# Greene's Theorem over a triangle

1. Sep 17, 2009

### manenbu

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

$$\oint_{C} (x+y)^2 dx - (x^2+y^2) dy$$
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,
$$\oint_{C} Pdx + Qdy = \iint_{D}(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dA$$
So in my case: $P = (x+y)^2 = x^2 + 2xy + y^2$ and $Q = -x^2 -y^2$
$$\frac{\partial Q}{\partial x} = -2x \\ \frac{\partial P}{\partial y} = 2x+2y$$
so:
$$\oint_{C} (x+y)^2 dx - (x^2+y^2) dy = \iint_{D} -2x -2x -2y dA = -2 \iint_{D} 2x + y dA$$

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).

Thanks!

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2. Sep 17, 2009

### LCKurtz

I don't see anything wrong.