📝 题目
例 13 试计算
$$ I = {\iiint }_{{x}^{2} + {y}^{2} + {z}^{2} = {2az}}\left( {{x}^{2} + {y}^{2} + {z}^{2}}\right) \mathrm{d}\left( {x,y,z}\right) . $$
💡 答案与解析
解 方法一 做通常的球坐标变换
$$ \left\{ \begin{array}{l} x = r\cos \theta \cos \varphi , \\ y = r\sin \theta \cos \varphi , \\ z = r\sin \varphi , \end{array}\right. $$
这里 $0 \leq \theta \leq {2\pi },0 \leq \varphi \leq \frac{\pi }{2},0 \leq r \leq {2a}\sin \varphi$ . 我们得到
$$ I = {\int }_{0}^{\frac{\pi }{2}}\mathrm{\;d}\varphi {\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{0}^{{2a}\sin \varphi }{r}^{4}\cos \varphi \mathrm{d}r $$
$$ = \frac{64}{5}\pi {a}^{5}{\int }_{0}^{\frac{\pi }{2}}{\sin }^{5}\varphi \cos \varphi \mathrm{d}\varphi = \frac{32}{15}\pi {a}^{5}. $$
方法二 (参看图 13-20)
从积分区域的形状得到启发,我们选用以(0,0, a)为极点的球坐标变换
$$ \left\{ \begin{array}{l} x = r\cos \theta \cos \varphi , \\ y = r\sin \theta \cos \varphi , \\ z = r\sin \varphi + a. \end{array}\right. $$
\begin{center} \includegraphics[max width=0.2\textwidth]{images/074.jpg} \end{center} \hspace*{3em}
图 13-20
这样求得
$$ I = {\int }_{0}^{a}\mathrm{\;d}r{\int }_{0}^{2\pi }\mathrm{d}\theta {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left( {{r}^{2} + {2ar}\sin \varphi + {a}^{2}}\right) {r}^{2}\cos \varphi \mathrm{d}\varphi $$
$$ = {4\pi }{\int }_{0}^{a}\left( {{r}^{2} + {a}^{2}}\right) {r}^{2}\mathrm{\;d}r = \frac{32\pi }{15}{a}^{5}. $$