Volume integral for vectors

[jmz34]

Homework Statement

I'm stuck on the following vector integral

Q(x)=INT[(p(y)/|x-y|)(dy)^3

For a sphere of uniform p (so it is not a function of y in this case). Where x is the position vector of a point lying outside the sphere and y is the position vector of a point lying inside the sphere.

Homework Equations

The Attempt at a Solution

I attempted this by taking advantage of the symmetry and picking x to lie along the z axis. I then attempted to integrate it in spherical polar coordinates and wrote the components of y in terms of theta and phi (the latter being the azimuthal angle). I wrote the volume element in spherical polars ignoring the vector notation for now. But after doing all this I get to an expression:

Q=2pi*p INT[(r^2*sin^2(theta))/(|x|^2-2|x|rcos(theta)+r^2)^1/2]drd(theta)

Which seem's like a very difficult integral to me.

Thanks in advance for your help.

 

[HallsofIvy]

That's NOT really a "vector" problem, because the length of a vector is a scalar. Taking your fixed vector, $\vec{x}$, outside the sphere, to be $<x_0, y_0, z_0>$, then your integration is
$$\int\int\int \frac{p}{\sqrt{(x-x_0)^2+ (y- y_0)^2+ (z- z_0)^2}}dxdydz$$

By the way, if you read the forum guidelines as you should have, then you know that "bumping" a thread may get you banned.

 

[jmz34]

If I attempt to do this in Cartesian coordinates the limits are:

Zmin(max)= -(+)R
Ymin(max) = -(+)SQRT(R^2-z^2)
Xmin(max) = -(+)SQRT(R^2-y^2-z^2)

Maybe I'm not seeing something obvious but the integral still seems difficult to me.