# Volume of the solid bounded by the sphere $x^2 + y^2 + z^2 = 9$ and paraboloid $8z = x^2 + y^2$

Find the volume of the solid bounded above by the sphere $x^2 + y^2 + z^2 = 9$ and below by the paraboloid $8z = x^2 + y^2$

I'm having some trouble finding the correct limits of integration in cylindrical coordinates. I appreciate some help. I'm prepping for my final and this is a question on an old final exam.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}\color{#66f}{\large V}& =\left.\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \dd x\,\dd y\,\dd z\right\vert_{\large{x^{2}\ +\ y^{2}\ +\ z^{2}\ <\ 9\atop z\ >\ \pars{x^{2}\ +\ y^{2}}/8}} =\left.\int_{0}^{2\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}r\,\dd r\,\dd z\,\dd\phi\right\vert _{{\large r^{2}\ +\ z^{2}\ <\ 9\atop z\ >\ r^{2}/8}} \\[5mm]&=2\pi\left.\int_{0}^{\infty}\int_{0}^{\infty}r\,\dd r\,\dd z\right\vert _{{\large r\ <\ \root{9 - z^{2}}\atop {r\ <\ \sqrt{8z}\atop 0\ <\ z\ <\ 3}}} =\color{#66f}{\large% 2\pi\int_{0}^{3}\int_{0}^{\min\pars{\root{9 - z^{2}},\root{8z}}}r\,\dd r\,\dd z} \end{align}

However, $$\root{9 - z^{2}}<\root{8z}\ \imp\ \pars{z - 1}\pars{z + 9}>0\ \imp\ \pars{~z<-9\ \mbox{or}\ z>1~}$$

\begin{align} \color{#66f}{\large V}&=2\pi\int_{0}^{1}\int_{0}^{\root{8z}}r\,\dd r\,\dd z +\int_{1}^{3}\int_{0}^{\root{9 - z^{2}}}r\,\dd r\,\dd z \\[5mm]&=2\pi\bracks{\int_{0}^{1}\half\,\pars{8z}\,\dd z +\int_{1}^{3}\half\,\pars{9 - z^{2}}\,\dd z} =2\pi\braces{\left. 2z^{2}\right\vert_{0}^{1} +\left. {9 \over 2}\,z - {1 \over 6}\,z^{3}\right\vert_{1}^{3}} \\[5mm]&=2\pi\bracks{2 + {27 \over 2} - {27 \over 6} - {9 \over 2} + {1 \over 6}} =\color{#66f}{\large{40 \over 3}\,\pi} \end{align}

Sphere: $$r^2+z^2=9\qquad z=\sqrt{9-r^2}.$$ Paraboloid: $$r^2=8z,\qquad z=\frac{r^2}8.$$ Intersection: $$\sqrt{9-r^2}=z=\frac{r^2}8,$$ $$9-r^2=\frac{r^4}{64},$$ Can you continue?

Integrate from $z=r^2/8$ to $z = \sqrt{(9-r^2)}$(positive because the volume lies above the z axis) Determine r as the radius for which the two objects intersect and integrate from 0 to the radius of intersection and from 0 to $2\pi$.

I had trouble doing these myself but the i came across this shadow method. In your head you imagine a light source on the object above the z axis. To find the limits of integration you just integrate z between the two given surfaces, that is write z as a function of x and y or r and theta. Finally you integrate over the shadow that object casts on the planar axis (in this case the circle with radius = r of intersection) and $2\pi$.

Sorry for my bad English and my inability to write the equation properly, hope this helps.