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Surface Integral over a Cone - Stokes?

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    zuKnP.png

    2. Relevant equations

    I'm guessing Stoke's Theorem? However, I'm not sure how to apply it exactly..

    3. The attempt at a solution

    Looking at Stoke's Theorem, I'm still not sure how to apply it. I'm really just lost as to where to begin; is there even a [itex]\grad F[/itex] to take? I know this is related to Stoke's (I hope), but not sure how to begin.
     
    Last edited: Nov 11, 2012
  2. jcsd
  3. Nov 11, 2012 #2

    Dick

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    It doesn't look like Stoke's to me. I don't see any vectors anywhere. Your first job is to figure out what dA is in terms of dx*dy. Then a change to polar coordinates might make it easier to do the dx*dy integral. It's actually pretty straight forward.
     
    Last edited: Nov 11, 2012
  4. Nov 11, 2012 #3
    I also am trying this formula:

    eq0002M.gif

    But I'm not getting to 9pi.. I'm not sure what the difference between f(x, y, z) and f(x, y, g(x, y)) is.
     
  5. Nov 11, 2012 #4

    Dick

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    That's the right direction. Your g(x,y) is just z. Solve your equation for z^2 for z to get g(x,y). It's sqrt(3)*sqrt(x^2+y^2), right?
     
  6. Nov 11, 2012 #5
    I did, that and my integrand on the right side came out to be 2(x^2 + y^2).

    But I don't see how to get to the correct answer from that.
     
  7. Nov 11, 2012 #6

    Dick

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    Ok. That's good. The easiest way to do it from here is to change to polar coordinates. Did you try that?
     
  8. Nov 11, 2012 #7
    Yes. But if I change to polar, I get 18pi, because of the 2 there.
     
  9. Nov 11, 2012 #8

    Dick

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    Which 2? Show your work, ok?
     
  10. Nov 11, 2012 #9
    Urgh, I just didn't want to type it out because I don't know how to use LaTeX and it looks messy.

    Integrand: 2(x^2 + y^2) dx dy

    Changed to polar w/:
    x = sqrt(3) * cos t
    y = sqrt(3) * sin t

    New Integrand: 6r dr dt [r from 0 to sqrt(3), t from 0 to 2pi]
     
  11. Nov 12, 2012 #10

    Dick

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    And there's your problem. The integrand is supposed to be integrated over r. You don't put a specific value of r in. x^2+y^2 is just r^2.
     
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