📝 题目
例 1 计算第二型曲面积分
$$ I = {\iint }_{S}\frac{x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y}{{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{3/2}}. $$
(1) $S : {x}^{2} + {y}^{2} + {z}^{2} = {\varepsilon }^{2}$ ;
(2) $S$ 是不含原点在其内部的光滑闭曲面；
(3) $S$ 是含原点在其内部的光滑闭曲面.
💡 答案与解析
解 (1) $\displaystyle{I = \frac{1}{{\varepsilon }^{3}}{\iint }_{S}x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y}$
$$ = \frac{1}{{\varepsilon }^{3}}{\iint }_{V}\left( {1 + 1 + 1}\right) \mathrm{d}x\mathrm{\;d}y\mathrm{\;d}z = \frac{3}{{\varepsilon }^{3}} \cdot \frac{4}{3}\pi {\varepsilon }^{3} = {4\pi }. $$
(2) $P = \frac{x}{{r}^{3}},Q = \frac{y}{{r}^{3}},R = \frac{z}{{r}^{3}},\mathbf{r} = x\mathbf{i} + y\mathbf{i} + z\mathbf{k},r = \left| \mathbf{r}\right|$ ,
$$ \frac{\partial P}{\partial x} = \frac{1}{{r}^{3}} - \frac{3{x}^{2}}{{r}^{5}},\;\frac{\partial Q}{\partial y} = \frac{1}{{r}^{3}} - \frac{3{y}^{2}}{{r}^{5}}, $$
$$ \frac{\partial R}{\partial z} = \frac{1}{{r}^{3}} - \frac{3{z}^{2}}{{r}^{5}}, $$
所以
$$ I = {\iint }_{S}\frac{x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y}{{r}^{3}} $$
$$ = {\iiint }_{V}\left( {\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}}\right) \mathrm{d}x\mathrm{\;d}y\mathrm{\;d}z $$
$$ = {\iiint }_{V}0\mathrm{\;d}x\mathrm{\;d}y\mathrm{\;d}z = 0. $$
(3)由于函数 $P,Q,R$ 及其偏导数在 $S$ 所围区域上不连续(在原点处不连续),为此作半径充分小的球面 ${S}_{\epsilon } : {x}^{2} + {y}^{2} + {z}^{2} = {\varepsilon }^{2}$ ,使 ${S}_{\epsilon }$ 在 $S$ 所围区域内,记 ${S}_{\varepsilon }$ 与 $S$ 间的区域为 ${V}_{\varepsilon }$ . 注意
$$ I = {\iint }_{S}\frac{x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y}{{r}^{3}} = {\iint }_{S}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{3}}\mathrm{\;d}S, $$
并用 ${S}_{\varepsilon }^{ - }$ 表示取内法线方向的定向曲面. 则由奥氏公式得
$$ {\iint }_{S}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{3}}\mathrm{\;d}S + {\iint }_{{S}_{e}^{ - }}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{3}}\mathrm{\;d}S = {\iiint }_{V}0\mathrm{\;d}x\mathrm{\;d}y\mathrm{\;d}z = 0, $$
所以
$$ I = {\iint }_{S}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{3}}\mathrm{\;d}S = - {\iint }_{{S}_{e}^{ - }}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{3}}\mathrm{\;d}S = {\iint }_{{S}_{e}}\frac{\mathbf{r} \cdot \mathbf{n}}{{r}^{3}}\mathrm{\;d}S = {4\pi }. $$