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Integrate this function over the volume of a sphere sphere

  1. Dec 23, 2011 #1
    1. The problem statement, all variables and given/known data


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

    2. Relevant equations

    [itex]\nabla^{2}[/itex] ([itex]\frac{1}{|\underline{r} - \underline{r}'|}[/itex]) = -4[itex]\pi[/itex][itex]\delta^{(3)}[/itex][itex](\underline{r} - \underline{r}')[/itex] (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[itex]\pi[/itex])


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


    [itex]\underline{r} = (x,y,z)[/itex]

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




    And the divergence theorem

    3. The attempt at a solution

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

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

    [itex] d\underline{S}' = \hat{\underline{n}} dS' [/itex]

    [itex] dS' = R^{2}sin(\theta)d\theta d\phi [/itex]

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

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

    Then, putting r' = R


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

    =


    [itex]\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[/itex]


    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?
     
  2. jcsd
  3. Dec 24, 2011 #2
    Whats the direction of the radius vector compared to the normal vector on the surface of the sphere? ;)
     
  4. Dec 25, 2011 #3
    Cheers. Clearly they are in the same direction, so I can make the simplification [itex] \underline{R} \circ \hat{\underline{n}} = R [/itex]

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

    Thus [itex] - (\underline{r} - \underline{R}) \circ \hat{\underline{n}} [/itex] becomes

    [itex] R - rcos(\theta) [/itex]

    Then what though? I guess I could turn the denominator in to

    [itex] (r^{2} + R^{2} - 2Rrcos(\theta))^{3/2} [/itex] by the cosine rule, but then I get, overall;



    [itex]\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[/itex]

    But how would I do that integral with respect to theta?
     
  5. Dec 25, 2011 #4
    yes very good. Now put [itex]x=\cos\theta[/itex]. And separate the integrands.

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