- #1

Gianmarco

- 42

- 3

## 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?