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Math Help - Greens theorem

  1. #1
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    Greens theorem

    Hi. I have a problem with this exercise. I wanted to verify the greens theorem for the vector field F(x,y)=(3x+2y,x-y) over the path \lambda[0,2\pi]\rightarrow{\mathbb{R}^2},\lambda(t)=(\cos t, \sin t)

    The Green theorem says: \displaystyle\int_{C^+}Pdx+Qdy=\displaystyle\int_{  }\int_{D}\left (\frac{{\partial Q}}{{\partial x}}-\frac{{\partial P}}{{\partial y}}\right ) dxdy
    and I have: P(x,y)=x-y,Q(x,y)=3x+2y

    So then I've made the line integral:
    \displaystyle\int_{0}^{2\pi}\left [-(3\cos t +2 \sin t)\sin t+ (cos t -\sin t)\cos t\right ]dt=-\pi
    Creo que esto esta bien, la integral la resolví con la computadora para no tener problemas

    And then the double integral:
    \displaystyle\int_{-1}^{1}\int_{-\sqrt[ ]{1-x^2}}^{\sqrt[ ]{1-x^2}}(1-2y)dydx=\pi

    The problem is clearly with the signi. The mistake I think I've committed was putting on the reverse the integral limits for the double integral. I think that x should go from 1 to -1, but the thing is I don't know why. So I'm not pretty sure on how to determine the integral limits on this cases. I've tried to think about the parametrization, but I don't know what to do. Its clear to me that the parametrization plays an important role in the sign of the integral. But I don't know how to reason this, so I wanted some help and suggestions ;D

    Bye there.
    Last edited by Ulysses; July 30th 2011 at 02:51 PM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Greens theorem

    Quote Originally Posted by Ulysses View Post
    Creo que esto esta bien, la integral la resolví con la computadora para no tener problemas
    This language sounds familiar to me.

    And then the double integral: \displaystyle\int_{-1}^{1}\int_{-\sqrt[ ]{1-x^2}}^{\sqrt[ ]{1-x^2}}(1-2y)dydx=\pi
    The left hand side is right so there must be a computation mistake. At any case you can find the double integral in the simple way:

    \iint_{D}(Q_x-P_y)\;dxdy=-\iint_{D}\;dxdy=-\textrm{Area\;}(D)=-\pi\cdot 1^2=-\pi
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