Multipole expansion - small problem

[Eats Dirt]

Homework Statement

Jackson 4.7

Given a localized charge distribution:

<br /> \rho(r)=\frac{1}{64\pi}r^{2} e^{-r} sin^{2}\theta<br />

make the multipole expansion of the potential due to this charge distribution and determine all nonvanishing moments. Write down the potential at large distances as a finite expansion in Legendre polynomials.

Homework Equations

<br /> \frac{1}{x-x&#039;}=4\pi\sum^{\inf}_{l=0}\sum^{l}_{m=-l}\frac{1}{2l+1}\frac{r^{l}_{&lt;}}{r^{l+1}_{&gt;}}Y^{*}_{l,m}(\theta&#039;,\phi&#039;)Y_{l,m}(\theta,\phi)<br />

The Attempt at a Solution

My main problem is with the \frac{r^{l}_{&lt;}}{r^{l+1}_{&gt;}} Term as I don't know what I should set the r values to in this case, my original idea was to use the r< term as some constant say R then proceed with the multipole expansion but I think the solution does not have this term in it.

 

[vela]

$r_< = \min\{r,r'\}$ and $r_> = \max\{r,r'\}$

 

[Eats Dirt]

Hey, I had another problem I input the multipole expansion into the integral

<br /> \frac{1}{\epsilon_o}\int^{r}_{0}\int^{2\pi}_{0}\int^{\pi}_{0}\frac{r&#039;^2 e^{-r&#039;}}{64\pi}4\pi\sin^2{\theta&#039;}\sum^{\inf}_{l=0}\sum^{l}_{m=-l}\frac{1}{2l+1}\frac{r&#039;^{l}}{r^{l+1}}Y^{*}_{l,m}(\theta&#039;,\phi&#039;)Y_{l,m}(\theta,\phi)r&#039;^2\sin\theta&#039;d\theta&#039;d\phi&#039;dr&#039;<br />

Now I have to integrate but have a problem on the sin\theta&#039; term, I know I can express the \sin^2\theta&#039;=1-\cos^2\theta&#039; which can then be used as a spherical harmonic but I do not know what to do with the sin(theta) term

 

[vela]

What $\sin\theta$ term?