• Support PF! Buy your school textbooks, materials and every day products Here!

Bounding surface

  • Thread starter rafaelpol
  • Start date
  • #1
17
0

Homework Statement



Say I have a sphere at the origin with radius "a". If I am integrating over a region in which r is greater than "a", how can the bounding surface of this volume be the spherical surface? This comes from an explanation in Greiner Classical Electrodynamics, in which he says the volume integral was done in a region such that r is greater than "a", but at the same time he says the bounding surface of this volume is the sphere...


The Attempt at a Solution



I feel like I am missing something on the defintion of a bounding surface, but the only definition I can find is the one that says that it is the surface that "binds" the whole volume. Hence, the spherical surface at "a" might be a part of the surface that is part of the volume "v", but there should be more, since the integration wasn't done over the sphere.

Any help will be appreciated.

Thanks
 

Answers and Replies

  • #2
2,544
2
Maybe you integrate from a<R<r , But im not sure
 

Related Threads for: Bounding surface

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
5
Views
4K
Replies
4
Views
468
Replies
2
Views
2K
Replies
4
Views
556
Replies
8
Views
2K
Top