第13章 重积分 · 第5题

例题

📝 题目

例 5 如下形状的集合被称为 $n$ 维单纯形:

$$ {C}_{n}\left( r\right) = \left\{ {\left( {{x}_{1},\cdots ,{x}_{n}}\right) \in {\mathbb{R}}^{n}\left| {\;\begin{array}{l} {x}_{1} \geq 0,\cdots ,{x}_{n} \geq 0, \\ {x}_{1} + \cdots + {x}_{n} \leq r \end{array}}\right. }\right\} . $$

试计算 ${C}_{n}\left( r\right)$ 的体积 ${W}_{n}\left( r\right)$ .

💡 答案与解析

解 我们来归纳 ${W}_{n}\left( r\right)$ 的一般公式. 首先, ${W}_{1}\left( r\right)$ 与 ${W}_{2}\left( r\right)$ 很容易求得:

$$ {W}_{1}\left( r\right) = {\int }_{0}^{r}\mathrm{\;d}{x}_{1} = r, $$

$$ {W}_{2}\left( r\right) = {\int }_{0}^{r}\mathrm{\;d}{x}_{2}{\int }_{{C}_{1}\left( {r - {x}_{2}}\right) }\mathrm{d}{x}_{1} $$

$$ = {\int }_{0}^{r}\left( {r - {x}_{2}}\right) \mathrm{d}{x}_{2} = \frac{1}{2}{r}^{2}. $$

假设对任何 $r \geq 0$ 已经求得

$$ {W}_{n - 1}\left( r\right) = \frac{1}{\left( {n - 1}\right) !}{r}^{n - 1}, $$

那么就有

$$ {W}_{n}\left( r\right) = {\int }_{0}^{r}\mathrm{\;d}{x}_{n}{\int }_{{C}_{n - 1}\left( {r - {x}_{n}}\right) }\mathrm{\;d}\left( {{x}_{1},\cdots ,{x}_{n - 1}}\right) $$

$$ = {\int }_{0}^{r}{W}_{n - 1}\left( {r - {x}_{n}}\right) \mathrm{d}{x}_{n} $$

$$ = {\int }_{0}^{r}\frac{1}{\left( {n - 1}\right) !}{\left( r - {x}_{n}\right) }^{n - 1}\mathrm{\;d}{x}_{n} $$

$$ = \frac{1}{n!}{r}^{n}\text{ . } $$