Stokes Theorem

  • Thread starter Gregg
  • Start date
  • #1
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Homework Statement



Prove that

## \oint_{\partial S} ||\vec{F}||^2 d\vec{F} = -\int\int_S 2 \vec{F}\times d\vec{A} ##

Homework Equations



Identities:

##\nabla \times (||\vec{F}||^2 \vec{k}) = 2\vec{F} \times \vec{k} ##

For ##\vec{k} ## constant i.e. ## \nabla \times \vec{k} = 0 ##

Stokes Theorem

##\oint_{\partial S} \vec{B} \cdot d\vec{x} = \int\int_S (\nabla \times \vec{B})\cdot d \vec{A} ##

The Attempt at a Solution



So I need to use that identity ##\nabla \times (||\vec{F}||^2 \vec{k}) = 2\vec{F} \times \vec{k} ##

The problem is that Stokes theorem is in a different form. The constant vector here I think is the k=dA.

I really can't think of what to do
 
Last edited:

Answers and Replies

  • #2
DryRun
Gold Member
838
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Well, [itex]\vec F[/itex] is the vector field, but i'm not sure what [itex]||\vec F||[/itex] represents in the original equation.
 
  • #3
459
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I made an error. It is a squared term.
 
  • #4
DryRun
Gold Member
838
4
Assuming that there are no more mistakes in your first post, [itex]\vec k[/itex] represents the outward unit normal vector from the surface, S.
 
  • #5
938
9
Maybe you should use Stokes' theorem for each component of the original vector-valued integral
[tex] \vec{I}= \oint ||\vec{F}||^2 d\vec{F} [/tex]
[tex] I_x = \oint (||\vec{F}||^2 \vec{e}_x) \cdot d\vec{F} [/tex]
etc. Now it's in the correct form.
 
  • #6
459
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Maybe you should use Stokes' theorem for each component of the original vector-valued integral
[tex] \vec{I}= \oint ||\vec{F}||^2 d\vec{F} [/tex]
[tex] I_x = \oint (||\vec{F}||^2 \vec{e}_x) \cdot d\vec{F} [/tex]
etc. Now it's in the correct form.

I dont understand this?
 
  • #7
938
9
I dont understand this?

[itex]\vec{e}_x [/itex] is the unit vector in x-direction. The lower integral is just the x-component of your full integral. You can calculate this by taking the dot product with [itex] \vec{e}_x [/itex].
 

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