📝 题目
例 4 计算第二型曲面积分
$$ I = {\oiint }_{{x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}}\frac{x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y}{{\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{3/2}}. $$
💡 答案与解析
解法 1 显然
$$ I = \frac{1}{{R}^{3}}{\iint }_{{x}^{2} + {y}^{2} + {x}^{2} = {R}^{2}}x\mathrm{\;d}y\mathrm{\;d}z + y\mathrm{\;d}z\mathrm{\;d}x + z\mathrm{\;d}x\mathrm{\;d}y. $$
因球面的外侧单位法向量为 $\left( {\frac{x}{R},\frac{y}{R},\frac{z}{R}}\right)$ 及
$$ \frac{x}{R}\mathrm{\;d}S = \mathrm{d}y\mathrm{\;d}z,\;\frac{y}{R}\mathrm{\;d}S = \mathrm{d}z\mathrm{\;d}x,\;\frac{z}{R}\mathrm{\;d}S = \mathrm{d}x\mathrm{\;d}y, $$
所以
$$ I = \frac{1}{{R}^{4}}{\iint }_{{x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}}\left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) \mathrm{d}S $$
$$ = \frac{1}{{R}^{2}}{\iint }_{{x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}}\mathrm{\;d}S = {4\pi }. $$
解法 2 令 $x = R\cos \theta \sin \varphi ,y = R\sin \theta \sin \varphi ,z = R\cos \varphi (0 \leq \theta \leq {2\pi }$ . $0 \leq \varphi \leq \pi )$ . 由