方企勤 第六章 多元函数积分学 第6.2题
📝 题目
6. 2.16 证明
$$ \begin{array}{l} \int \cdots \int f\left( {{x}_{1} + {x}_{2} + \cdots + {x}_{n}}\right) \mathrm{d}{x}_{1}\mathrm{\;d}{x}_{2}\cdots \mathrm{d}{x}_{n} \\ \mathop{\sum }\limits_{\substack{{i = 1} \\ {{x}_{i} \geq 0\;\left( {i = 1,\cdots ,n}\right) } }}^{n}{x}_{i} \leq a \end{array} $$
$$ = {\int }_{0}^{a}f\left( x\right) \frac{{x}^{n - 1}}{\left( {n - 1}\right) !}\mathrm{d}x. $$
💡 答案解析
例 4 (2) $\pi {R}^{2}\ln \frac{1}{\sqrt{{a}^{2} + {b}^{2}}}$ .
6.2.6 $I = 0$ (先对 $\theta$ 积分).
6.2.8 (1) $\frac{5}{4}\pi$ ; (2) $\frac{3367}{3}\pi$ ; (3) ${40\pi }$ .
6.2.9 (1) $\frac{1}{192}$ ; (2) $\frac{64}{9}\pi$ ; (3) $\frac{{128}\left( {4\sqrt{2} - 5}\right) }{15}\pi$ .
6.2.10 (1) $\frac{1}{40}\left( {\frac{1}{{a}^{2}} - \frac{1}{{b}^{2}}}\right) \left( {{d}^{5} - {c}^{5}}\right) \cdot \left\lbrack {\left( {{\beta }^{2} - {\alpha }^{2}}\right) \left( {1 + \frac{1}{{\alpha }^{2}{\beta }^{2}}}\right) + 4\ln \frac{\beta }{\alpha }}\right\rbrack$ ;
(2) $\frac{2}{65}\left( {\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}}}\right) \left( {\frac{1}{{\alpha }^{5}} - \frac{1}{{\beta }^{5}}}\right) {h}^{6}\sqrt{h}$ ;
(3) $4\left( {\mathrm{e} - 2}\right) {\pi abc}$ .
6.2.11 (1) $F\left( t\right) = {4\pi abc}{\int }_{0}^{t}f\left( {r}^{2}\right) {r}^{2}\mathrm{\;d}r$ ; (2) ${F}^{\prime }\left( t\right) = {4\pi abc}{t}^{2}f\left( {t}^{2}\right)$ .
6.2.12 作变量代换 $x = r{\sin }^{2}\varphi {\cos }^{2}\theta ,y = r{\sin }^{2}\varphi {\sin }^{2}\theta ,z = r{\cos }^{2}\varphi$ .
6.2.13 $\frac{{x}_{1} + {x}_{2} + {x}_{3} + {x}_{4}}{4}V$ .
6.2.14 $\frac{{\pi }^{\frac{n}{2}}}{\Gamma \left( {\frac{n}{2} + 1}\right) }{R}^{n}$ .
6.2.15 $\frac{{a}^{n}}{n!}$ .
6.2.16 先用数学归纳法,然后再作二重积分变换. 或作 $n$ 重积分变换:
$$ \left\{ \begin{array}{l} {y}_{i} = {x}_{i}\;\left( {i = 1,2,\cdots ,n - 1}\right) , \\ {y}_{n} = {x}_{1} + \cdots + {x}_{n}, \end{array}\right. $$
然后应用上一题结果.