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I am currently studying this problem: 14 b)

There you see an integral $$A(r) = \int f(\theta) (-\sin(\phi), \cos(\phi),0) d \Omega$$ where $f$ is the function containing all the rest of the integrand that you see there and I don't see why this integral is not zero? (I am especially referring to the integral in the second row of part b) . I mean clearly: $$\int_{0}^{2\pi} (-\sin(\phi), \cos(\phi),0) d\phi = 0$$ so why does this integral not vanish completely?

Also I don't get why there is this $\phi'$ in the denominator? I mean, don't we have $$||r-r'||= \sqrt{r^2+r'^2-2rr'\cos(\theta-\theta')}$$ so this should not depend on $\phi'$? (This is the reason why I said that $f$ only depends on $\theta$.)

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  • $\begingroup$ If an integral vanishes and you multiply the integrand with an arbitrary function, the integral changes, why should it still vanish? The position vector on a spherical shell depends on $(\theta, \phi)$, the angle in your equation is not the same as the $\theta$ in my equation. If you make a diagram, everything should be clear. $\endgroup$
    – auxsvr
    Jun 13, 2014 at 16:42

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Your statement about $|| r-r'||$ is true only if $r$ and $r'$ have the same $\phi$ coordinate. (same "longitude") The denominator does have a $\phi '$ dependance. The value of that modulus will be larger when $\phi \neq \phi '$.

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  • $\begingroup$ can you say what the exact equation is for $||r-r'||$? $\endgroup$
    – Xin Wang
    Jun 13, 2014 at 14:54
  • $\begingroup$ In the context of this problem, it looks like the expression you are looking for appears in the second-to-last line of the solution. It appears that the denominator is expanded in spherical harmonics. (It might be the Green's function for the problem?) I think a general answer to your last question is difficult, because you'll might want different expressions in different contexts. For the distance alone you could start with the distance in Cartesian coordinates and then sub in the spherical coords. $\endgroup$
    – garyp
    Jun 13, 2014 at 15:21
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    $\begingroup$ It's the expansion of $\frac{1}{|r-r'|}$ in spherical harmonics. See eq (105) in this document $\endgroup$
    – garyp
    Jun 13, 2014 at 15:28
  • $\begingroup$ I see, but probably there is a way to write this without using spherical harmonics or do you think that's the only way? Cause I have the feeling that they use the spherical harmonics, just because this makes the integration easier. $\endgroup$
    – Xin Wang
    Jun 13, 2014 at 15:29
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    $\begingroup$ Yes, I think that's exactly why it was done. Other ways would be harder, and perhaps impossible to integrate. $\endgroup$
    – garyp
    Jun 13, 2014 at 15:30

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