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?
 A: 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:
$$\int_{-4}^{4} \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} 32 - 2 \left( x^2 +y^2\right) \,\mbox{d}y\,\mbox{d}x = \int_{0}^{4} \int_{0}^{2\pi} r \left( 32 - 2r^2\right)  \,\mbox{d}t\,\mbox{d}r = 256\pi$$
A: 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.
$x = r \cos\theta\\
y = r \sin \theta\\
dy\;dx = r\;dr\;d\theta$
$\int_0^{2\pi}\int_{0}^{4} 2(16 -r^2)r \;dr\;d\theta\\
u = (16-r^2), du = -2r \;dr\\
\int_0^{2\pi}\int_{16}^{0} -u \;du\;d\theta\\
\int_0^{2\pi} 16^2 \;d\theta\\
512 \pi$
A: There is a different path by considering this question in a mostly geometrical way.
The general shape of such a volume is that of a cocktail shaker made of two paraboloidal-shaped "bowls" (see below a sectional figure along plane $y=0$). These bowls intersect along a horizontal circle with radius $a=4$ at $z=14$ as a results from equation 
$$z=x^2+y^2-2=30-(x^2+y^2) \ \ \Leftrightarrow \ \ x^2+y^2=4^2 \ \ \text{giving} \ z=14.$$
These bowls are clearly symmetrical with respect to plane with equation $z=14$.  It suffices to compute the content of one of these bowls. It is given by formula: $\dfrac{1}{2}\pi a^2 h^2$ where $a$ is the upper radius and $h$ its height $h$(see this). The final result is thus twice the following value:
$$\dfrac{1}{2}\pi 4^2 16^2=128\pi.$$  
giving the desired result.

