下册 9.3 第二型曲面积分及高斯公式 第37题

\section*{解题过程：} （1）如图 9.172 所示，记 $S$ 所用成的区域为 $\Omega . \Omega$ 在 $y z$ 平面的投影为 $\displaystyle y^{2}+z^{2} \leqslant \frac{3}{4} R^{2}$ ．由高斯公式得 $$ \begin{aligned} \iint_{S} x z^{2} \mathrm{~d} y \mathrm{~d} z+y z^{2} \mathrm{~d} z \mathrm{~d} x+z^{3} \mathrm{~d} x \mathrm{~d} y & =\iiint_{\Omega}\left(z^{2}+z^{2}+3 z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =5 \iiint_{\Omega} z^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \end{aligned} $$ 作柱面坐标变换 $y=r \cos \theta, z=r \sin \theta, x=x$ ，则 \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-361.jpg?height=1071&width=1237&top_left_y=2970&top_left_x=4365} \captionsetup{labelformat=empty} \caption{图 9.172} \end{figure} $\Omega$ 变成 $\displaystyle 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant \frac{\sqrt{3}}{2} R, R-\sqrt{R^{2}-r^{2}} \leqslant x \leqslant \sqrt{R^{2}-r^{2}}$. 于是 $$ \begin{aligned} \iiint_{\Omega} z^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z & =\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\sqrt{3}}{2} R} r(r \sin \theta)^{2} \mathrm{~d} r \int_{R-\sqrt{R^{2}-r^{2}}}^{\sqrt{R^{2}-r^{2}}} \mathrm{~d} x \\ & =\int_{0}^{2 \pi} \sin ^{2} \theta \mathrm{~d} \theta \int_{0}^{\frac{\sqrt{3}}{2} R} r^{3}\left(2 \sqrt{R^{2}-r^{2}}-R\right) \mathrm{d} r \\ & =\pi \int_{0}^{\frac{\sqrt{3}}{2} R}\left(2 \sqrt{R^{2}-r^{2}}-R\right) r^{3} \mathrm{~d} r=2 \pi \int_{0}^{\frac{\sqrt{3}}{2} R} \sqrt{R^{2}-r^{2}} r^{3} \mathrm{~d} r-\frac{9}{2^{6}} \pi R^{5} \end{aligned} $$ 令 $r=R \sin t$ ，则 $$ \begin{aligned} \int_{0}^{\frac{\sqrt{3}}{2} R} \sqrt{R^{2}-r^{2}} r^{3} \mathrm{~d} r & =\int_{0}^{\frac{\pi}{3}} R \cos t \cdot R^{3} \sin ^{3} t \cdot R \cos t \mathrm{~d} t=R^{5} \int_{0}^{\frac{\pi}{3}} \cos ^{2} t \cdot \sin ^{3} t \mathrm{~d} t \\ & =\left.R^{5}\left(-\frac{1}{3} \cos ^{3} t+\frac{1}{5} \cos ^{5} t\right)\right|_{0} ^{\frac{\pi}{3}}=R^{5}\left[\frac{1}{3}\left(1-\frac{1}{2^{3}}\right)-\frac{1}{5}\left(1-\frac{1}{2^{5}}\right)\right]=\frac{47}{15 \cdot 2^{5}} . \end{aligned} $$ 于是 $$ \iiint_{\Omega} z^{2} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=\pi R^{5}\left(2 \cdot \frac{47}{15 \cdot 2^{5}}-\frac{9}{2^{6}}\right)=\frac{53}{15 \cdot 2^{6}} \pi R^{5} $$ （2）如图 9.173 所示，补面 $\displaystyle S^{\prime}: z=0, \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leqslant 1$ ，方向取下侧．由高斯公式得 $$ \iint_{S+S^{\prime}} y^{2} z^{2} \mathrm{~d} y \mathrm{~d} z+z^{2} x^{2} \mathrm{~d} z \mathrm{~d} x+x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y=\iiint_{\Omega} 0 \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=0 $$ 又 $$ \iint_{S^{\prime}} x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y=-\iint_{D_{\mathrm{y}}} x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y=-a b \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{1} r^{5} \sin ^{2} \theta \cos ^{2} \theta \mathrm{~d} r=-\frac{1}{24} a b \pi $$ 所以 $$ \iint_{S} y^{2} z^{2} \mathrm{~d} y \mathrm{~d} z+z^{2} x^{2} \mathrm{~d} z \mathrm{~d} x+x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y=-\iint_{S^{\prime}} y^{2} z^{2} \mathrm{~d} y \mathrm{~d} z+z^{2} x^{2} \mathrm{~d} z \mathrm{~d} x+x^{2} y^{2} \mathrm{~d} x \mathrm{~d} y=\frac{1}{24} a b \pi $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-362.jpg?height=1189&width=1527&top_left_y=1913&top_left_x=1084} \captionsetup{labelformat=empty} \caption{图9．173} \end{figure} \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-362.jpg?height=1320&width=1237&top_left_y=1782&top_left_x=3536} \captionsetup{labelformat=empty} \caption{图9．174} \end{figure} （3）如图 9.174 所示，补面 $S^{\prime}: z=0, x^{2}+y^{2} \leqslant 4$ ，方向向下；补面 $S^{\prime \prime}: x+z=2, x^{2}+y^{2} \leqslant 4$ ，法向量为 $\displaystyle \frac{1}{\sqrt{2}}(1,0,1)$ ．记 $\Omega$ 为 $S+S^{\prime}+S^{\prime \prime}$ 围成的立体，在 $x O y$ 平面的投影区域为 $D_{x y}: x^{2}+y^{2} \leqslant 4$ 。由高斯公式得 $$ \iint_{S+S^{+}+S^{*}}-y \mathrm{~d} z \mathrm{~d} x+(z+1) \mathrm{d} x \mathrm{~d} y=\iiint_{\Omega}(-1+1) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=0 $$ 又 $$ \begin{aligned} \iint_{S^{\prime}}-y \mathrm{~d} z \mathrm{~d} x+(z+1) \mathrm{d} x \mathrm{~d} y & =\iint_{S^{\prime}} \mathrm{d} x \mathrm{~d} y=-\iint_{D_{\mathrm{v}}} \mathrm{~d} x \mathrm{~d} y=-4 \pi \\ \iint_{S^{\prime}}-y \mathrm{~d} z \mathrm{~d} x+(z+1) \mathrm{d} x \mathrm{~d} y & =\iint_{S^{\prime}}(z+1) \mathrm{d} x \mathrm{~d} y=\iint_{S^{*}}(z+1) \mathrm{d} x \mathrm{~d} y=\iint_{D_{\mathrm{v}}}(2-x+1) \mathrm{d} x \mathrm{~d} y \\ & =\iint_{D_{\mathrm{v}}}(3-x) \mathrm{d} x \mathrm{~d} y=\iint_{D_{\mathrm{v}}} 3 \mathrm{~d} x \mathrm{~d} y-\iint_{D_{\mathrm{v}}} x \mathrm{~d} x \mathrm{~d} y=12 \pi \end{aligned} $$ 于是 $$ \iint_{S+S^{\prime}+S^{\prime}}-y \mathrm{~d} z \mathrm{~d} x+(z+1) \mathrm{d} x \mathrm{~d} y=-12 \pi+4 \pi=-8 \pi $$