Originally Posted by

**ThePerfectHacker** The $\displaystyle \phi$ angle opens up from $\displaystyle 0$ to $\displaystyle \tfrac{\pi}{4}$.

The $\displaystyle \theta$ angles travels a complete revolution.

What is trickier is finding the bounds on $\displaystyle \rho$.

If $\displaystyle z=1$ then the plane cut is $\displaystyle 1$ unit above the $\displaystyle xy$ plane, but the side of the cone makes an angle of $\displaystyle \tfrac{\pi}{4}$, therefore the length of the side of the cone is $\displaystyle \cos \tfrac{\pi}{4}=\tfrac{\sqrt{2}}{2}$. For $\displaystyle z=2$ by similar argument we get $\displaystyle \rho = 2\cdot \tfrac{\sqrt{2}}{2} = \sqrt{2}$. Therefore, the integral becomes,

$\displaystyle \int_{\sqrt{2}/2}^{\sqrt{2}} \int_0^{2\pi} \int_0^{\pi/4}\rho^2 \sin \phi ~ d\phi ~ d\theta ~ d\rho$