Evaluating an integral over the hypersphere I am working on the follwoing exercise:

For $\alpha_1, \alpha_2, \ldots, \alpha_m > 0$ evaluate 
  $$\int \cdots \int_{{x_1^2+\cdots + x_m^2 \le 1}\\{x_1,\ldots,x_m \ge 0}} x_1^{\alpha_1-1}x_2^{\alpha_2-1} \cdot \ldots \cdot x_m^{\alpha_m-1} \ dx_1 dx_2 \ldots dx_m.$$

I have no idea how to do that. Could you help me?
 A: Introduce $m-1$ angles $\theta_i,\,0\le i\le m-2$, viz.$$d^mx=r^{m-1}dr\prod_i\sin^i\theta_id\theta_i.$$Whereas the full sphere runs over $\theta_0\in[0,\,2\pi]$ and $\theta_i\in[0,\,\pi]$ for $i\ge1$, our integration range here is the first orthant, for which $\theta_i\in[0,\,\pi/2]$ for $i\ge0$ [sic], dividing the volume by $4\cdot 2^{m-2}=2^m$ as expected. For $0\le i\le m-1$,$$x_{i+1}=r\cos^{[i\ne m-1]}\theta_{i}\prod_{j=0}^{i-1}\sin\theta_j$$(see here - it denotes $\theta_i$ as $\varphi_{i+1}$ - and here.) So the integral is$$\begin{align}I&:=\int_0^1r^{m-1}dr\int_{[0,\,\pi/2]^{m-1}}d^{m-1}\theta\prod_{i=0}^{m-2}\left(r\cos^{[i\ne m-1]}\theta_{i}\prod_{j=0}^{i-1}\sin\theta_j\right)^{\alpha_{i+1}-1}\\&=\int_0^1r^{\sum_{i=1}^m\alpha_i-1}dr\prod_{i=0}^{m-2}\int_0^{\pi/2}\cos^{[i\ne m-1](\alpha_{i+1}-1)}\theta_i\sin^{\sum_{j=i+2}^{m}(\alpha_j-1)}\theta_id\theta_i.\end{align}$$You can do the rest with Beta functions, obtaining$$I=\frac{1}{2^{m-1}}\frac{1}{\sum_{i=1}^m\alpha_i}\frac{\prod_{i=1}^m\Gamma\left(\frac{\alpha_i}{2}\right)}{\Gamma\left(\sum_{i=1}^m\frac{\alpha_i}{2}\right)}.$$The $\alpha_i=1$ special case gives$$2^mI=\frac{\pi^{m/2}}{\Gamma\left(\frac{m}{2}+1\right)},$$famously the $m$-ball's volume.
