# Inconsistent integral and distance in spherical coordinates

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 $\theta$.)

• 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. – auxsvr Jun 13 '14 at 16:42

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 $\phi \neq \phi '$.
• can you say what the exact equation is for $||r-r'||$? – Xin Wang Jun 13 '14 at 14:54
• It's the expansion of $\frac{1}{|r-r'|}$ in spherical harmonics. See eq (105) in this document – garyp Jun 13 '14 at 15:28