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Use Stokes Theorem to show a relationship

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the Stokes' Theorem to show that

    [itex]\int[/itex]f(∇ X A) dS = [itex]\int[/itex](A X ∇f) dS + [itex]\oint[/itex]f A dl

    2. Relevant equations

    Use vector calculus identities. Hint given : Start with the last integral in the above relation.

    3. The attempt at a solution
    To be honest, I really don't know how to start doing this. I could understand Stokes theorem when needed to evaluate the circulation, vector and surface, but I don't have any idea on how to start to prove this relation. Please help?
     
    Last edited: May 10, 2013
  2. jcsd
  3. May 10, 2013 #2
    Do you know any identities? What does stokes theorem state?
     
  4. May 11, 2013 #3
    The ones that I know are like the ones states in this webpage : http://hyperphysics.phy-astr.gsu.edu/hbase/vecal2.html

    Stokes theorem is [itex]\oint[/itex]F.dr = [itex]\int[/itex]∇ X F . n dS
     
  5. May 11, 2013 #4

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    Did you know that by convention ##\mathbf n dS## is the same as ##d\mathbf S##?

    So what do you get if you set ##\mathbf F = f \mathbf A## and ##\mathbf r = \mathbf l##?

    Is there another identity on your web page that you can apply then?
     
  6. May 11, 2013 #5
    Sorry but I still don't get it..quite confused. How should I actually begin proving it?
     
  7. May 11, 2013 #6

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    Start with [itex]\oint[/itex]F.dr = [itex]\int[/itex]∇ X F . dS

    What do you get if you substitute ##\mathbf F = f \mathbf A## and ##\mathbf r = \mathbf l##?
     
  8. May 11, 2013 #7
    I'll get [itex]\oint[/itex]fA.dl = [itex]\int[/itex]∇ X F . dS
     
  9. May 11, 2013 #8

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    Good!
    But there's still an ##\mathbf F## in there.
    Can you replace that too?
     
  10. May 11, 2013 #9
    So it will be [itex]\oint[/itex]fA.dl = [itex]\int[/itex]∇ X fA . dS
     
  11. May 11, 2013 #10

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    Right!
    Actually, let's make that [itex]\oint[/itex](fA).dl = [itex]\int[/itex]∇ X (fA) . dS

    Do you have another identity on your web page for something like that?
    Perhaps something like ##\nabla \times (u\mathbf A)##?
     
  12. May 11, 2013 #11
    Okay!

    Yeah ∇ x (uA) = u∇ x A - A x ∇u
     
  13. May 11, 2013 #12

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    Good!

    Substitute?
     
  14. May 11, 2013 #13
    [itex]\oint[/itex](fA).dl = [itex]\int[/itex] (u∇ x A - A x ∇u ) dS ?

    I can see where is this going now..But the original form they wanted is f(∇ x A) dS . so u∇ x A dS is equivalent to that form?
     
  15. May 11, 2013 #14

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    Yes.
    You were supposed to replace u by f as part of the substitution.
     
  16. May 11, 2013 #15
    alright. So the it doesn't matter that the f in the original form is outside the bracket and not just stuck with ∇?
     
  17. May 11, 2013 #16

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    In this case it doesn't matter. Both forms are the same.
    But as a rule, you should read ##f\nabla \times \mathbf A## as ##f(\nabla \times \mathbf A)##.
     
  18. May 11, 2013 #17
    Okay thank you for the great help!
     
  19. May 15, 2013 #18
    Hi there. I have another question. I realised that I made a mistake on my Stokes' Theorem equation at first. The surface integral should be a double integral. But for the one that I need to prove, it is just a single integral. Will that make any difference to my answer?
     
  20. May 15, 2013 #19

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    That's not really a mistake.
    You are integrating with ##d\mathbf S## which is an infinitesimal surface element.
    You can either use one or two integral symbols - it means the same thing.

    Formally, it should be a single integral symbol, since there is only one infinitesimal specified.
    It should only be a double integral if your integrating with the infinitesimals ##dx dy## or some such.
     
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