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[Calc 3] Verifying Stokes Theorem

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data

    V.Field F(x,y,z)=<x^2 z, xy^2, z^2> where S is part of the plane x+y+z=1 inside cylinder x2 + y2 =9

    2. Relevant equations

    Line integrals, Stokes Theorem, Parametrizing intersections...

    3. The attempt at a solution

    I found the answer to be 81pi/2 using stoke's theorem and the double integral, but now I have to verify it using a line integral and I'm stuck. I think r(t) is supposed to = <3cos(t), 3sin(t), 1-3cos(t)-3sin(t)> but it just turns out so nasty I'm wondering if there are pieces I'm missing. Any help is appreciated.
     
  2. jcsd
  3. Dec 8, 2011 #2

    vela

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    You're doing it right. It's just kind of tedious. This is what Mathematica came up with for the integrand, in case you want to check your algebra:
    [tex]27\,\sin^3 t - 18\,\sin^2 t + 3 \sin t - 27\,\cos^3 t +
    18\,\cos^2 t - 3 \cos t + 81 \sin t\,\cos^3 t + 162\,\sin^2 t\,\cos^2 t -
    54 \sin t\,\cos^2 t + 27 \sin^2 t\,\cos t[/tex]
    Integrate over the interval [itex][-\pi,\pi][/itex]. You can avoid doing most of the integrals by arguing they're going to be 0 for various reasons. Two will cancel with each other.
     
  4. Dec 8, 2011 #3
    Mind explaining why the interval is -pi to pi? Thought its be 0 to 2pi. Appreciate the help!
     
  5. Dec 8, 2011 #4
    Do it from 0 to 2pi, you'll see much easier which terms go to zero.

    [Hint] - any term which is going to leave you with just sine or cosine raised to some power will go to zero over that interval, can you see why?
     
  6. Dec 8, 2011 #5

    vela

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    You can use either one, but with the symmetric interval, you can use the fact that an odd integrand will result in the integral being 0 without actually having to do the integral. For example, the integral of sin3 t will be 0. If you're clever, you can argue why this result implies the integral of cos3 t is also 0.

    If you instead just grind out the integrals, which actually aren't too bad, it doesn't make a difference what interval you use. You should also recognize it shouldn't make a difference what interval you use as long as you end up going around the loop once.
     
    Last edited: Dec 8, 2011
  7. Dec 8, 2011 #6
    It's a matter of choice of which terms you want to make go to zero I suppose, but I like using 0 to 2Pi because for the trig functions it's equivalent to integrating over an interval with both end points being the same.
     
  8. Dec 8, 2011 #7

    vela

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    Which is also true of [itex][-\pi, \pi][/itex]. I don't see how 0 to [itex]2\pi[/itex] makes it any easier.
     
  9. Dec 8, 2011 #8
    Thanks so much guys, it helps a lot to know that I'm going down the right bath, even if the integral is a pain in the a**.
     
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