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Math Help - Verifying Green theorem

  1. #1
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    Verifying Green theorem

    Hi there. I have the next exercice: Verify the Green theorem with the field F(x,y)=(x+y,2x-y), and the path which its image is the boundary positively oriented for the region between x^2+y^2=9 y |x|+|y|=1

    At first I've drawn this:


    Then I've parametrized the path
    \lambda_1[0,2\pi]\rightarrow{\mathbb{R}^2}:\lambda_1(t)=(3\cos t,3\sin t)
    \lambda_2[-1,1]\rightarrow{\mathbb{R}^2}:\lambda_2(t)=(t,1-|t|)
    \lambda_3[1,-1]\rightarrow{\mathbb{R}^2}:\lambda_3(t)=(t,|t|-1)
    \lambda=\lambda_1+\lambda_2+\lambda_3

    I_1=\displaystyle\int_{0}^{2\pi}(-(3\cos t+3 \sin t)3\sin t+(6\cos t-3\sin t)3\cos t)dt=9\pi
    I_2=\displaystyle\int_{-1}^{1}[t+1-|t|+(2t-1+|t|)sg(t)]dt=3
    I_2=\displaystyle\int_{1}^{-1}[t+|t|-1+(2t-|t|+1)sg(t)]dt=-1

    \displaystyle\int_{\lambda}F\cdot{d\lambda}=I_1+I_  2+I_3=9\pi+2

    Then I made the double integral:
    \displaystyle\int \int_{D} \left ( \frac{{\partial Q}}{{\partial x}}-\frac{{\partial P}}{{\partial y}} \right )dxdy=\displaystyle\int_{-3}^{3}\displaystyle\int_{-\sqrt[ ]{9-x^2}}^{\sqrt[ ]{9-x^2}}dydx-\displaystyle\int_{-1}^{1}\displaystyle\int_{|x|-1}^{1-|x|}dydx=\displaystyle\int_{0}^{r} \int_{0}^{2\pi}r d \theta dr - \displaystyle\int_{-1}^{1}\displaystyle\int_{-1}^{1}\displaystyle\frac{1}{2}dvdu=9\pi-2
    In the last integral I've made the change of variables y+x=u,y-x=v. As you can see I have a problem with a sign, in the 2. Perhaps I should take the opposite signs because lambda2 and lambda3 are "negative" oriented? I thought the sign should came naturally given by the parametrization, so I'm not sure what I did wrong here.

    Bye and thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Verifying Green theorem

    Quote Originally Posted by Ulysses View Post
    \displaystyle\int_{\lambda}F\cdot{d\lambda}=I_1+I_  2+I_3=9\pi+2
    According to Green's theorem for non simply connected regions it should be

    \displaystyle\int_{\lambda}F\cdot{d\lambda}=I_1-I_2-I_3=9\pi-2

    P.S. You can compute the double integral in a simple way taking into account that \iint_D\;dxdy is just the area of D .
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