Using Green's Theorem to evaluate the line integral.

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Homework Statement



Green's Theorem to evaluate the line following line integral, oriented clockwise.
∫xydx+(x^2+x)dy, where C is the path though points (-1,0);(1,0);(0,1)

Homework Equations


Geen's theorem: ∫F°DS=∫∫ \frac{F_2}{δx}-\frac{F_1}{δy}


The Attempt at a Solution


attachment.php?attachmentid=53123&stc=1&d=1353355503.jpg


What would I use for the bounds. I know it has to do with the triangle. But how would I find the bounds?
 

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You've pretty much written it down yourself: the lower limit is clearly y= 0. The upper limits are those two other sides of the triangle: if x< 0, y= x+ 1, if x> 0, y= -x+1.
So you can divide this into two sets of integrals: for x from -1 to 1, y goes from 0 to x+ 1, for x from 0 to 1, y goes from 0 to 1- x.
 
Thanks... Halls. I wasn't sure what to do, but I understand now. It's like you've divided the triangle into 2 pieces.
 
And made both of the equations x+1 and 1-x satisfy their respective points... but I think the upper and lower bound of x+1 is -1 and 0 respectively since it goes from 0 to satisfy point (0,1) and -1 for (-1,0) and goes from the first point to the other.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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