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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
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

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

FibonacciMariachiBest 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

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

TuringTestBest 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

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

TuringTestBest 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

FibonacciMariachiBest 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

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

FibonacciMariachiBest 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
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