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The triple integral given is:

$$\int\int\int \frac1{\sqrt{(x-a)^2 + y^2 + z^2}}dV ,a\ge 2 $$

The region of integration given is $ 0\le x^2 + y^2 + z^2\le 1$

The region of integration is clearly spherically symmetrical so I tried to use spherical coordinates. However, I ended up with a triple integral that seemed unsolvable. Both the region and the integrand seem to have cylindrical symmetry over the x-axis. Sadly, I got stuck trying to solve that integral as well. Any hints on how to proceed with this question would be appreciated!!

P.S This is my first post so please give me some advice on formatting!

EDITS:

1) Using cylindrical coordinates I get:

$$\int\int\int \frac1{\sqrt{r^2 + z^2 -2arcos(\theta) +a^2 }}rdzdrd\theta $$

With the region:
$-\sqrt{1-r^2}\le z \le \sqrt{1-r^2}$
$ 0\le r\le 1$
$ 0\le \theta\le 2\pi$

2) Using spherical coordinates I get:

$$\int\int\int \frac1{\sqrt{r^2 -2rsin(\phi)cos(\theta) +a^2 }}r^2sin(\phi)drd\phi d\theta $$

With the region:
$ 0\le r\le 1$
$ 0\le \theta\le 2\pi$
$ 0\le \phi\le \pi$

EDIT NUMBER 2:

Following Martin's advice and using cylindrical coordinates I get:

$$\int\int\int \frac1{\sqrt{r^2 + (z-a)^2}}rdzdrd\theta $$

With the region:
$-\sqrt{1-r^2}\le z \le \sqrt{1-r^2}$
$ 0\le r\le 1$
$ 0\le \theta\le 2\pi$

While this does seem solvable, does anyone have any hints on where to start. The issue now it that the first integral's antiderivative gives: $$\int\int r[ln\left|(z-a) + \sqrt{r^2 + (z-a)^2}\right|]_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}$$ I have no idea how to integrate this complicated expression further. Any help is greatly appreciated!

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    $\begingroup$ Would you mind posting your work on the conversion of the integral from Cartesian to Spherical and your conversion to Cylindrical? $\endgroup$ – AlgorithmsX Nov 24 '16 at 18:38
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    $\begingroup$ Did you use this(or some mosification of this) substitution? $$x=r\cos a \cos b\ ;\ y=r\cos a\sin b\ ;\ z=r\sin a$$ ? If so, it would be helpful if you post your work. $\endgroup$ – Qwerty Nov 24 '16 at 18:42
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    $\begingroup$ I suggest you go through the tour section under the help tab for more information on how to format and post your answer. Generally, if we can see the work you yourself have done, we either can use it to help answer your problem and see where your problem lies. $\endgroup$ – AlgorithmsX Nov 24 '16 at 19:34
  • $\begingroup$ @Qwerty, the $a$ is already used in the integral. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 24 '16 at 20:40
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    $\begingroup$ you can kill the $\theta$ dependence by following Martin's advice $\endgroup$ – tired Nov 24 '16 at 21:20
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Idea, too long for a comment. By the obvious symmetry: $$\iiint\frac1{\sqrt{(x-a)^2 + y^2 + z^2}}dV = \iiint\frac1{\sqrt{x^2 + y^2 + (z-a)^2}}dV$$ and try cylindrical coordinates.

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  • $\begingroup$ The right side seems doable but, as seen above, resulted in a very complicated double integral after computing the dz integral first. Thanks! $\endgroup$ – Akarsh Verma Nov 24 '16 at 23:36

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