下册 9.3 第二型曲面积分及高斯公式 第25题

\section*{解题过程：} 由高斯公式得： （1）如图 9.158 所示， $$ \begin{aligned} & \iint_{S} x z \mathrm{~d} y \mathrm{~d} z+y z \mathrm{~d} z \mathrm{~d} x+z \sqrt{x^{2}+y^{2}} \mathrm{~d} x \mathrm{~d} y \\ & =\iiint_{\Omega}\left(2 z+\sqrt{x^{2}+y^{2}}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =\int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{4}} \mathrm{~d} \varphi \int_{a}^{2 a}(2 r \cos \varphi+r \sin \varphi) \cdot r^{2} \sin \varphi \mathrm{~d} r \\ & =2 \pi \int_{0}^{\frac{\pi}{4}} \mathrm{~d} \varphi \int_{a}^{2 a} r^{3}(2 \cos \varphi+\sin \varphi) \sin \varphi \mathrm{d} r \\ & =2 \pi \frac{1}{4}\left[(2 a)^{4}-a^{4}\right] \int_{0}^{\frac{\pi}{4}}(2 \cos \varphi+\sin \varphi) \sin \varphi \mathrm{d} \varphi \\ & =2 \pi \frac{1}{4}\left[(2 a)^{4}-a^{4}\right] \int_{0}^{\frac{\pi}{4}}\left(2 \sin \varphi \cos \varphi+\frac{1-\cos 2 \varphi}{2}\right) \mathrm{d} \varphi \\ & =2 \pi \frac{1}{4}\left[(2 a)^{4}-a^{4}\right] \int_{0}^{\frac{\pi}{4}}\left(2 \sin \varphi \cos \varphi+\frac{1-\cos 2 \varphi}{2}\right) \mathrm{d} \varphi \\ & =2 \pi \frac{1}{4}\left[(2 a)^{4}-a^{4}\right]\left(\frac{1}{2}+\frac{1}{2} \frac{\pi}{4}-\frac{1}{4}\right)=\frac{15(\pi-2) \pi a^{4}}{16} . \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-352.jpg?height=1230&width=1486&top_left_y=2500&top_left_x=4144} \captionsetup{labelformat=empty} \caption{图 9.158} \end{figure} （2） $\displaystyle \iint_{S} x^{3} \mathrm{~d} y \mathrm{~d} z+\left(\frac{y}{z^{2}}+y^{3}\right) \mathrm{d} z \mathrm{~d} x+\left(\frac{1}{z}+z^{3}\right) \mathrm{d} x \mathrm{~d} y$ $$ \begin{aligned} & =\iiint_{\Omega}\left(3 x^{2}+3 y^{2}+3 z^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z \\ & =3 \int_{0}^{2 \pi} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{4}} \mathrm{~d} \varphi \int_{1}^{2} r^{4} \sin \varphi \mathrm{~d} r=\frac{186}{10}(2-\sqrt{2}) \pi \end{aligned} $$ （3）如图 9.159 所示， $$ \begin{aligned} & \iint_{S} y^{2} z \mathrm{~d} x \mathrm{~d} y+x z \mathrm{~d} y \mathrm{~d} z+x^{2} y \mathrm{~d} x \mathrm{~d} z \\ & =\iiint_{\Omega}\left(z+x^{2}+y^{2}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=\int_{0}^{\frac{1}{2} \pi} \mathrm{~d} \theta \int_{0}^{1} r \mathrm{~d} r \int_{0}^{r^{2}}\left(z+r^{2}\right) \mathrm{d} z=\frac{1}{8} \pi \end{aligned} $$ \begin{figure} \includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/468aa2e6-2fe6-41d5-b96d-487ad792954d-352.jpg?height=1272&width=1182&top_left_y=6375&top_left_x=4323} \captionsetup{labelformat=empty} \caption{图 9.159} \end{figure}