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Volume integral for vectors

  • Thread starter jmz34
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  • #1
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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.
 

Answers and Replies

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

By the way, if you read the forum guidelines as you should have, then you know that "bumping" a thread may get you banned.
 
  • #3
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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.
 

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