Another triple integral problem

[Gianmarco]

Homework Statement

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

Homework Equations

In spherical coordinates:
[tex]
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
[/tex]

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:
[tex]
x=\rho cos\theta sin\phi\\y=\rho sin\theta sin\phi\\ z=\rho cos\phi + A
[/tex]
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! :)