I'd like to evaluate $\int \int \int_B x^2+y^2 \, dxdydz$ where $B$ is the area enclosed by $x^2+y^2=2z$ and $z=2$ but I'm not sure about the bounds. I've thought something like this... $$\int_0^2 \int_{-\sqrt{2z-x^2}}^{\sqrt{2z-x^2}} \int_{-\sqrt{2z-y^2}}^{\sqrt{2z-y^2}} x^2+y^2 dxdydz$$
but obviously, it won't work because I reintroduce $x$ in the second integral. Any suggestions on the bounds?
From symmetry of the region, it is best to integrate in cylindrical coordinates $( r, \phi, z ) $
The bounds are $ z = \dfrac{1}{2} r^2 $ and $ z = 2 $
So $ r $ will range from $0$ to $ 2$.
$I = \displaystyle \int_{\phi = 0 }^{2\pi} \int_{r = 0}^{2} \int_{ z = \dfrac{1}{2} r^2 }^{2} dz r^3 dr d\phi $
Integrating with respect to $\phi$ and $z$ simplifies the above to
$I = \displaystyle 2 \pi \int_{r = 0 }^2 \left( 2 r^3 - \dfrac{1}{2} r^5 \right) dr $
And this evaluates to
$ I = \displaystyle 2 \pi \left( \dfrac{1}{2} r^4 - \dfrac{1}{12} r^6 \right)\biggr| _{0}^{2} = \dfrac{16 \pi}{3} $