Find the Volume of: $x^2+y^2\le z\le \sqrt{6-x^2-y^2}$
My Attempt:
So I need to find $\iint_D(\int^{\sqrt{6-x^2-y^2}}_{x^2+y^2}dz)dxdy$, while $D$ is the area of the intersection between the two sufaces.
Finding $D$:
$x^2+y^2=z \Longrightarrow\sqrt{6-(x^2+y^2)}=\sqrt{6-z} \Longrightarrow z^2=6-z \Longrightarrow z^2 +z-6=0 \Longrightarrow \frac{-1 \pm\sqrt{1-4*1*-6}}{2}=-3,2 \Longrightarrow z=2$.
Substitute $z=2$ to find the intersection curve:
$1=\frac{x^2}{2}+\frac{y^2}{2}$.
Moving to elliptical coordinates:
$x=\sqrt{2}rcos(a)$, $y=\sqrt{2}rsin(a)$, $|J|=2r$
And here I got a little stuck, I usually substitute in the inequalities to get the possible values of $a, r$, but here I'm not sure what to do, feels like I've missed $z$ since if I substitute $r^2\le z$, and in the same time I don't think an ellipsoidal coordinates fits this problem.
Any help and feedback is really appreciated, thanks in advance!