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anonymous
 5 years ago
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)
anonymous
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0∫[0,1]∫[x,1]x dydx+∫[0,1]∫[1,x+2]x dydx ∫[0,1]x(1x)dx+∫[0,1]x(x+1)dx 2∫[0,1]xx^2dx 2(x^2/2x^3/3)[0,1] 2(1/21/3) 12/3 1/3 should be your final answer or 1/3 and thats for the FLOW

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OOPPPS 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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