# Gravitational Potential

#### Shackleford

I'm not understanding this integral quite fully. Of course G is the constant and rho is also a constant since the sphere is homogeneous. And pi*r'^2 dr' gives the area of circle with respect to r' with limits of a and b. However, I'm not understanding how theta comes into it and the ignoring of phi. I understand the rest of the derivation.

Of course, the integral from a to b of pi*r'^2 dr' gives an area for the annulus in the plane. I guess I'm having trouble seeing the three-dimensional derivation.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-03144234.jpg?t=1286135265 [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-03144253.jpg?t=1286135285 [Broken]

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#### Quinzio

They use spherical coordinates.
A unit of volume in spherical c. is
$$V=\int \int \int \rho^2\ \sin\phi\ dp\ d\phi\ d\theta$$

phi and theta may be swapped in your text.

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#### Shackleford

Crap. Duh. I forgot about the transformation. Thanks.

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