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Stokes Thm - don't undestand this question fully

  1. Mar 17, 2006 #1
    Stokes Thm - I don't undestand this question fully

    In Stoke's law, let [itex] v_1 = -y [/itex] and [itex] v_2 = 0 [/itex] to show that the area of S equals the line integral [itex] -\int_C y\,\,\,dx [/itex] . Find the area of an ellipse ([itex] x = a \cos t [/itex], [itex] y = b \sin t [/itex], [itex] x^2/a^2+y^2/b^2 = 1 [/itex], [itex] 0 \leq t \leq 2\pi [/itex]).

    It's asking me to do the following:
    1) Compute the line integral [itex] -\int_C y\,\,\,dx [/itex] along an ellipse.
    2) Compute [itex] \int \int_S curl \vec v\,\,\, dx\,dy [/itex] with the conditions: [itex] v_1 = -y [/itex] and [itex] v_2 = 0 [/itex] along an ellipse.

    And (1) should equal (2)?

    This is how I'm interpretting the question, but my calculations are not agreeing. If my idea of how to interpret the question is correct I'll post my work. Thanks :)
    Last edited: Mar 17, 2006
  2. jcsd
  3. Mar 17, 2006 #2


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    Yes, that's what it says. And it works out.
  4. Mar 17, 2006 #3


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    Show us your calculations.
  5. Mar 17, 2006 #4
    Shoot. I was hoping I didn't understand the question :)


    [tex] -\int_C y\,\,dx = -\int_C \vec F \cdot d\vec r [/tex]

    [tex] \vec F = (y,0,0) [/tex]

    [tex] \vec r = (a \cos t , b \sin t , 0 ) [/tex]

    [tex] 0 \leq t \leq 2 \pi [/tex]

    [tex] d\vec r = (-a \sin t , b \cos t , 0 ) dt [/tex]

    -[tex] \int_0^{2\pi} (y)(-a \sin t )\,\, dt = \int_0^{2\pi}(b \sin t)(a \sin t)\,\, dt[/tex]

    Damn. I should stop right there :( !!!!
    Last time I substituted [itex] y = a \sin t [/itex] so I was getting [itex] a^2 \pi [/itex] from the integral. Now it should come out to [itex] a b \pi [/itex] which IS the area of an ellipse, and is the same answer I'm geting for the double integral using Stoke's Thm...

    well thanks for looking the question over for me :)
    Last edited: Mar 17, 2006
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