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Math Help - Green's Theorem - where did I go wrong?

  1. #1
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    Green's Theorem - where did I go wrong?

    Let R be the region bounded by the curves:

    y=x^2-1 and y=-x^2+1

    Verify Green's Theorem in the plane for:

     \oint_{\partial R} (y^2+x)dx + (xy+1)dy=\int_{-1}^{1}\int_{x^2-1}^{-x^2+1}-ydydx

    \int_{-1}^{1}-\frac{(-x^2+1)^2}{2}-(-\frac{(x^2-1)^2}{2})dx=\int_{-1}^{1}0dx

    Umm... not sure if I was supposed to end up here.
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  2. #2
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    Would this make more sense:

    \int_{-1}^{1}\int_{0}^{-x^2+1}-ydydx+\int_{-1}^{1}+\int_{x^2-1}^{0}-ydydx

    or just:

    2\int_{-1}^{1}\int_{0}^{-x^2+1}-ydydx
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  3. #3
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    Quote Originally Posted by MathSucker View Post
    Let R be the region bounded by the curves:

    y=x^2-1 and y=-x^2+1

    Verify Green's Theorem in the plane for:

     \oint_{\partial R} (y^2+x)dx + (xy+1)dy=\int_{-1}^{1}\int_{x^2-1}^{-x^2+1}-ydydx

    \int_{-1}^{1}-\frac{(-x^2+1)^2}{2}-(-\frac{(x^2-1)^2}{2})dx=\int_{-1}^{1}0dx

    Umm... not sure if I was supposed to end up here.

    You are right !


     \frac{\partial Q}{\partial x} - \frac{ \partial P }{\partial y}

     = y - 2y  = -y

    since the path is closed , the line integral  = \int\int_D -ydydx

    the region is  x^2 - 1 \leq y \leq -x^2 + 1  ,  -1 \leq x \leq 1

    Consider  y-coordinate ~of~ the ~center~ of ~mass~ = \frac{\int\int_D ydydx }{ area ~ of ~the ~region }

    But we know the y-coordinate of the center of mass of the region is zero

    so the integral is zero too !
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  4. #4
    Member billym's Avatar
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    When these questions ask you to "verify green's theorem." Is it enough to just run the two equations through \oint_{\partial R} P \, dx + Q \, dy and say "the answer is zero."

    Or does it require some kind of explanation?
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