I'm currently learning how to calculate the volume of a 3D surface expressed in spherical coordinates using triple integrals.
There was this exercice (from here) which asked me to find the volume of the region described by those two equations:
$x^2+y^2=1$
$x^2+y^2+z^2=4$
(The image of the region from the website)
From this I've found that:
$csc\phi\le \rho\le2$ and $0\le \theta\le 2\pi$
I was also able to find a value of $\phi$ which was $\frac{\pi}{6}$ (I wasn't able to deduce if it corresponded to the lower or higher bound).
And from here I didn't knew what to do next, so I assumed that $0\le \phi\le \frac{\pi}{6}$ because I was stucked.
And the actual right setup of the integral in spherical coordinates was:
$\int_{\frac{\pi}{6}}^{5\frac{\pi}{6}}\int_{0}^{2\pi}\int_{csc(\phi)}^{2}\rho^2sin(\phi) d\rho d\theta d\phi$
So my brain died.
I couldn't figure why first of all $d\theta$ was located before $d\phi$ (what constraint would have caused this ?).
And also why the other bound of $\phi$ was $5\frac{\pi}{6}$, how was I supposed to find this value ?
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\underline{"Outside"}\,\,\, Volume\ {\Large\cal V}_{\mrm{out}} \pars{~\mbox{Cylindrical Coordintaes}~}}$.
Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket.
Namely, $\ds{\bracks{P} = \left\{\begin{array}{rcl} \ds{1} & \mbox{if}\ P\ \mbox{is}\ true. \\[1mm] \ds{0} & \,\,\,\,\,\mbox{if}\ P\ \mbox{is}\ false. \end{array}\right.}$
\begin{align} {\Large\cal V}_{\mrm{out}} & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}\bracks{x^{2} + y^{2} > 1}\bracks{x^{2} + y^{2} + z^{2} < 4} \dd^{3}\vec{r}} \\[5mm] & = \iiint_{\mathbb{R}^{3}}\bracks{\rho^{2} > 1} \bracks{\rho^{2} + z^{2} < 4} \rho\,\dd\rho\,\dd \phi\,\dd z \\[5mm] & = \iiint_{\mathbb{R}^{3}} \bracks{1 < \rho^{2} < 4 - z^{2}} \rho\,\dd\rho\,\dd \phi\,\dd z \\[5mm] & = \iiint_{\mathbb{R}^{3}} \bracks{1 < 4 - z^{2}} \bracks{1 < \rho < \root{4 - z^{2}}} \\[2mm] &\ \rho\,\dd\rho\,\dd \phi\,\dd z \\[5mm] & = \iiint_{\mathbb{R}^{3}} \bracks{\verts{z} < \root{3}} \bracks{1 < \rho < \root{4 - z^{2}}} \\[2mm] & \rho\,\dd\rho\,\dd \phi\,\dd z \\[5mm] & = 2\pi\int_{-\root{3}}^{\root{3}} \int_{1}^{\root{4 - z^{2}}}\rho\,\dd\rho\,\dd z \\[5mm] & = \pi\int_{-\root{3}}^{\root{3}}\pars{3 - z^{2}}\dd z \\[5mm] & = \bbx{4\root{3}\pi} \approx 21.7656 \\ & \end{align}