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Stokes' Theorem

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose we want to verify Stokes' theorem for a vector field F = <y, -x, 2z + 3> (in cartesian basis vectors), where the surface is the hemispherical cap +sqrt(a^2 - x^2 - y^2)

    3. The attempt at a solution

    Why is it that if I substitute spherical coordinates x = asinθcosΦ, y = asinθsinΦ, z = acosθ, into F, and then take the curl where del = <d/dr, d/dθ, d/dΦ> (note that r = a in this case, so if the term involves only a, the d/dr of that term is 0), when I do double integral over curl F * dS with θ and Φ as parameters, I get 0?

    However, if I first take the curl of F where del = <d/dx, d/dy, d/dz> and THEN substitute spherical coordinates into the curl F, and dot with dS, and do double integral, I get the right answer ( i know the "right answer" because of the simplicity of the line integral).

    Why can I not substitute spherical coordinates into F and then take the curl where del = <d/dr, d/dθ, d/dΦ>?

    Why must I take the curl F where del = <d/dx, d/dy, d/dz> and then substitute spherical coordinates?
     
  2. jcsd
  3. Oct 31, 2009 #2

    Pengwuino

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    The curl in spherical coordinates is not something like [tex]
    \frac{\partial }{{\partial r}},\frac{\partial }{{\partial \theta }},\frac{\partial }{{\partial \varphi }}[/tex]

    It's more complicated then that. Whatever text you are using should have the curl in spherical coordinates. It's quite long.

    Edit: Ah, thank you wikipedia: http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates you're looking for the curl in spherical coordinates.
     
  4. Oct 31, 2009 #3
    Oh sorry. I was parametrizing in spherical coordinates but not necessarily using spherical basis vectors, which explains why I took the curl in cartesian coordinates (and did not need scale factors)
     
  5. Oct 31, 2009 #4

    Pengwuino

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    hmm? You didn't attempt to take the curl in cartesian though, you attempted to take them in spherical.
     
  6. Oct 31, 2009 #5
    I used spherical coordinate parameters but wrote in terms of x, y and z (cartesian coordinates).

    Ah. I figured out my mistake now. Since I did not rewrite F in spherical coordinates, I *cannot* take the curl with del as <d/dr, d/dtheta, d/dphi>--that would not work since F was expressed in cartesian coordinates.

    Thanks.
     
  7. Oct 31, 2009 #6

    Pengwuino

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    Well, remember, even if you put F in terms of spherical coordinates in the cartesian basis, the curl isn't simply [tex]\frac{\partial }{{\partial r}}\hat x + \frac{\partial }{{\partial \theta }}\hat y + \frac{\partial }{{\partial \phi }}\hat z[/tex] either.
     
  8. Oct 31, 2009 #7
    is it even possible to compute the curl of F of in terms of spherical coordinates in the cartesian basis, or would you have to convert F to spherical basis if F is in terms of spherical coordinates to compute the curl?

    (of course, the curl of F could be computed in cartesian basis in terms of cartesian coordinates, but just wondering)
     
  9. Oct 31, 2009 #8

    Pengwuino

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    Yes it's possible but kinda ugly I can imagine since you'd have to determine things like [tex]\hat x \cdot \hat \theta [/tex] that'll add to your derivatives. It's best to simply convert everything to the proper basis and parametrization for what you want to do.
     
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