Triple integral in spherical coordinates in an example I am not sure how to do this. I am given a function in spherical coordinates.
$C$ is a normalization constant given by the triple integral.
How can I find C and use that to do part (b)?

 A: $$\begin{align}
\int_{\mathscr{R}^3}\psi_{210}^2(\vec r)\,dV&=\int_0^{2\pi}\int_0^\pi\int_0^\infty C^2e^{-r/a_0}\cos^2(\phi)r^2\,dr\sin^2(\phi)\,d\phi\,d\theta\\\\
&=C^2\,\left(\int_0^{2\pi}(1)\,d\theta\right)\,\left(\int_0^\pi \cos^2(\phi)\sin^2(\phi)\,d\phi\right)\,\left(\int_0^\infty r^2e^{-r/a_0}\,dr\right)\\\\
&=C^2\,(2\pi)\,\left(\frac{\pi}{8}\right)\,\left(2a^3\right)\\\\
&=\frac{\pi^2 a^3}{2}\,C^2 \tag 1
\end{align}$$
Therefore, setting the right-hand side of $(1)$ equal to $1$ yields
$$\bbox[5px,border:2px solid #C0A000]{C=\sqrt{\frac{2}{\pi^2 a^3}}}$$
To find the mean distance $\bar r$, we have
$$\begin{align}
\bar r&=\int_0^{2\pi}\int_0^\pi\int_0^\infty r\,C^2e^{-r/a_0}\cos^2(\phi)r^2\,dr\sin^2(\phi)\,d\phi\,d\theta\\\\
&=C^2\,\left(\int_0^{2\pi}(1)\,d\theta\right)\,\left(\int_0^\pi \cos^2(\phi)\sin^2(\phi)\,d\phi\right)\,\left(\int_0^\infty r^3e^{-r/a_0}\,dr\right)\\\\
&=\frac{2}{\pi^2 a^3}\,(2\pi)\,\left(\frac{\pi}{8}\right)\,(6a^4)\\\\
&=3a
\end{align}$$

NOTE:
We did not actually need to find $C$ to determine $\bar r$.  Rather, since the integrand is separable in terms of the product of functions of $r$, $\theta$, and $\phi$, $bar r$ is simply given by
$$\begin{align}
\bar r&=\frac{\int_{\mathscr{R}^3}r\psi_{210}^2(\vec r)\,dV}{\int_{\mathscr{R}^3}\psi_{210}^2(\vec r)\,dV}\\\\
&=\frac{\int_0^\infty r^3e^{-r/a_0}\,dr}{\int_0^\infty r^2e^{-r/a_0}\,dr}\\\\
&=\frac{6a^4}{2a^3}\\\\
&=3a
\end{align}$$
