我觉得我上次讲的不太友好,题目太难了.
我们应该从简单点的开始讲,比如:
$$\int_{{{\left[ {a,b} \right]}^n}} {{\text{d}}{x_1}{\text{d}}{x_2} \ldots {\text{d}}{x_n}} = {\left( {b - a} \right)^n}$$
嗯,就是求超立方体体积...
然后我们来求超球体的体积,当然是用球坐标咯.
$$\begin{aligned}
dV&=r^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,dr\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1}\\
V_{n}(R)&=\int _{0}^{R}\int _{0}^{\pi }\cdots \int _{0}^{2\pi }r^{n-1}\sin ^{n-2}(\varphi _{1})\cdots \sin(\varphi _{n-2})\,d\varphi _{n-1}\cdots d\varphi _{1}\,dr\\
&=\left(\int _{0}^{R}r^{n-1}\,dr\right)\left(\int _{0}^{\pi }\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(\int _{0}^{2\pi }d\varphi _{n-1}\right)\\
&={\frac {R^{n}}{n}}\left(2\int _{0}^{\frac {\pi }{2}}\sin ^{n-2}(\varphi _{1})\,d\varphi _{1}\right)\cdots \left(4\int _{0}^{\frac {\pi }{2}}d\varphi _{n-1}\right)\\
&={\frac {R^{n}}{n}}\textstyle \mathrm {B} \left({\frac {n-1}{2}},{\frac {1}{2}}\right)\mathrm {B} \left({\frac {n-2}{2}},{\frac {1}{2}}\right)\cdots \mathrm {B} \left(1,{\frac {1}{2}}\right)\cdot 2\mathrm {B} \left({\frac {1}{2}},{\frac {1}{2}}\right)\\
&={\frac {R^{n}}{n}}{\frac {\Gamma \left({\frac {n-1}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left({\frac {n}{2}}\right)}}{\frac {\Gamma \left({\frac {n-2}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left({\frac {n-1}{2}}\right)}}\cdots {\frac {\Gamma \left(1\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left({\frac {3}{2}}\right)}}\cdot 2{\frac {\Gamma \left({\frac {1}{2}}\right)\Gamma \left({\frac {1}{2}}\right)}{\Gamma \left(1\right)}}\\
&={\frac {2\pi ^{\frac {n}{2}}R^{n}}{n\Gamma \left({\frac {n}{2}}\right)}}={\frac {\pi ^{\frac {n}{2}}R^{n}}{\Gamma \left({\frac {n}{2}}+1\right)}}
\end{aligned}$$Read More
