Integrate this function over the volume of a sphere sphere

[Silversonic]

Homework Statement

Integrate \nabla^{2} (\frac{1}{|\underline{r} - \underline{r}'|}) over the volume of a sphere using the divergence theorem.

Homework Equations

\nabla^{2} (\frac{1}{|\underline{r} - \underline{r}'|}) = -4\pi\delta^{(3)}(\underline{r} - \underline{r}') (i.e. it's a dirac delta function, this just tells me that the answer I should get if I integrate over the volume is -4\pi)


\nabla^{} (\frac{1}{|\underline{r} - \underline{r}'|}) = -\frac{(\underline{r} - \underline{r}')}{|\underline{r} - \underline{r}'|^{3}}


\underline{r} = (x,y,z)

\underline{r'} = (x',y',z') (note that the dashes do not mean the differential of, just a different point in space from x,y,z).




And the divergence theorem

The Attempt at a Solution

\int_{V'}\nabla^{2} (\frac{1}{|\underline{r} - \underline{r}'|} )dV'

=
\int_{S'}\nabla (\frac{1}{|\underline{r} - \underline{r}'|} ) \circ d\underline{S}' (by the divergence theorem)

d\underline{S}' = \hat{\underline{n}} dS'

dS' = R^{2}sin(\theta)d\theta d\phi

I guess R should be the radius of the sphere. In which case I choose r' = R?

= \int_{S'}-\frac{(\underline{r} - \underline{r}')}{|\underline{r} - \underline{r}'|^{3}} \circ d\underline{S}'

Then, putting r' = R


\int_{S'}-\frac{(\underline{r} - \underline{R})}{|\underline{r} - \underline{R}|^{3}} \circ \hat{\underline{n}} dS'

=


\int^{2\pi}_{0} \int^{\pi}_{0} -(\frac{(\underline{r} - \underline{R})}{|\underline{r} - \underline{R}|^{3}} \circ \hat{\underline{n}}) R^{2}sin(\theta)d\theta d\phi


Am I right so far? This looks like a complete mess, and I'm not sure what to do with the dot product of the vector and the normal vector?

 

[Thaakisfox]

Whats the direction of the radius vector compared to the normal vector on the surface of the sphere? ;)

 

[Silversonic]

Whats the direction of the radius vector compared to the normal vector on the surface of the sphere? ;)

Cheers. Clearly they are in the same direction, so I can make the simplification \underline{R} \circ \hat{\underline{n}} = R

Then maybe with \underline{r} \circ \hat{\underline{n}} I can turn it into rcos(\theta)

Thus - (\underline{r} - \underline{R}) \circ \hat{\underline{n}} becomes

R - rcos(\theta)

Then what though? I guess I could turn the denominator into

(r^{2} + R^{2} - 2Rrcos(\theta))^{3/2} by the cosine rule, but then I get, overall;
\int^{2\pi}_{0} \int^{\pi}_{0} (\frac{R - rcos(\theta)}{(r^{2} + R^{2} - 2Rrcos(\theta))^{3/2}}) R^{2}sin(\theta)d\theta d\phi

But how would I do that integral with respect to theta?

 

[Thaakisfox]

yes very good. Now put x=\cos\theta. And separate the integrands.

Note that: \frac{d}{dx}(r^2+R^2-2Rrx)^{-1/2}) =\frac{-Rr}{(r^2+R^2-2Rrx)^{3/2}}