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Homework Help: Identity proof using Stoke's Theorem

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Show using Stoke's Theorem that gif.latex?\bg_white%20\iint_S%20(\nabla%20f)\times%20d\vec%20S%20=%20-\oint_C%20fd%20\vec%20r.gif

    S is an open surface with boundary C (a space curve). [tex]f(\vec r)[/tex] is a scalar field.

    2. Relevant equations

    Stoke's theorem [tex]\iint_S (\nabla\times \vec F) \cdot d\vec S = \int_C F \cdot d\vec r[/tex]

    3. The attempt at a solution

    Thus far I've tried evaluating [tex] (\nabla f)\times d\vec S[/tex] by taking components in various ways and then trying to force it to become something that I can take the curl of... with no success at all. I'm completely stumped now.
  2. jcsd
  3. May 16, 2010 #2


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    Try applying Stokes' theorem to the vector function [itex]\textbf{F}\equiv f\textbf{c}[/itex], where [itex]\textbf{c}[/itex] is any constant vector.:wink:
  4. May 16, 2010 #3
    Well, what I get here is

    [tex] \iint_S \vec c\cdot\nabla f \times d\vec S = -\oint \vec c \cdot fd\vec r[/tex]

    And I'm not entirely sure whether I can just "cancel" the c dot product...

    (should be a vector c there, by the way)
    Last edited: May 16, 2010
  5. May 16, 2010 #4


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    Well, since [itex]\textbf{c}[/itex] is a constant vector, you can certainly pull it out of the integrals and say

    [tex]\vec c\cdot\left(\iint_S \nabla f \times d\vec S\right) = \vec c\cdot\left(-\oint fd\vec r\right)[/tex]

    [tex]\implies \vec c\cdot \left(\iint_S \nabla f \times d\vec S+\oint fd\vec r\right)=0[/tex]

    And since this is true for arbitrary, constant [itex]\vec{c}[/itex], what must you conclude?
  6. May 16, 2010 #5
    Yeah, those last steps were fine for me, I just wasn't sure if that even with c a constant vector that it was "allowed" to just pull it out like so. Thanks for the help.
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