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!
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.