can someone help me with setting up the required integral?
The solid is bounded by $z = 4x^2 + 4y^2 -27$ and by $z = x^2+y^2$. Should I project this on $x-y$ plane or use cylindrical or spherical coordinates or do something else? I am a bit lost and could use an example to crystalize how to approach these types of problems.
Using cylindrical coordinates, you have
$ \rho^2 = x^2 + y^2 $
So the two paraboloids are $ z = 4 \rho^2 - 27 $ and $ z = \rho^2 $
These are equal when $ 3 \rho^2 = 27 $, i.e. $\rho = 3 $
Now the volume integral is
$ V = \displaystyle \int_{\phi = 0}^{2 \pi} \int_{\rho = 0 }^{\rho = 3 } \int_{z = 4 \rho^2 - 27} ^ { \rho^2 } d z \ \rho \ d \rho \ d \phi $
This simplifies to
$ V = 2 \pi \displaystyle \int_{\rho = 0 }^{\rho = 3 } \rho(27 - 3 \rho^2) d \rho $
And finally,
$ V = 2 \pi \left( \dfrac{243}{2} - \dfrac{243}{4} \right) = \dfrac{243 \pi}{2} $