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Homework Help: Greene's Theorem over a triangle

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\oint_{C} (x+y)^2 dx - (x^2+y^2) dy[/tex]
    C is the edge of the triangle ABD on the positive direction with A(1,1), B(3,2), C(2,5).

    2. Relevant equations

    Greene's Theorem, Double Integral.

    3. The attempt at a solution
    According to the theorem,
    [tex]\oint_{C} Pdx + Qdy = \iint_{D}(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dA[/tex]
    So in my case: [itex]P = (x+y)^2 = x^2 + 2xy + y^2[/itex] and [itex]Q = -x^2 -y^2[/itex]
    \frac{\partial Q}{\partial x} = -2x \\ \frac{\partial P}{\partial y} = 2x+2y
    \oint_{C} (x+y)^2 dx - (x^2+y^2) dy = \iint_{D} -2x -2x -2y dA = -2 \iint_{D} 2x + y dA

    Now I need to find D, so if we got a triangle I need to find the equations for each of the 3 sides of the triangle. They are:
    AB - y = x/2 + 1/2
    BD - y = -3x+11
    DA - y = 4x-3

    I need to split it into 2 domains, one for 1<x<2 and the other for 2<x<3 like this:
    \iint_{D} -2x -2x -2y dA = -2 \iint_{D} 2x + y dA = -2 \left(\int_{1}^{2} \int_{\frac{x}{2} + \frac{1}{2}}^{4x-3} 2x+y dy dx + \int_{2}^{3} \int_{\frac{x}{2} + \frac{1}{2}}^{-3x+11} 2x+y dy dx \right)

    Was I doing anything wrong so far?
    Evaluating this integral gives me the wrong answer and I've tried a few times (both in the computer and by hand).


    Attached Files:

  2. jcsd
  3. Sep 17, 2009 #2


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    Science Advisor
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    Gold Member

    I don't see anything wrong.
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