方企勤 第六章 多元函数积分学 第3题
📝 题目
例 3 (1)求积分
$$ I = {\iint }_{{x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}}\frac{\mathrm{d}S}{\sqrt{{x}^{2} + {y}^{2} + {\left( z - h\right) }^{2}}}\;\left( {h \neq R}\right) ; $$
(2)求球面的单层位势
$$ u\left( {a,b,c}\right) = {\iint }_{{x}^{2} + {y}^{2} + {z}^{2} = {R}^{2}}\frac{\mathrm{d}S}{\sqrt{{\left( x - a\right) }^{2} + {\left( y - b\right) }^{2} + {\left( z - c\right) }^{2}}} $$
$$ \left( {{a}^{2} + {b}^{2} + {c}^{2} \neq {R}^{2}}\right) \text{ . } $$
💡 答案解析
解 (1) 令 $x = R\cos \theta \sin \varphi ,y = R\sin \theta \sin \varphi ,z = R\cos \varphi (0 \leq \theta \leq {2\pi }$ , $0 \leq \varphi \leq \pi )$ .
$$ \left\lbrack \begin{matrix} {x}_{\varphi }^{\prime } & {y}_{\varphi }^{\prime } & {z}_{\varphi }^{\prime } \\ {x}_{\theta }^{\prime } & {y}_{\theta }^{\prime } & {z}_{\theta }^{\prime } \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} R\cos \theta \cos \varphi & R\sin \theta \cos \varphi & - R\sin \varphi \\ - R\sin \theta \sin \varphi & R\cos \theta \sin \varphi & 0 \end{matrix}\right\rbrack , $$
得
$$ A = {R}^{2}\cos \theta {\sin }^{2}\varphi ,\;B = {R}^{2}\sin \theta {\sin }^{2}\varphi ,\;C = {R}^{2}\sin \varphi \cos \varphi , $$
$$ \mathrm{d}S = \sqrt{{A}^{2} + {B}^{2} + {C}^{2}}\mathrm{\;d}\theta \mathrm{d}\varphi = {R}^{2}\sin \varphi \mathrm{d}\theta \mathrm{d}\varphi , $$
所以
$$ I = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{\pi }\frac{{R}^{2}\sin \varphi \mathrm{d}\varphi }{\sqrt{{R}^{2} - {2Rh}\cos \varphi + {h}^{2}}} $$
$$ = \frac{\pi }{h}{\int }_{0}^{\pi }\frac{R\mathrm{\;d}\left( {{R}^{2} - {2Rh}\cos \varphi + {h}^{2}}\right) }{\sqrt{{R}^{2} - {2Rh}\cos \varphi + {h}^{2}}} $$
$$ = {\left. \frac{2\pi R}{h}\sqrt{{R}^{2} - {2Rh}\cos \varphi + {h}^{2}}\right| }_{0}^{\pi } $$
$$ = \frac{2\pi R}{h}\left\lbrack {\left( {R + h}\right) - \left| {R - h}\right| }\right\rbrack $$
$$ = \left\{ \begin{array}{ll} \frac{{4\pi }{R}^{2}}{h}, & h > R, \\ {4\pi R}, & 0 < h < R. \end{array}\right. $$
(2)作坐标系的正交变换,使新坐标 ${O\xi \eta \zeta }$ 的 $\zeta$ 轴通过(a, b, c) 点. 则由 (1) 知
$$ u\left( {a,b,c}\right) = {\iint }_{{\xi }^{2} + {\eta }^{2} + {\zeta }^{2} = {R}^{2}}\frac{\mathrm{d}S}{\sqrt{{\xi }^{2} + {\eta }^{2} + {\left( \zeta - h\right) }^{2}}}\left( {{h}^{2} = {a}^{2} + {b}^{2} + {c}^{2}}\right) $$
$$ = \left\{ \begin{array}{ll} \frac{{4\pi }{R}^{2}}{\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}}, & {a}^{2} + {b}^{2} + {c}^{2} > {R}^{2}, \\ {4\pi R}, & 0 < {a}^{2} + {b}^{2} + {c}^{2} < {R}^{2}. \end{array}\right. $$
评注 球面的单层位势相当于把球面质量(或电量)集中于原点时的位势一样, 在球面内部各点位势为常数.