Another triple integral problem

[Gianmarco]

Homework Statement

Calculate
$$\int_D \frac{dxdydz}{\sqrt{x^2+y^2+(z-A)^2}}, \: A>R$$
on $D = {(x,y,z)\: s.t. x^2+y^2+z^2 \leq R^2}$.

Homework Equations

In spherical coordinates:
$$x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\ z=cos\phi\\dxdydz=\rho^2sin\phi d\theta d\rho d\phi$$

The Attempt at a Solution

Since the integrand is $\frac{1}{\rho}$ of a sphere centered in $(0,0,A)$, I think the most logical change of coordinates is:
$$x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\ z=\rho cos\phi + A$$
I am centered at a point $(0,0,A)$, so nothing changed with respect to the xy-plane, therefore $0 \leq \theta \leq 2\pi$; that much is obvious. And, since A>R, the boundaries for $\rho$ are $A-R \leq \rho \leq A+R$. I found those geometrically.
Changing the variables to$\rho, \theta, \phi$ in D we get:
$\rho^2 + 2\rho Acos\phi + A^2 \leq R^2$
And at this point I have no idea. Solving for $\phi$ would give me an upper limit, but it would be a an arccos of the form $arccos(\frac{a}{\rho} - b\rho)$ which I'd have to integrate in $\rho$ afterwards. Any ideas?

 

[blue_leaf77]

I have got the impression that the cylindrical coordinate is more helpful instead. In this case, the integral will be
$$
\int_0^{2\pi} \int_{-R}^R \int_0^{r(z)} \frac{rdrd\phi dz}{\sqrt{r^2+(z-A)^2}}
$$
Performing the integral over $r$ first should be easy.

 

[Gianmarco]

I was able to solve it, thank you for the hint blue leaf! :)