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Stokes Theorem

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, Cis its boundary, and the vector field F = 3yi - xzj + yzk.


    3. The attempt at a solution

    I had found (∇xF) = (z+x)i + (-z-3)k

    r = [u, v, 0.5(u^2 + v^2)]
    Therefore N= ru X rv = -ui -uj +k
    Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
    After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
    Thus ∫(0-2)∫(0-2pi) (∇xF).Nr dθdr

    But I cant seems to get the answer. Can anyone help? Help would greatly appreciated :)
     
  2. jcsd
  3. May 11, 2010 #2

    vela

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    Looks okay. What did you get for the answer or where are you getting stuck?
     
  4. May 11, 2010 #3
    ok so its ∫(0-2)∫(0-2pi) (∇xF).Nr dθdr

    Therefore its:
    ∫(0-2)∫(0-2pi) -xz-x^2-z-3 dA (Using x = rcos(θ), y = rsin(θ), z = 0.5r^2)
    = ∫(0-2)∫[(0-2pi) -0.5r^3cosθ - r^2cos^2(θ) - 0.5r^2 - 3]r dθdr

    Im not sure is my steps correct? Jus a little problem with integrating cos^2(θ).

    Furthermore, I've used ∫F.dr to do a check and the answer seems to be 0. Is this correct?
     
  5. May 11, 2010 #4

    vela

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    Looks good, but I got -20 pi for the answer both ways.

    To integrate cos^2(θ), use the trig identity cos^2(θ)=[1+cos(2θ)]/2.
     
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