# 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.

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}

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

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