Volume inside paraboloid and sphere 
Find the volume inside the paraboloid $az=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2a^2$

Since we have $x^2+y^2$ I thought about polar coordinates , but before that I showed the functions as $z_1=\frac {x^2+y^2}{a}$ for the paraboloid and $z_2=\sqrt{2a^2-x^2-y^2}$ and I am not sure if this is a right way to see it but if $x$ and $y$ are close to zero we can then see that the sphere is above.
$\iint z_2(x,y)-z_1(x,y)dA$ should be the integral that is the volume
$\iint(\sqrt{2a^2-x^2-y^2})-(\frac {x^2+y^2}{a})dA$ here it seems like the right choice to go with polar coordinates but I got stuck in this part , $\iint(\sqrt{2a^2-(x^2+y^2)})-(\frac {x^2+y^2}{a})dA$
we know that
\begin{cases}
 x=rcos\theta\\
 y=rsin\theta\\
 r^2=x^2+y^2\\    
tan\theta=\frac{y}{x}
\end{cases}
$\iint_{r\theta}(\sqrt{2a^2-r^2})-(\frac {r^2}{a})rdrd\theta$ according to the way I drew it I think that $\theta$ should be $0\leq \theta \leq \pi$ But I cannot find $r$.
Would appreciate any help and tips , other approaches are always great thank you!

The intersection between the sphere and the paraboloid takes place when $2a^2-z^2=az$, that is, when $z=a$. So, if $(x,y,z)$ belongs to the region whose volume we are trying to compute, then:

*

*if $z\leqslant a$, then $0\leqslant x^2+y^2\leqslant az$;

*if $z\geqslant a$, then $0\leqslant x^2+y^2\leqslant 2a^2-z^2$.

So, compute$$\int_0^{2\pi}\int_0^a\int_0^{\sqrt{az}}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta+\int_0^{2\pi}\int_a^{\sqrt2a}\int_0^{\sqrt{2a^2-z^2}}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta\left(=\frac{ \left(8 \sqrt{2}-7\right) \pi  a^3}6\right).$$
