Formulate and verify Green theroem when [P,Q]=[xy,1] and the curve (gama) is positively oriented circle containing the points: A(0,0) ----> B(0,2) ----> C(1,1) ---> A(0,0)

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Formulate and verify Green theroem when [P,Q]=[xy,1] and the curve (gama) is positively oriented circle containing the points: A(0,0) ----> B(0,2) ----> C(1,1) ---> A(0,0)

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any help?
so looks like you going positivly oriented around a triangle in the xy plane. Lets call the boundary T \[\int\limits_{∂T}^{}(xy)dx+1(dy)\] which according to brother green thats nothing more than \[\int\limits_{T}^{}\int\limits_{}^{}∂(1)/∂x-∂(xy)/∂ydA\] which is nothing more than\[\int\limits_{T}^{}\int\limits_{}^{}-xdxdy\] then you set up your paramerization A(0,0) ----> B(0,2) ----> C(1,1) ---> A(0,0) |dw:1315809779023:dw| now since you go clockwise in the A-->B-->C-->A and in the drawing i go counterclockwise(positivly oriented) just put a (-) in front of the whole integral -∫∫(-x)dxdy=∫∫xdxdy you may have to slipt it up into two different integrals |dw:1315809979957:dw|
∫[0,1]∫[x,1]x dydx+∫[0,1]∫[1,-x+2]x dydx ∫[0,1]x(1-x)dx+∫[0,1]x(-x+1)dx 2∫[0,1]x-x^2dx 2(x^2/2-x^3/3)|[0,1] 2(1/2-1/3) 1-2/3 1/3 should be your final answer or -1/3 and thats for the FLOW

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OOPPPS i did a triangle when it was a circle. same concept ill draw and you can evaluate |dw:1315810765086:dw| those should be your bounds and evaluates the same but change it to rdrd(theta) because that integral looks like its from hell. Polars a lot nicer

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