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FibonacciMariachi
 3 years 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)
FibonacciMariachi
 3 years 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)

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1so, where are you stuck?

FibonacciMariachi
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

FibonacciMariachi
 3 years ago
Best ResponseYou've already chosen the best response.0I do have the right answer if you want it.

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

FibonacciMariachi
 3 years ago
Best ResponseYou've already chosen the best response.0oh? How would I do that? That sounds a lot easier.

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, it will most likely be a square in the u v plane

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