下册 9.3 第二型曲面积分及高斯公式 第48题

\section*{解题过程：} 由 $\displaystyle \cos (\boldsymbol{r}, \boldsymbol{n})=\frac{\boldsymbol{r} \cdot \boldsymbol{n}}{|\boldsymbol{n} \| \boldsymbol{r}|}=\frac{1}{r}(x \cos \alpha+y \cos \beta+z \cos \gamma)$ 知 $$ \iint_{S} \frac{1}{r^{2}} \cos (r, n) \mathrm{d} S=\iint_{S} \frac{1}{r^{3}}(x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y) $$ 作小球 $S^{\prime}: x^{2}+y^{2}+z^{2}=\varepsilon^{2}$ ，使其完全包含在 $S$ 内，取外侧，记 $S_{-}^{\prime}$ 为小球内侧，$\Omega$ 为 $S$ 与 $S_{-}^{\prime}$ 之间的区域．$\Omega^{\prime}$ 为 $S_{-}^{\prime}$ 为边的球域． 因为 $\displaystyle \frac{\partial}{\partial x}\left(\frac{x}{r^{3}}\right)=\frac{r^{2}-3 x^{2}}{r^{5}}, \frac{\partial}{\partial y}\left(\frac{y}{r^{3}}\right)=\frac{r^{2}-3 y^{2}}{r^{5}}, \frac{\partial}{\partial z}\left(\frac{z}{r^{3}}\right)=\frac{r^{2}-3 z^{2}}{r^{5}}$ ，所以 $$ \frac{\partial}{\partial x}\left(\frac{x}{r^{3}}\right)+\frac{\partial}{\partial y}\left(\frac{y}{r^{3}}\right)+\frac{\partial}{\partial z}\left(\frac{z}{r^{3}}\right)=\frac{r^{2}-3 x^{2}}{r^{5}}+\frac{r^{2}-3 y^{2}}{r^{5}}+\frac{r^{2}-3 z^{2}}{r^{5}}=0 . $$ 由高斯公式得 $\displaystyle \oiint_{S^{+} S^{\prime}} \frac{\cos (r, n)}{r^{2}} \mathrm{~d} S=0$ 。于是 $$ \begin{aligned} \oiint_{S} \frac{\cos (r, n)}{r^{2}} \mathrm{~d} S & =-\oiint_{S^{\prime}} \frac{\cos (r, n)}{r^{2}} \mathrm{~d} S=\oiint_{S^{\prime}} \frac{\cos (r, n)}{r^{2}} \mathrm{~d} S \\ & =\frac{1}{\varepsilon^{3}} \iint_{S^{\prime}}(x \mathrm{~d} y \mathrm{~d} z+y \mathrm{~d} z \mathrm{~d} x+z \mathrm{~d} x \mathrm{~d} y)=\frac{1}{\varepsilon^{3}} \iiint_{\Omega^{\prime}} 3 \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z=4 \pi \end{aligned} $$