Triple integral in spherical coordinates in an example

Question

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)?

Answer 1

$$\begin{align} \int_{\mathscr{R}^3}\psi_{210}^2(\vec r)\,dV&=\int_0^{2\pi}\int_0^\pi\int_0^\infty C^2e^{-r/a_0}\cos^2(\phi)r^2\,dr\sin^2(\phi)\,d\phi\,d\theta\\\\ &=C^2\,\left(\int_0^{2\pi}(1)\,d\theta\right)\,\left(\int_0^\pi \cos^2(\phi)\sin^2(\phi)\,d\phi\right)\,\left(\int_0^\infty r^2e^{-r/a_0}\,dr\right)\\\\ &=C^2\,(2\pi)\,\left(\frac{\pi}{8}\right)\,\left(2a^3\right)\\\\ &=\frac{\pi^2 a^3}{2}\,C^2 \tag 1 \end{align}$$

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

$$\begin{align} \bar r&=\int_0^{2\pi}\int_0^\pi\int_0^\infty r\,C^2e^{-r/a_0}\cos^2(\phi)r^2\,dr\sin^2(\phi)\,d\phi\,d\theta\\\\ &=C^2\,\left(\int_0^{2\pi}(1)\,d\theta\right)\,\left(\int_0^\pi \cos^2(\phi)\sin^2(\phi)\,d\phi\right)\,\left(\int_0^\infty r^3e^{-r/a_0}\,dr\right)\\\\ &=\frac{2}{\pi^2 a^3}\,(2\pi)\,\left(\frac{\pi}{8}\right)\,(6a^4)\\\\ &=3a \end{align}$$

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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

$$\begin{align} \bar r&=\frac{\int_{\mathscr{R}^3}r\psi_{210}^2(\vec r)\,dV}{\int_{\mathscr{R}^3}\psi_{210}^2(\vec r)\,dV}\\\\ &=\frac{\int_0^\infty r^3e^{-r/a_0}\,dr}{\int_0^\infty r^2e^{-r/a_0}\,dr}\\\\ &=\frac{6a^4}{2a^3}\\\\ &=3a \end{align}$$