# Calculating multipole expansion outside uniformly charged sphere

When calculating the potential outside a uniformly charged sphere with charge density $$\rho$$ and radius R using the multipole expansion, it makes sense that only the monopole term survives (since it looks like a point charge).

However, mathematically speaking, I am unsure why the dipole and higher-order terms vanish. Just by looking at the multipole expansion equation

$$V(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \sum_{n=0}^{\infty} \frac{1}{r^{(n+1)}} \int\left(r^{\prime}\right)^{n} P_{n}(\cos \alpha) \rho\left(\mathbf{r}^{\prime}\right) d \tau^{\prime}$$

I believe that $$r'$$ must be zero for this to work, where $$r'$$ is the magnitude of the vector from some reference point to some infinitesimal volume element. However, I cannot seem to intuit why this would be the case.

It has nothing to do with $$r’$$ being zero. The only place $$r’$$ is zero is at the origin.
When $$\rho$$ is spherically symmetric, it is independent of the angular coordinates, say the usual ones $$\theta’$$ and $$\phi’$$. Then, if one takes the $$z$$-axis along $$\mathbf r$$ so that the angle $$\alpha$$ between $$\mathbf r$$ and $$\mathbf{r}’$$ is just $$\theta’$$, the $$\theta’$$ integral is
$$\int_0^\pi P_n(\cos\theta’)\sin\theta’\,d\theta’=\int_{-1}^1 P_n(u)\,du =0$$
when $$n\ne 0$$, by the orthogonality of Legendre polynomials.