S}\left(x^{3}+a z^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y^{3}+a x^{2}\right) \mathrm{d} z \mathrm{~d} x+\left(z^{3}+a y^{2}\right) \mathrm{d} x \mathrm{~d} y$ ． 又 $$ \iint_{S+S^{\prime}}\left(x^{3}+a z^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y^{3}+a x^{2}\right) \mathrm{d} z \mathrm{~d} x+\left(z^{3}+a y^{2}\right) \mathrm{d} x \mathrm{~d} y=3 \iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{6}{5} \pi a^{5} $$ $$ \begin{aligned} & \iint_{S^{\prime}}\left(x^{3}+a z^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y^{3}+a x^{2}\right) \mathrm{d} z \mathrm{~d} x+\left(z^{3}+a y^{2}\right) \mathrm{d} x \mathrm{~d} y=\iint_{S^{\prime}} a y^{2} \mathrm{~d} x \mathrm{~d} y=a \iint_{S^{\prime}} y^{2} \mathrm{~d} x \mathrm{~d} y=-a \iint_{D_{\mathrm{n}}} y^{2} \mathrm{~d} x \mathrm{~d} y \\ & \iint_{D_{\mathrm{n}}} y^{2} \mathrm{~d} x \mathrm{~d} y=\iint_{D_{\mathrm{w}}}(r \sin \theta)^{2} \cdot r \mathrm{~d} r \mathrm{~d} \theta=\iint_{D_{\mathrm{r}}} r^{3} \sin ^{2} \theta \mathrm{~d} r \mathrm{~d} \theta=\int_{0}^{2 \pi} \sin ^{2} \theta \mathrm{~d} \theta \cdot \int_{0}^{a} r^{3} \mathrm{~d} r \\ &=\left.\left.\frac{1}{2}\left(\theta-\frac{1}{2} \sin 2 \theta\right)\right|_{0} ^{2 \pi} \cdot\left(\frac{1}{4} r^{4}\right)\right|_{0} ^{a}=\frac{1}{4} \pi a^{4} \end{aligned} $$ 故 $$ I=\frac{6}{5} \pi a^{5}-\iint_{S_{1}}\left(x^{3}+a z^{2}\right) \mathrm{d} y \mathrm{~d} z+\left(y^{3}+a x^{2}\right) \mathrm{d} z \mathrm{~d} x+\left(z^{3}+a y^{2}\right) \mathrm{d} x \mathrm{~d} y=\frac{6}{5} \pi a^{5}+\frac{1}{4} \pi a^{5}=\frac{29}{20} \pi a^{5} $$ $$ \begin{equation*} \iint_{S+S^{\prime}} x z^{2} \mathrm{~d} y \mathrm{~d} z+\left(x^{2} y-z^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(2 x y+y^{2} z\right) \mathrm{d} x \mathrm{~d} y=\iiint_{\Omega}\left(z^{2}+x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{5} \pi a^{5} \tag{9} \end{equation*} $$ 又 $$ \begin{aligned} & \iint_{S^{\prime}} x z^{2} \mathrm{~d} y \mathrm{~d} z+\left(x^{2} y-z^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(2 x y+y^{2} z\right) \mathrm{d} x \mathrm{~d} y \\ & =\iint_{S^{\prime}} 2 x y \mathrm{~d} x \mathrm{~d} y=-\iint_{D_{x y}} 2 x y \mathrm{~d} x \mathrm{~d} y=-\int_{0}^{2 \pi} \mathrm{~d} \varphi \int_{0}^{a} r \cos \varphi r \sin \varphi r \mathrm{~d} r=0 \end{aligned} $$ 所以 $\displaystyle \iint_{S} x z^{2} \mathrm{~d} y \mathrm{~d} z+\left(x^{2} y-z^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(2 x y+y^{2} z\right) \mathrm{d} x \mathrm{~d} y=\frac{2}{5} \pi a^{5}$ ． （10） $\displaystyle \iint_{S+S^{\prime}}\left(x y^{2}-z^{3}\right) \mathrm{d} y \mathrm{~d} z+\left(y z^{2}-z^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(z x^{2}+a^{3}\right) \mathrm{d} x \mathrm{~d} y=\iiint_{\Omega}\left(y^{2}+z^{2}+x^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{5} \pi a^{5}$ ． 又 $\quad \iint_{S^{\prime}}\left(x y^{2}-z^{3}\right) \mathrm{d} y \mathrm{~d} z+\left(y z^{2}-z^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(z x^{2}+a^{3}\right) \mathrm{d} x \mathrm{~d} y=\iint_{S^{\prime}} a^{3} \mathrm{~d} x \mathrm{~d} y=-\iint_{D_{\mathrm{vy}}} a^{3} \mathrm{~d} x \mathrm{~d} y=-\pi a^{5}$ ． 所以 $\displaystyle \iint_{S}\left(x y^{2}-z^{3}\right) \mathrm{d} y \mathrm{~d} z+\left(y z^{2}-z^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(z x^{2}+a^{3}\right) \mathrm{d} x \mathrm{~d} y=\frac{2}{5} \pi a^{5}+\pi a^{5}=\frac{7}{5} \pi a^{5}$ ． （11） $\iint_{S+S^{\prime}} x\left(x^{2}+1\right) \mathrm{d} y \mathrm{~d} z+y\left(y^{2}+2\right) \mathrm{d} z \mathrm{~d} x+z\left(z^{2}+3\right) \mathrm{d} x \mathrm{~d} y$ $$ \begin{aligned} & =\iiint_{\Omega}\left(3 x^{2}+1+3 y^{2}+2+3 z^{2}+3\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=3 \iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z+6 \iiint_{\Omega} \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & \xlongequal{a=1} \frac{6 \pi}{5}+6 \cdot \frac{2 \pi}{3}=\frac{26 \pi}{5} \end{aligned} $$ 又 $\quad \iint_{S^{\prime}} x\left(x^{2}+1\right) \mathrm{d} y \mathrm{~d} z+y\left(y^{2}+2\right) \mathrm{d} z \mathrm{~d} x+z\left(z^{2}+3\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D_{\mathrm{w}}} 0 \mathrm{~d} x \mathrm{~d} y=0$ ． 所以 $\displaystyle \iint_{S} x\left(x^{2}+1\right) \mathrm{d} y \mathrm{~d} z+y\left(y^{2}+2\right) \mathrm{d} z \mathrm{~d} x+z\left(z^{2}+3\right) \mathrm{d} x \mathrm{~d} y=\frac{26 \pi}{5}$ ． （12） $\displaystyle \iint_{S+S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+\left(x^{2}+y^{2}+z^{3}\right) \mathrm{d} x \mathrm{~d} y=3 \iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{6}{5} \pi a^{5}$ ． 又 $\quad \iint_{S^{\prime}} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+\left(x^{2}+y^{2}+z^{3}\right) \mathrm{d} x \mathrm{~d} y$ $$ =\iint_{S^{\prime}}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y=-\iint_{D_{y}}\left(x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y=-\int_{0}^{2 \pi} \mathrm{~d} \theta \cdot \int_{0}^{a} r^{3} \mathrm{~d} r=-\frac{1}{2} \pi a^{4} . $$ 所以 $\displaystyle \quad \iint_{S} x^{3} \mathrm{~d} y \mathrm{~d} z+y^{3} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y=\frac{6}{5} \pi a^{5}+\frac{1}{2} \pi a^{4}$ ． （13） $\displaystyle \iint_{S+S^{\prime}}\left(x y^{2}+\frac{1}{3} x^{3}\right) \mathrm{d} y \mathrm{~d} z+y z^{2} \mathrm{~d} z \mathrm{~d} x+a^{3} \mathrm{~d} x \mathrm{~d} y=\iiint_{\Omega}\left(x^{2}+y^{2}+z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\frac{2}{5} \pi a^{5}$ ． 又 $$ \iint_{S^{\prime}}\left(x y^{2}+\frac{1}{3} x^{3}\right) \mathrm{d} y \mathrm{~d} z+y z^{2} \mathrm{~d} z \mathrm{~d} x+a^{3} \mathrm{~d} x \mathrm{~d} y=\iint_{S^{\prime}} a^{3} \mathrm{~d} x \mathrm{~d} y=-\iint_{D_{v}} a^{3} \mathrm{~d} x \mathrm{~d} y=-\pi a^{5} $$ 所以 $$ \iint_{S}\left(x y^{2}+\frac{1}{3} x^{3}\right) \mathrm{d} y \mathrm{~d} z+y z^{2} \mathrm{~d} z \mathrm{~d} x+a^{3} \mathrm{~d} x \mathrm{~d} y=\frac{2}{5} \pi a^{5}+\pi a^{5}=\frac{7}{5} \pi a^{5} . $$