Convergence of multiple integral in $\mathbb R^4$ Denote $(x,y,z,w)$ the euclidean coordinates in $\mathbb R^4$. I am trying to study the convergence of the integral $$\int \frac{1}{(x^2+y^2)^a}\frac{1}{(x^2+y^2+z^2+w^2)^b} dx\,dy\, dz\, dw$$
over a disk (or cube, or any open set) containing the origin
in terms of the real parameters $a>0$ and $b>0$. If needed we can assume that $b$ is fixed and get a bound on $a$ in terms of $b$.
Does anyone have an idea how could I proceed? I tried using cylindrical coordinates (in $(x,y)$ and $(z,w)$ separately) but I couldn't get anywhere.
Thank you in advance.

As DiegoMath suggests, using polar coordinates for $(x,y)$ and $(z,w)$ is efficient, more precisely, define
$$(x,y,z,w):=(r\cos\theta,r\sin\theta,s\cos\phi,s\sin\phi).$$
Then the integral
$$I(a,b):=\int_{B(0,1)} \frac{1}{(x^2+y^2)^a}\frac{1}{(x^2+y^2+z^2+w^2)^b} dx\,dy\, dz\, dw$$
converges if and only if the integral
$$\int_0^1\int_0^1 \frac{rs}{r^{2a}(r^2+s^2)^b}\mathrm ds\mathrm dr$$
converges. Using again polar coordinates, this time with $(r,s)$, that is, $r=R\cos t$, $s=R\sin t$, we derive that $I(a,b)$ converges if and only if 
$$\int_0^1\int_0^{2\pi}\frac{R^3\cos t\sin t }{R^{2a+2b}(\cos t)^{2a}}   \mathrm dt\mathrm dR .$$
One can compute explicitely the integral in $t$ on $(0,2\pi)\setminus (\pm\pi/2-\delta,\pm\pi/2+\delta)$  in order to get a condition on $a$; the integral on $R$ yields a conditionon $a$ and $b$.
