Integration of function $\sqrt{1 - \sum_{i}x_{i}^{a_{i}}}$ by using spherical coordinates in D dimensions Let's have integral
$$
I = \int \limits_{0}^{\text{T.P.}}dx_{1}...dx_{D}\sqrt{1 - \sum_{i}x_{i}^{a_{i}}}, \quad \text{T.P.}: \quad 1 - \sum_{i}x_{i}^{a_{i}} = 0
$$
By rewriting it in coordinates $x_{i}^{a_{i}} = y_{i}^{2}$ we have
$$
\tag 1 I \sim \int \limits_{0}^{T.P.}\sqrt{1 - \sum_{i}y_{i}^{2}}\prod_{i = 1}^{D}dy_{i}y_{i}^{\frac{2 - a_{i}}{a_{i}}}
$$
Is it possible to evaluate it by performing integration in n-spherical coordinate system? The problem is that after introducing them I get set of integrals like
$$
\int \limits_{0}^{\pi}\sin^{\alpha_{j}}(\varphi_{j})cos^{\beta_{j}}(\varphi_{j})d\varphi_{j}
$$
which are complex, so it seems that I have missed something.
Or, maybe, integral $I$ in new variables given by Eq. $(1)$ isn't written correctly, and I need to write something like $|x_{i}|^{a_{i}} = y_{i}^{2}$? Finally, is integral $I$ ill defined in zone where $x_{i} < 0$?
 A: Yes you can evaluate it using spherical coordinates, but it is a long and tedious calculation. The final result is fairly simple though
$$\color{red}{\int\limits_{0\leq x_i,~\sum_{i=1}^D x_i^{a_i}\leq 1}\sqrt{1-\sum_{i=1}^Dx_i^{a_i}}{\rm d}V = \frac{\sqrt{\pi}}{2}\frac{\prod_{i=1}^{D}\Gamma\left(a_i^{-1}\right)a_i^{-1}}{\Gamma\left(\frac{3}{2} + \sum_{i=1}^Da_i^{-1}\right)}}$$
so there are probably easier ways to derive this (see Did's comments).

Starting with the substitution you propose $x_i = y_i^{\frac{2}{a_i}}\implies {\rm d}x_i = \frac{2}{a_i}y_i^{\frac{2-a_i}{a_i}}{\rm d}y_i$ we get
$$I = \frac{2^D}{\prod_{i=1}^Da_i}\int\sqrt{1-\sum_{i=1}^D y_i^2}\prod_{i=1}^Dy_i^{\frac{2-a_i}{a_i}}{\rm d}y_i$$
Now changing to (hyper)spherical coordinates we find
$$\prod_{i=1}^{D}{\rm d}y_i = r^{D-1}{\rm d}r\prod_{i=1}^{D-1}\sin^{D-1-i}(\theta_i){\rm d}\theta_i$$
$$\prod_{i=1}^{D}y_i^{z_i} = r^{\sum_{j=1}^D z_j}\prod_{i=1}^{D-1}\cos^{z_i}(\theta_i)\sin^{\sum_{j=i+1}^{D}z_j}(\theta_i)$$
Since you only want to integrate over the first quadrant the integration limits for the angular variables are $\theta_i\in[0,\pi/2]$. This gives us the integral(s)
$$I = \frac{2^D}{\prod_{j=1}^Da_j}\int_0^1\sqrt{1-r^2}r^{\left(\sum_{i=1}^D \frac{2}{a_i}\right)-1}\,{\rm d}r\prod_{i=1}^{D-1}\int_0^{\frac{\pi}{2}}\cos^{\frac{2}{a_i}-2}(\theta_i)\sin^{\left(\sum_{j=i+1}^{D}\frac{2}{a_j}\right)-1}(\theta_i)\,{\rm d}\sin\theta_i$$
Perform a change of variables $\sin\theta_i\to z_i$ in the last $D-1$ integrals and evaluate using the $\beta$-function to get
$$I = \frac{2^D}{\prod_{j=1}^Da_j}\left[\frac{\Gamma\left(\frac{3}{2}\right)\Gamma \left(\sum_{i=1}^D \frac{1}{a_i}\right)}{2\Gamma\left(\sum_{i=1}^D \frac{1}{a_i}+\frac{3}{2}\right)}\right]\prod_{i=1}^{D-1}\frac{\Gamma\left(\frac{1}{a_i}\right) \Gamma \left(\sum_{j=i+1}^{D}\frac{1}{a_j}\right)}{2\Gamma\left(\sum_{j=i}^{D}\frac{1}{a_j}\right)}$$
which can be simplified down to
$$I= \frac{\sqrt{\pi}}{2}\frac{\prod_{i=1}^{D}\Gamma\left(a_i^{-1}\right)a_i^{-1}}{\Gamma\left(\frac{3}{2} + \sum_{i=1}^Da_i^{-1}\right)}$$
As a consistency check lets take $a_1=a_2=\ldots=a_D=2$. Then the integral is just $1/2^{D+1}$ of the volume of a $D+1$ dimensional unit sphere. The formula above gives $2^{D+1}I = \frac{\pi^{\frac{D+1}{2}}}{\Gamma\left(\frac{D+1}{2}+1\right)}$ which agrees with the known expression for the volume.
