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Stokes Theorem

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that

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

    2. Relevant equations


    ##\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} ##

    3. 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: May 10, 2012
  2. jcsd
  3. May 10, 2012 #2


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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.
  4. May 10, 2012 #3
    I made an error. It is a squared term.
  5. May 10, 2012 #4


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    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.
  6. May 10, 2012 #5
    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.
  7. May 10, 2012 #6
    I dont understand this?
  8. May 10, 2012 #7
    [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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