方企勤 第六章 多元函数积分学 第4题
📝 题目
例 4 (1) 计算重积分
$$ I = {\iiint }_{{x}^{2} + {y}^{2} + {z}^{2} \leq {R}^{2}}\frac{\mathrm{d}x\mathrm{\;d}y\mathrm{\;d}z}{\sqrt{{x}^{2} + {y}^{2} + {\left( z - h\right) }^{2}}}\;\left( {h > R}\right) ; $$
(2)写出球的单层位势
$$ u\left( {a,b,c}\right) = {\iiint }_{{x}^{2} + {y}^{2} + {z}^{2} \leq {R}^{2}} - \frac{\mathrm{d}x\mathrm{\;d}y\mathrm{\;d}z}{\sqrt{{\left( x - a\right) }^{2} + {\left( y - b\right) }^{2} + {\left( z - c\right) }^{2}}} $$
$$ \left( {{a}^{2} + {b}^{2} + {c}^{2} > {R}^{2}}\right) \text{ . } $$
💡 答案解析
解 (1) 令 $x = r\cos \theta ,y = r\sin \theta ,z = z$ ,则
$$ I = {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{-R}^{R}\mathrm{\;d}z{\int }_{0}^{\sqrt{{R}^{2} - {z}^{2}}}\frac{r\mathrm{\;d}r}{\sqrt{{r}^{2} + {\left( z - h\right) }^{2}}} $$
$$ = \pi {\int }_{-R}^{R}\mathrm{\;d}z{\int }_{0}^{\sqrt{{R}^{2} - {z}^{2}}}\frac{\mathrm{d}{r}^{2}}{\sqrt{{r}^{2} + {\left( z - h\right) }^{2}}} $$
$$ = {2\pi }{\int }_{-R}^{R}\left\lbrack {\sqrt{{R}^{2} + {h}^{2} - {2hz}} - \left( {h - z}\right) }\right\rbrack \mathrm{d}z $$
$$ = {\left. 2\pi \left\lbrack -\frac{1}{3h}{\left( {R}^{2} + {h}^{2} - 2hz\right) }^{3/2} + \frac{{\left( h - z\right) }^{2}}{2}\right\rbrack \right| }_{-R}^{R} $$
$$ = \frac{{4\pi }{R}^{3}}{3h}\text{ . } $$
(2)取坐标系 ${O\xi \eta \zeta }$ ,使 $\zeta$ 轴过(a, b, c)点,且使坐标系 ${O\xi \eta \zeta }$ 到坐标系 ${Oxyz}$ 之间的变换为正交变换,变换的行列式为 1 . 显然该变换把半径为 $R$ 的球仍变为半径为 $R$ 的球. 令 ${a}^{2} + {b}^{2} + {c}^{2} = {h}^{2}$ ,则由 (1) 知
$$ u\left( {a,b,c}\right) = {\iiint }_{{\xi }^{2} + {\eta }^{2} + {\zeta }^{2} \leq {R}^{2}}\frac{\mathrm{d}\xi \mathrm{d}\eta \mathrm{d}\zeta }{\sqrt{{\xi }^{2} + {\eta }^{2} + {\left( \zeta - h\right) }^{2}}} $$
$$ = \frac{{4\pi }{R}^{3}}{3 \cdot \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}}. $$
评注 球的单层位势相当于把球的质量集中于球心时产生的位势. 从物理意义看,设在球体均匀分布密度为 $\rho$ 的电荷,然后有一单位正电荷从点(a, b, c)移至无穷,带电球体对它所作的功等于把电荷集中于球心时对它所作的功一样.