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Homework Help: Stokes' Theorem problem

  1. Mar 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Use stokes' theorem to find I = [tex]\int\int (\nabla x F) n dS[/tex] where D is the part of the sphere [tex]x^2 + y^2 + (z-2)^2 = 8[/tex] that lies above the xy plane, and

    [tex]F=ycos(3xz^2)i + x^3e^[-yz]j - e^[zsinxy]k[/tex]


    Attempt at solution:
    I want to use the line integral [tex]\int F dr[/tex] to solve this.

    I parametrize the boundary [tex]r(t) = (2cos\theta)i + (2sin\theta)j + 0k)[/tex] (the sphere on the xy plane)

    Then F dotted with r'[t] is [tex]\int -4(sin\theta)^2 + (2cos\theta)^4 [/tex]

    and 0 < theta < 2pi

    And the answer that I get is 8pi. Is this correct? More specifically is my boundary curve legit, considering that the sphere "bulges" out above this boundary? :S
     
    Last edited: Mar 21, 2010
  2. jcsd
  3. Mar 20, 2010 #2
    I just realized that on the xy plane the radius would be 2, not sqrt(8). Nonetheless, is this otherwise okay?
     
  4. Mar 20, 2010 #3

    ideasrule

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    This is not what I got. Did you get dr/d(theta)=-2sin(theta)i + 2cos(theta)j?

    That's not a problem. The boundary can curve in or out as much as it wants, but Stokes' theorem always holds true.
     
  5. Mar 20, 2010 #4
    Is the answer you got 0?
     
  6. Mar 21, 2010 #5

    ideasrule

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    Me? Yes, I got 0.
     
  7. Mar 21, 2010 #6
    Since stoke's theorem is an alt of green's, wouldn't it be easier to just compare the 6 partial derivatives? Because the integral will be 0 if it is conservative which occurs if they are all equal, correct?
     
  8. Mar 21, 2010 #7
    I changed the r[t] to [tex]2cos\theta,2sin\theta,0[/tex]. For the F dot r'[t] I got [tex]\int -4(sin\theta)^2 + (2cos\theta)^4[/tex] with theta from 0 to 2pi. Why is this not giving me zero :S ?
     
  9. Mar 21, 2010 #8

    ideasrule

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    Try it. I don't think the field is conservative.
     
  10. Mar 21, 2010 #9

    ideasrule

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    You've done everything right so far. If you don't get 0, it's just a mistake in your integration somewhere.
     
  11. Mar 21, 2010 #10
    I double checked on maple and 8pi seems correct, but I get zero when I directly compute the double integral without using Stokes'...
     
    Last edited: Mar 21, 2010
  12. Mar 21, 2010 #11

    gabbagabbahey

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    I get [itex]8\pi[/itex] using both methods, so I'm pretty sure you are just doing something wrong when you calculate the double integral directly.
     
  13. Mar 21, 2010 #12
    Can anyone confirm this? We have two homework helpers with two different answers!
     
  14. Mar 21, 2010 #13
    Shameless bump
     
  15. Mar 21, 2010 #14

    gabbagabbahey

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    I've asked in the homework helpers forum for another HH to take a look when they get time, so hopefully someone will reply.
     
  16. Mar 22, 2010 #15

    vela

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    I get [itex]8\pi[/itex] for the line integral. Is there a trick to evaluating the double integral? I get a mess for curl F.
     
  17. Mar 22, 2010 #16
    When I found the curl I took z to equal 0, and so the curl ends up being 3x2-1.
    I then did [tex]\int\int3x^{2}-1 dA = \int_{0}^{2\pi}\int_{0}^{2}(r^{3}cos^2 {\phi} - r) drd{\phi} [/tex]
    this gives me zero. There is a similar problem in my book and this seems to be how it's done...
     
  18. Mar 22, 2010 #17

    ideasrule

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    Are we all integrating [tex]
    \int -4(sin\theta)^2 + (2cos\theta)^4
    [/tex] from 0 to 2pi?
     
  19. Mar 22, 2010 #18
    I did my integral a different way.
     
  20. Mar 22, 2010 #19

    ideasrule

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    Which way?
     
  21. Mar 22, 2010 #20
    I took the curl on the cirlce in the xy plane, so z=0.
     
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