Surface Integral: Right Side = Left Side?

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Discussion Overview

The discussion revolves around the application of Stokes's Theorem in the context of electrodynamics, specifically addressing the relationship between surface integrals and line integrals. Participants explore the implications of choosing different surfaces that share the same boundary and the conditions under which the theorem holds.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions which surface to choose when multiple surfaces share the same boundary, suggesting that while the right side of the equation may equal the left, the reverse may not always hold.
  • Another participant asserts that it does not matter which surface is chosen, emphasizing the flexibility provided by the theorem and mentioning that different surfaces, such as a flat disk or half-sphere, can be used depending on symmetry.
  • A later reply reiterates the idea that the choice of surface is not critical, referencing Stokes's Theorem as a foundational principle.

Areas of Agreement / Disagreement

Participants express differing views on the implications of surface choice in relation to Stokes's Theorem. While some assert that the choice is flexible, others raise concerns about the conditions under which the theorem applies, indicating that the discussion remains unresolved.

Contextual Notes

The discussion highlights potential limitations regarding the assumptions made about the surfaces and the specific conditions under which Stokes's Theorem is applicable. There is an acknowledgment of the need for a proof, which remains unprovided.

pardesi
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May be this should have been in math section but since this came out while studying Electrodynamics i put it here
we have
[tex]\boxed{\int_{S} \nabla \times \vec{B}.d\vec{a}=\oint \vec{B}.d\vec{l}}[/tex]

Q.well there are many areas with the same boundary which one to choose from?

well if we know the area the boundary is fixed but not vice-versa does only the right side equal left but nor always the left side equals right.
Can someone explain
 
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It doesn't matter which surface you take, that's the beauty of it. For example, if the curve on the right hand side is a circle, depending on the symmetry it may be easiest to take either a flat disk, or a half-sphere, or anything else.

There's probably a nice proof for it too, but I wouldn't be able to give you that by heart.
 
wow
tat's gr8
 
CompuChip said:
It doesn't matter which surface you take, that's the beauty of it. For example, if the curve on the right hand side is a circle, depending on the symmetry it may be easiest to take either a flat disk, or a half-sphere, or anything else.

There's probably a nice proof for it too, but I wouldn't be able to give you that by heart.

Stokes's Theorem.
 

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