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FibonacciMariachi
Group Title
Evaluate the integral by using the following transformation:
∫∫(R) x*y^2 dA, where R is the region bounded by the lines xy=2, xy=1, 2x+3y=1, and 2x+3y=0; let x=1/5(3u+v), y=1/5(v2u)
 one year ago
 one year ago
FibonacciMariachi Group Title
Evaluate the integral by using the following transformation: ∫∫(R) x*y^2 dA, where R is the region bounded by the lines xy=2, xy=1, 2x+3y=1, and 2x+3y=0; let x=1/5(3u+v), y=1/5(v2u)
 one year ago
 one year ago

This Question is Closed

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so, where are you stuck?
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Well, I tried doing the problem and I didn't get the right answer. Are these the correct steps to get to the answer? 1. Plot x/y coordinate graph. 2. Convert critical xy points to uv points. 3. find the Jacobian. 4. Set up the integral in terms of u and v with the jacobian.
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
I do have the right answer if you want it.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
your steps are fine, but I would just convert the lines to u and v instead of the "critical xy points"
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
oh? How would I do that? That sounds a lot easier.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
just sub in x=1/5(3u+v), y=1/5(v2u) into each equation for the boundary of R
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Ohhhh, alright. So once I converted to uv form would I just replot the graphs and look for my bounds?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yeah, it will most likely be a square in the u v plane
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Gotcha, Ill rework the problem. That sounds a lot better than just looking for the points which takes forever.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
good luck!
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
thanks
 one year ago
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