Let the cross section of the paraboloid at z=4 be our region R.
Thus, the x and y limits of integration become pretty obvious: $\displaystyle \left\{R: \ -2\leq x \leq 2, \ -2\leq y \leq 2\right\}$.
(this is the region R)
Thus, our volume integral would be $\displaystyle \int_{-2}^2\int_{-2}^2 x^2+y^2 \,dx\,dy$.
We can also represent this integral in polar coordinates. Note that the limits of integration change!
This time we see : $\displaystyle \left\{R: \ 0\leq r \leq 2, \ 0\leq \theta \leq 2\pi\right\}$.
As a result, the integral becomes $\displaystyle \int_0^{2\pi}\int_0^2 r^3 \,dr\,d\theta$.
Hope that this makes sense!
--Chris