📝 题目
例 6 设 ${B}_{n}\left( r\right)$ 表示半径为 $r$ 的 $n$ 维闭球体,试计算 ${B}_{n}\left( r\right)$ 的体积 ${V}_{n}\left( r\right)$ .
💡 答案与解析
解 考察 $n = 1,2,3$ 的情形,可以猜测 ${V}_{n}\left( r\right)$ 具有如下形式的表示:
$$ {V}_{n}\left( r\right) = {\alpha }_{n}{r}^{n}. $$
我们来证明这公式并推导系数 ${\alpha }_{n}$ 的递推关系. 显然有
$$ {V}_{1}\left( r\right) = {2r},\;{\alpha }_{1} = 2. $$
对一般情形有
$$ {V}_{n}\left( r\right) = {\int }_{-r}^{r}\mathrm{\;d}{x}_{n}\mathop{\int }\limits_{{{B}_{n - 1}\left( \sqrt{{r}^{2} - {x}_{n}^{2}}\right) }}\mathrm{\;d}\left( {{x}_{1},\cdots ,{x}_{n - 1}}\right) $$
$$ = {\int }_{-r}^{r}{\alpha }_{n - 1}{\left( {r}^{2} - {x}_{n}^{2}\right) }^{\frac{n - 1}{2}}\mathrm{\;d}{x}_{n} $$
$$ = 2{\alpha }_{n - 1}{\int }_{0}^{r}{\left( {r}^{2} - {x}_{n}^{2}\right) }^{\frac{n - 1}{2}}\mathrm{\;d}{x}_{n} $$
$$ = \left( {2{\alpha }_{n - 1}{\int }_{0}^{\frac{\pi }{2}}{\sin }^{n}t\mathrm{\;d}t}\right) {r}^{n} $$
$$ = {\alpha }_{n}{r}^{n}\text{ . } $$
我们看到: 系数 ${\alpha }_{n}$ 满足递推关系
$$ {\alpha }_{n} = 2{\alpha }_{n - 1}{\int }_{0}^{\frac{\pi }{2}}{\sin }^{n}t\mathrm{\;d}t,\;{\alpha }_{1} = 2. $$
从这递推关系可以求得
$$ {\alpha }_{n} = {2}^{n}{\beta }_{n}{\beta }_{n - 1}\cdots {\beta }_{1}, $$
其中 ${\beta }_{m}$ 表示积分
$$ {\int }_{0}^{\frac{\pi }{2}}{\sin }^{m}t\mathrm{\;d}t $$
根据第九章 §6 中的计算,
$$ {\beta }_{m} = \left\{ \begin{array}{ll} \frac{\left( {m - 1}\right) !!}{m!!}, & \text{ 对奇数 }m, \\ \frac{\left( {m - 1}\right) !!}{m!!} \cdot \frac{\pi }{2}, & \text{ 对偶数 }m. \end{array}\right. $$
我们求得:
$$ {\alpha }_{1} = 2,\;{\alpha }_{2} = \pi ,\;{\alpha }_{3} = \frac{4}{3}\pi , $$
$$ {\alpha }_{4} = \frac{1}{2}{\pi }^{2},\;{\alpha }_{5} = \frac{8}{15}{\pi }^{2},\;\cdots ; $$
$$ {V}_{1}\left( r\right) = {2r},\;{V}_{2}\left( r\right) = \pi {r}^{2},\;{V}_{3}\left( r\right) = \frac{4}{3}\pi {r}^{3}, $$
$$ {V}_{4}\left( r\right) = \frac{1}{2}{\pi }^{2}{r}^{4},\;{V}_{5}\left( r\right) = \frac{8}{15}{\pi }^{2}{r}^{5},\;\cdots . $$
${\alpha }_{n}$ 与 ${V}_{n}\left( r\right)$ 的一般表示式为:
$$ {\alpha }_{2k} = {2}^{2k}\frac{1}{\left( {2k}\right) !!}{\left( \frac{\pi }{2}\right) }^{k} = \frac{{\pi }^{k}}{k!}, $$
$$ {\alpha }_{{2k} + 1} = {2}^{{2k} + 1}\frac{1}{\left( {{2k} + 1}\right) !!}{\left( \frac{\pi }{2}\right) }^{k} $$
$$ = \frac{{2}^{k + 1}{\pi }^{k}}{\left( {{2k} + 1}\right) !!}; $$
$$ {V}_{2k}\left( r\right) = \frac{{\pi }^{k}}{k!}{r}^{2k}, $$
$$ {V}_{{2k} + 1}\left( r\right) = \frac{{2}^{k + 1}{\pi }^{k}}{\left( {{2k} + 1}\right) !!}{r}^{{2k} + 1}. $$