Patch of a surface in spherical coordinates?

  • Thread starter theBEAST
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  • #1
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Homework Statement


I am currently trying to prove:

S = ∫∫a2sinΦdΦdθ

Here is my work (note that in my work I use dS instead of S, this is an accident):
92zNaWh.jpg


I end up with:

S = ∫∫a*da2sinΦdΦdθ

Where da is the infinitesimal thickness of the surface.

Why am I getting the wrong answer?
 

Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement


I am currently trying to prove:

S = ∫∫a2sinΦdΦdθ

Here is my work (note that in my work I use dS instead of S, this is an accident):
92zNaWh.jpg


I end up with:

S = ∫∫a*da2sinΦdΦdθ

Where da is the infinitesimal thickness of the surface.

Why am I getting the wrong answer?
You haven't computed S. You've computed the volume V=S*da. To get S, divide by the da.
 
  • #3
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You haven't computed S. You've computed the volume V=S*da. To get S, divide by the da.
I thought that if the volumetric shell thickness approaches zero you end up with the surface area?
 
  • #4
Dick
Science Advisor
Homework Helper
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I thought that if the volumetric shell thickness approaches zero you end up with the surface area?
If the shell thickness approaches zero then you end up with zero volume. To get the surface area divide by the thickness as it approaches zero. Maybe that's what 'volumetric shell' means.
 
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