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?

This is where it goes wrong:

The two outer integrals both had the bounds from $-4$ to $4$.

If you let $x$ and $y$ run from $-4$ to $4$, you are integrating over a square in the $xy$-plane, but you want to integrate over the projection of the given region onto the $xy$-plane. Observe that the two surfaces intersect at:
$$x^2+x^2-2 = 30-x^2-y^2 \iff x^2+y^2 = 16$$
and this is a circle of radius $4$, centered at the origin.
You can keep $x:-4\to4$ but then you need $y$ as a function of $x$ to integrate over this circle (or vice versa). It works, but the calculations can become a bit messy.
Hint: polar coordinates.
Can you take it from here?

Addendum after comments.
In polar coordinates, you have $r^2=x^2+y^2$ so for the integrand:
$$\left( 30-x^2-y^2 \right) - \left( x^2+x^2-2 \right) = 32 - 2 \left( x^2 +y^2\right)
\to 32-2r^2$$
Integrating over the circle $x^2+y^2 = 4^2$ is done by letting $r:0\to 4$ and $t:0\to 2\pi$. Note though that $\mbox{d}x\,\mbox{d}y \to \color{red}{r}\,\mbox{d}r\,\mbox{d}t$.
You can verify that:
$$\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} 32 - 2 \left( x^2 +y^2\right) \,\mbox{d}y\,\mbox{d}x = \int_{0}^{4} \int_{0}^{2\pi} r \left( 32 - 2r^2\right)  \,\mbox{d}t\,\mbox{d}r = 256\pi$$
