Finding the bounds of a triple integral (spherical coordinates) 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 ?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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$\ds{\underline{"Outside"}\,\,\, Volume\
{\Large\cal V}_{\mrm{out}} \pars{~\mbox{Cylindrical Coordintaes}~}}$.
Hereafter, $\ds{\bracks{\cdots}}$ is an
Iverson Bracket.
Namely, $\ds{\bracks{P} =
\left\{\begin{array}{rcl}
\ds{1} &amp; \mbox{if}\ P\ \mbox{is}\ true.
\\[1mm]
\ds{0} &amp; \,\,\,\,\,\mbox{if}\ P\ \mbox{is}\ false.
\end{array}\right.}$

\begin{align}
{\Large\cal V}_{\mrm{out}} &amp; \equiv
\bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}}\bracks{x^{2} + y^{2} &gt; 1}\bracks{x^{2} + y^{2} + z^{2} &lt; 4}
\dd^{3}\vec{r}}
\\[5mm] &amp; =
\iiint_{\mathbb{R}^{3}}\bracks{\rho^{2} &gt; 1}
\bracks{\rho^{2} + z^{2} &lt; 4}
\rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] &amp; =
\iiint_{\mathbb{R}^{3}}
\bracks{1 &lt; \rho^{2} &lt; 4 - z^{2}}
\rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] &amp; =
\iiint_{\mathbb{R}^{3}}
\bracks{1 &lt; 4 - z^{2}}
\bracks{1 &lt; \rho &lt; \root{4 - z^{2}}}
\\[2mm] &amp;\ \rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] &amp; =
\iiint_{\mathbb{R}^{3}}
\bracks{\verts{z} &lt; \root{3}}
\bracks{1 &lt; \rho &lt; \root{4 - z^{2}}}
\\[2mm] &amp; \rho\,\dd\rho\,\dd \phi\,\dd z
\\[5mm] &amp; =
2\pi\int_{-\root{3}}^{\root{3}}
\int_{1}^{\root{4 - z^{2}}}\rho\,\dd\rho\,\dd z
\\[5mm] &amp; =
\pi\int_{-\root{3}}^{\root{3}}\pars{3 - z^{2}}\dd z
\\[5mm] &amp; =
\bbx{4\root{3}\pi} \approx 21.7656 \\ &amp;
\end{align}
It does not matter what order $\rho, \theta, \phi$ are integrated in, as long as the limits on each are expressed only in terms of variables yet to be integrated. Since $\theta, \phi$ have constant limits, they can be interchanged freely. Since the limits for $\rho$ and $\theta$ are independent of each other, they can also be exchanged. But to exchange the order of the integrations of $\rho$ and $\phi$, you would need to express the limits of $\phi$ in terms of $\rho$ instead of $\rho$ in terms of $\phi$. However, between the two, your approach is simpler. That is:
$$\begin{align}\int_{\frac{\pi}6}^{\frac{5\pi}6}\int_0^{2\pi}\int_{\csc \phi}^2\rho^2\sin \phi\, d\rho\, d\theta\, d\phi &amp;= \int_{\frac{\pi}6}^{\frac{5\pi}6}\int_{\csc \phi}^2\int_0^{2\pi}\rho^2\sin \phi\,d\theta\,d\rho\,d\phi \\
&amp;= \int_0^{2\pi}\int_{\frac{\pi}6}^{\frac{5\pi}6}\int_{\csc \phi}^2\rho^2\sin \phi\,d\rho\,d\phi\,d\theta\\
&amp;=\int_0^{2\pi}\,d\theta\int_{\frac{\pi}6}^{\frac{5\pi}6}\int_{\csc \phi}^2\rho^2\sin \phi\,d\rho\,d\phi\\
&amp;=2\pi\int_{\frac{\pi}6}^{\frac{5\pi}6}\left(\sin \phi\int_{\csc \phi}^2\rho^2\,d\rho\right)\,d\phi\end{align}$$
since the integrand and limits are constant with respect to $\theta$, the $\sin \phi$ factor of the integrand does not depend on $\rho$.
Now $\int_{\csc \phi}^2\rho^2\,d\rho$ is not hard to figure out. And once you do, the integrand for $\phi$ may not be as bad as you think.