calculate the integral $\int (x_1^2+x_2^2+x_3^2+x_4^2)$

Question

Calculate the following integral over $G = \left\{\sqrt{x_1^2+x_2^2}+\sqrt{x_3^2+x_4^2}<2\right\}$ $$\int_G (x_1^2+x_2^2+x_3^2+x_4^2)dx_1dx_2dx_3dx_4$$ My try:
I did some coordiantes change along the way and I got to this $$ 64\pi^2\int_B(q+s)dqds $$ $B = \left\{q+s\leq 1,q,s\geq 0\right\}$ and from here I wanted to use Gamma function with Dirichlet's formula but I'm new to gamma function so I got confused on how to set it right.

Answer 1

Letting $(x_1,x_2)=2\rho(\cos\theta,\sin\theta)$ and $(x_3,x_4)=2\tau(\cos\phi,\sin\phi)$ we are left with

$$\begin{eqnarray*} 256\pi^2 \int_{0\leq \rho+\tau < 1}\rho\tau(\rho^2+\tau^2) \,d\mu &\stackrel{\text{symmetry}}{=}&512\pi^2\int_{0\leq \rho+\tau < 1}\rho^3\tau\,d\mu\\&=&512\pi^2\int_{0}^{1}\int_{0}^{1-\rho}\rho^3\tau\,d\tau\,d\rho\end{eqnarray*}$$ and the RHS equals $$ 256\pi^2 \int_{0}^{1}\rho^3(1-\rho)^2\,d\rho = 256\pi^2 B(4,3) = 256\pi^2 \frac{\Gamma(3)\Gamma(4)}{\Gamma(7)}=256\pi^2\frac{2!3!}{6!} =\color{red}{\frac{64}{15}\pi^2}.$$

Answer 2

Let's take $x_1=r \cos \phi, x_2=r \sin \phi, x_3=\rho \cos \theta, x_4=\rho \sin \theta$. Jakobian will be $J = r \rho.$ Integral comes to $$4 \pi^2\int_{0}^{2}\int_{0}^{2-x}xy(x^2+y^2)dxdy$$ Where $x,y$ are $r, \rho.$