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.
 A: 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
$$
\begin{align}
V & = \iint\limits_{R} 2-x^2-y^2 - \sqrt{x^2+y^2} \, dA \\
 & = \int_0^{2\pi} \hspace{-5pt} \int_0^1 (2-r^2 -r) r \, dr \, d \theta \\
 & = 2 \pi \int_0^1 2r -r^3 -r^2 \, dr \\
 & = 2 \pi \left( 1 - \frac{1}{4} - \frac{1}{3} \right) \\
 & = \frac{5 \pi}{6}.
\end{align}
$$
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
V =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\int_{-\infty}^{\infty}
\Theta\pars{z - \root{x^{2} + y^{2}}}
\Theta\pars{2 - x^{2} - y^{2} - z}\,\dd x\,\dd y\,\dd z
$$

In polar coordinates $\pars{\rho,z,\theta}$, $x \equiv \rho\cos\pars{\theta}$ and
  $y \equiv \rho\sin\pars{\theta}$ where $\rho \geq 0$, $0 \leq \theta < 2\pi$ and
  $z\ \in\ {\mathbb R}$:
  \begin{align}
V & =
\int_{0}^{\infty}\dd\rho\,\rho\int_{-\infty}^{\infty}\dd z
\int_{-\infty}^{\infty}\dd\theta
\,\Theta\pars{z - \rho}\Theta\pars{2 - \rho^{2} - z}
\\[5mm] & =
2\pi\int_{0}^{\infty}\dd\rho\,\rho
\int_{-\infty}^{\infty}\dd z\,
\Theta\pars{z - \rho}\Theta\pars{2 - \rho^{2} - z}
\\[5mm] & =
2\pi\int_{0}^{\infty}\dd\rho\,\rho\,\Theta\pars{\bracks{2 - \rho^{2}} - \rho}
\int_{\rho}^{2 - \rho^{2}}\dd z
\\[5mm] & =
2\pi\int_{0}^{\infty}\dd\rho\,\rho
\,\Theta\pars{2 - \rho - \rho^{2}}
\bracks{\pars{2 - \rho^{2}} - \rho}
\end{align}

$2 - \rho - \rho^{2} > 0\quad\imp\quad -2 < \rho < 1$. However, $\rho > 0$ such that
$$
\color{#00f}{\large V}
=2\pi\quad\overbrace{\int_{0}^{1}\pars{2\rho - \rho^{2} - \rho^{3}}\,\dd\rho}
^{\ds{5 \over 12}}\
=\color{#00f}{\large{5 \over 6}\,\pi} \approx 2.6180
$$
