I am trying to solve integral

$$I=\int \frac{dS}{\sqrt{\frac{x^2}{a^2}+\frac{y^2}{a^2}+\frac{z^2}{b^2}}}$$ over a sphere, where $r = \sqrt{\frac{x^2}{a}+\frac{y^2}{a}+\frac{z^2}{b}}$.

I thought about using generalized spherical coordinates like $\frac{x}{\sqrt{a}}=r \cos{\phi}\sin{\theta}$ etc. but $a$ and $b$ are squared in the integral...

How to solve this integral?

  • $\begingroup$ by sphere here you mean the ball (the volume of the sphere) or just the surface? $\endgroup$ – Masacroso Apr 21 at 12:58
  • $\begingroup$ just the surface $\endgroup$ – Andrej Apr 21 at 16:10

Ok, sorry about the last comment. I tried crossing things out, it didn't work.

I think that the best that can be done is the following substitution:

$$ \left\{ \begin{array}{c} x=arcos(\theta) \\ y=arsin(\theta) \\ z=bv \end{array} \right. $$ And then integrating with respect to r first then z then $\theta$

Because the determinant of the determinant of the Jacobian has an r in it.

And then you need to separate the two the integrals, and do a trigonometric substitution to each one alone.

Final note: the initial integrand is independent of $\theta$ so you can multiply by 2$\pi$ from the beginning.


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