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Greens theorem

  1. Aug 1, 2014 #1
    1. The problem statement, all variables and given/known data

    ##\mathscr{C}## is an ellipse ##\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1##
    and ##\vec{F}(x,y) = <xy^2, yx^2>##

    write ##\displaystyle \int_\mathscr{C} \vec{F} \cdot d\vec{s}## as a double integral using greens theorem and evaluate
    2. Relevant equations

    ##\displaystyle \int_\mathscr{C} (Pdx+Qdy) = \iint_\mathscr{C} \Bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigg)dA##

    3. The attempt at a solution

    seems to be I need to use ##\nabla \times \vec{F} = \Bigg(0,0, \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}\Bigg) = (2xy-2yx)=0##


    not sure about the double integral though, figured maybe this

    ##\displaystyle \int_{-a}^{a} \int_{-\sqrt{1-\frac{x^2}{a^2}-b^2}}^{\sqrt{1-\frac{x^2}{a^2}-b^2}}0dydx=0##
     
    Last edited: Aug 1, 2014
  2. jcsd
  3. Aug 1, 2014 #2

    Orodruin

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    Your problem is two-dimensional so the double integral should be over the interior of the ellipse.

    If you necessarily want to see it as a 3D problem and apply Stoke's theorem, then the surface can be any surface with the ellipse as its boundary.
     
  4. Aug 1, 2014 #3
    this was close to an example in my book, so where is it incorrect?

    the book showed taking curl F then integrating over the boundary of the region in the given problem.

    I don't know Stoke's theorem yet so I'd rather not try to apply that to this
     
  5. Aug 1, 2014 #4

    Orodruin

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    You need to rethink your limits for the y-integral, but they do not really matter since the integrand is zero.

    Stoke's theorem (in the simple classical form) relates the integral of the curl of a vector field over a surface with the line integral of the same field along the border of that surface.
     
  6. Aug 1, 2014 #5
    ah okay thanks, ##-b \le y \le b## in similar fashion to how x was treated?
     
  7. Aug 1, 2014 #6

    vela

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    No, you need to check your algebra when you solved for ##y##.
     
  8. Aug 1, 2014 #7
    ##y=\pm \frac{b\sqrt{a^2-x^2}}{a}##


    ##\displaystyle\int_{-a}^{a}\int_{-\frac{b\sqrt{a^2-x^2}}{a}}^{\frac{b\sqrt{a^2-x^2}}{a}} 0 dydx##
     
  9. Aug 2, 2014 #8

    HallsofIvy

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    What is the integral of 0 over any region?
     
  10. Aug 2, 2014 #9
    It would be zero
     
  11. Aug 3, 2014 #10
  12. Aug 3, 2014 #11

    HallsofIvy

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    Yes, so there was no reason to worry about the limits on the integral to begin with. That was what Orodruin meant when he said "they do not really matter since the integrand is zero".
     
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