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求解 $\iint_{\Sigma} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\left(\sqrt{x^2+y^2+z^2}\right)^3}$, 其中 $\Sigma: x^2+y^2+\frac{z^2}{2}=1$ 当中 $z \geqslant-\frac{1}{2}$ 的部分, 取外侧。

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答案：

解析:
记 $\Sigma_1:\left\{\begin{array}{l}x^2+y^2 \leqslant \frac{7}{8} \\ z=-\frac{1}{2}\end{array}\right.$, 取下侧; $\Sigma_2: x^2+y^2+z^2=\delta^2\left(\delta \rightarrow 0^{+}\right)$, 取内侧。并且记 $\Sigma, \Sigma_1, \Sigma_2$ 围成区域 $\Omega$ 的外侧。
并记 $(P, Q, R)=\frac{(x, y, z)}{\left(\sqrt{x^2+y^2+z^2}\right)^3} \stackrel{r=\sqrt{x^2+y^2+z^2}}{=} \frac{(x, y, z)}{r^3}$

所以 $\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=\frac{1 \cdot r^3-x \cdot 3 r^2 \cdot \frac{x}{r}}{r^6}+\frac{1 \cdot r^3-y \cdot 3 r^2 \cdot \frac{y}{r}}{r^6}+\frac{1 \cdot r^3-z \cdot 3 r^2 \cdot \frac{z}{r}}{r^6}=0$ 所以由高斯公式得 $\iint_{\Sigma+\Sigma_1+\Sigma_2} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}}=\iiint_{\Omega}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right) \mathrm{d} x \mathrm{~d} y \mathrm{~d} z=0$ 所以原式 $=\iint_{\Sigma+\Sigma_1+\Sigma_2} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}}-\iint_{\Sigma_1} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}}$
$$
\begin{aligned}
& -\iint_{\Sigma_2} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}} \\
= & -\iint_{\Sigma_1} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}}-\iint_{\Sigma_2} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}} \\
= & -\iint_{\Sigma_1} \frac{x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+z^2\right)^3}}+\iiint_{x^2+y^2+z^2 \leqslant \delta^2} \frac{1+1+1}{\delta^3} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \\
= & \iint_{x^2+y^2 \leqslant \frac{7}{8}} \frac{\left(-\frac{1}{2}\right) \mathrm{d} x \mathrm{~d} y}{\sqrt{\left(x^2+y^2+\frac{1}{4}\right)^3}}+4 \pi=-\frac{1}{2} \int_0^{2 \pi} \mathrm{d} \theta \int_0^{\sqrt{\frac{7}{8}}} \frac{r}{\sqrt{\left(r^2+\frac{1}{4}\right)^3}} \mathrm{~d} r+4 \pi \\
= & -\frac{\pi}{2} \int_0^{\sqrt{\frac{7}{8}}} \frac{\mathrm{d}\left(r^2+\frac{1}{4}\right)}{\sqrt{\left(r^2+\frac{1}{4}\right)^3}+4 \pi}=\left.\pi \cdot \frac{1}{\sqrt{r^2+\frac{1}{4}}}\right|_{r=0} ^{r=\sqrt{\frac{7}{8}}}+4 \pi=\pi\left(\frac{1}{\sqrt{\frac{9}{8}}}-\frac{1}{\frac{1}{2}}\right)+4 \pi \\
= & \frac{2 \sqrt{2}}{3} \pi+2 \pi
\end{aligned}
$$

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