<s>[INST] Volume above cone and below paraboloid. I need to find the volume above the cone $z=\sqrt{x^2+y^2}$ and below the paraboloid $z=2-x^2-y^2$. I thought about using spherical coordinates and finding $p$, which would be (before simplification) :
$$-z=-2+x^2+y^2$$
$$-p \cos(\theta) =-2+p^2 \sin^2 (\theta) \cos^2(\theta)+p^2 \sin^2(\theta) \sin^2(\theta)$$
But I can't seem to be able to isolate $p$ even knowing that $\sin^2(\theta)+\cos^2(\theta)=1$.
Any hint would be greatly appreciated!
Thanks.
 [/INST] I believe cylindrical coordinates will be the easiest. When setting it up as a triple integral you will have
$$V = \iiint \, dV = \iint\limits_{R} \hspace{-5pt} \int_{\sqrt{x^2+y^2}}^{2 -x^2-y^2} \, dz \, dA = \iint\limits_{R} 2-x^2-y^2 - \sqrt{x^2+y^2} \, dA.$$
These surfaces intersect each other when $x^2+y^2=1$, therefore the projected region in the $xy$ plane is a disk of radius 1. Performing this coordinate change we have
$$
\begin{align}
V & = \iint\limits_{R} 2-x^2-y^2 - \sqrt{x^2+y^2} \, dA \\
 & = \int_0^{2\pi} \hspace{-5pt} \int_0^1 (2-r^2 -r) r \, dr \, d \theta \\
 & = 2 \pi \int_0^1 2r -r^3 -r^2 \, dr \\
 & = 2 \pi \left( 1 - \frac{1}{4} - \frac{1}{3} \right) \\
 & = \frac{5 \pi}{6}.
\end{align}
$$
</s>