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I have the formula that $$\int\int_D \frac{\partial M}{\partial x} - \frac{\partial N}{\partial y} dx dy = \oint_\Gamma [Mdy + Ndx].$$ I want to calculate the following:

let $D$ be the triangle bounded by $(1,0),(4,0)$ and $(1,10)$.

Note for later that $1\leq x \leq 4$ and $\displaystyle 0\leq y \leq \frac{-10x}{3} + \frac{40}{3}.\;$ Let $C = \partial D$, that is C is the boundary of D oriented counterclockwise with by some parameterization. Compute $\displaystyle \oint_C (y^2 dx - xydy)$.

So what I have thus far: Since $D$ is a Greens domain we can apply the formula. Take $M(x,y) = -xy$ and $N(x,y) = y^2$ which gives $\displaystyle\frac{\partial M}{\partial x} = -y$ and $\displaystyle\frac{\partial N}{\partial y} = 2y$. So we can shovel this into the format of $\displaystyle \int_1^4 \int_0^{\large\frac{-10x}{3} + \frac{40}{3}} -y - 2y\ dy\ dx $ which becomes $$\int_1^4 \int_0^{\frac{-10x}{3} + \frac{40}{3}} -3y\ dy\ dx $$ the inner integral then evaluates to give $$\int_1^4 \frac{-50 x^2}{3} + \frac{400 x}{3} - \frac{800}{3}\ dx $$ and this integral gives $\displaystyle \left[\frac{-50 x^3}{9} + \frac{200 x}{3} - \frac{800x}{3}\right]^4_1 = -150$.

Can anybody take a look and tell me if there are any obvious problems with it? We only got to Green & Stokes on the last day of classes so I have to learn it on my own.

Thanks in advance.

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Looks good to me. Instead of just saying that its a greens domain you may want to say you have a domain over an closed simply connected region.

Looks to me like you made a careless error in the calculating part of the integration, the lower bound of the $x$ integration is being evaluated at $0$ when it should be at $1$. rest should be good.

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  • $\begingroup$ good catch on the Integral evaluation. But if that's the only error then I'm feeling pretty good about this. $\endgroup$ – AvatarOfChronos Dec 11 '12 at 4:32

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