Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$ 
Find the volume of the regions enclosed by $z = x^2+y^2-2$ and $z = 30-x^2-y^2$

I set up a triple integral with the bounds of the inmost as $x^2 + y^2 - 2$ to $30 - x^2 - y^2$. The two outer integrals both had the bounds from $-4$ to $4$. When I solved it I got $1024$ as the volume, but this isn't correct. Can someone please show me the steps to finding the bounds, and the correct answer?

Set $z = z$ in the two equations to find the circle of intersection:
$x^2 + y^2 - 2 = 30 - x^2 - y^2\\
2x^2 +2y^2 = 32\\
x^2+ y^2 = 16$
$\iiint \;dz\;dy\;dx\\
\iint z\;dy\;dx$
$\int_{-4}^{4}\int_{-\sqrt{16-x^2}}^{\sqrt {16-x^2}} 32 - 2x^2- 2y^2\;dy\;dx$
Now we could keep cranking through that in Cartesian, but I think a conversion to cylindrical coordinates will make this easier.
$x = r \cos\theta\\
y = r \sin \theta\\
dy\;dx = r\;dr\;d\theta$
$\int_0^{2\pi}\int_{0}^{4} 2(16 -r^2)r \;dr\;d\theta\\
u = (16-r^2), du = -2r \;dr\\
\int_0^{2\pi}\int_{16}^{0} -u \;du\;d\theta\\
\int_0^{2\pi} 16^2 \;d\theta\\
512 \pi$
