Dipole Moment of a Hollow Sphere, simplify calculation

[LeoJakob]

Homework Statement

Consider the non-uniformly charged hollow sphere with radius $R$ and charge density$\rho(\vec{r})=\rho_{0} \cos \theta \delta(r-R) .$
Calculate the dipole moment $\vec{p}$

Relevant Equations

$\vec{p}=\int \vec{r} \rho(\vec{r})\mathrm{d}^{3} r .$

is there an easier way to calculate the dipole moment? I described $\vec r$ in spherical coordinates. I thought at first that due to the symmetry I can assume that dipole-moment only points in the $z$-direction, but the charge distribution is inhomogeneous, so I made the following calculation:
My calculation results in $$\vec{p}=\int \vec{r} \rho(\vec{r}) \mathrm{d}^{3} r=\int \limits_{0}^{2 \pi} \int \limits_{0}^{\pi} \int \limits_{0}^{\infty} R\left(\sin \theta \cos \phi \vec{e}_{x}+\sin \theta \sin \phi \vec{e}_{y}+\cos \theta \vec{e}_{z}\right)r^{2} \cdot \delta(r-R) \rho_{0} \sin \theta \cos \theta d r d \theta d \phi =\frac{4}{3} \rho_{0} R^{3} \overrightarrow{e_{z}}$$

Which is the correct result but the calculation of the integrals took quite some time. In the end I realized that the dipole moment,indeed, only has a $z$-component, could I have recognized this earlier and thus simplified my calculation? I'm unsure because the charge density is not homogenous.

 

[PhDeezNutz]

Weird, I would think that it was only necessary to do a double integral instead of a triple integral to calculate the dipole moment of this specific distribution. My interpretation is that there is only charge on the surface. Ah I see you used the delta function never mind.


I presume $\theta$ refers to the polar angle. $\cos \theta$ is even about the $z-axis$. If you go $\theta$ clockwise from the z-axis and $\theta$ counterclockwise from the z-axis , draw the vectors, you will see the components in the xy-plane cancel out.