anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
any help?
anonymous
  • anonymous
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|
anonymous
  • anonymous
∫[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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.