Gravitational Potential

1,642
2
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]
 
Last edited by a moderator:
556
1
They use spherical coordinates.
A unit of volume in spherical c. is
[tex]
V=\int \int \int \rho^2\ \sin\phi\ dp\ d\phi\ d\theta

[/tex]

phi and theta may be swapped in your text.
 
Last edited:
1,642
2
Crap. Duh. I forgot about the transformation. Thanks.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top