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Line integral

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve:
    [tex]\oint x^{99}y^{100}dx + x^{100}y^{99}dy[/tex]

    Assuming that it satisfies the conditions for Green's theroem, and:

    [tex] y = \sin{t} + 2, x = \cos{t}, 0 \leq t \leq 2\pi[/tex]

    2. Relevant equations

    Green's theorem.

    3. The attempt at a solution

    [tex]\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = 100(xy)^{99}[/tex]
    Which means that the integral is 0.
    Is this right? It's that simple, or am I missing something here?
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2

    Dick

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    Did you check that the contour was closed before you applied Greens theorem?
     
  4. Sep 22, 2009 #3
    Sorry, it should be [itex]2\pi[/itex]. Yes, it is closed.
     
  5. Sep 22, 2009 #4

    arildno

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    Rewrite the line integral's integrand as:
    [tex]\sin^{99}t\cos^{99}(t)(\cos^{2}t-\sin^{2}t)=\frac{1}{2^{99}}\sin^{99}(2t)\cos(2t)[/tex]

    Observe that Green's theorem can be verified in this particular case.
     
  6. Sep 22, 2009 #5
    I'm sorry, but I didn't understand how did you make it look like that?
    y = sint + 2, how did you get rid of the "+2"?

    About Green's theorem - that was how I did it the first time. Differentiating each of the terms and observing that they are equal, thus I get [itex]/iint_{D}0 dA[/itex].
    Someone else also pointed it out to me, that since this is a conservative field (with the potential being sin(xy)/100), a line integral of a conservative field in a closed circuit equals 0. Which is the same I get using Green's theorem.

    I would still like your explanation as for how you got the integrand in this form.
     
  7. Sep 22, 2009 #6

    arildno

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    Oops, I forgot about that!
    Nonsense on my part..

    The integrand will be slightly more complicated, I'll post a proper line integral solution later on
     
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