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Given a localized distribution of charge with a charge density: $$\rho(\vec{r})=\rho(r,\theta)=\frac{1}{64\pi}r^2\frac{1}{e^r}\sin^2\theta$$

And I know that:

$$ \phi(\vec{r})=\sum_{l=0}^{\infty} \frac{1}{2l+1}\int_{V'}\frac{r_<^l}{r_>^{l+1}}\sum_{m =-l}^l Y^*_{l,m}(\Omega_<)Y_{l,m}(\Omega_>)\rho(\vec{r'})d^3r'$$

Where $$Y_{l,m}(\Omega)=Y_{l,m}(\theta,\varphi)=\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos\theta)e^{im\varphi}$$

Is it obvious that, due to the fact that $\rho\neq\rho(\varphi)$ (i.e. it is an azimuthally symmetric charge distribution), and the definition of $\phi$, only the $m=0$ moments will be nonvanishing??

Why?

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Yes it is obvious. Since $Y_{\ell m}\sim f(\theta)e^{im\varphi}$ and $\rho$ does not depend on $\varphi$, the integration over $\varphi$ will give $0$ unless $m=0$; in other words: $$ \int\, d\varphi \, e^{im\varphi}=2\pi\delta_{m0}\, . $$

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