# Jacobians

by AeroFunk
Tags: jacobians
 P: 40 Let R be the region bounded by the graphs of x+y=1, x+y=2, 2x-3y=2, and 2x-3y+5. Use the change of variables: $$x=1/5(3u+v)$$ $$y=1/5(2u-v)$$ to evaluate the integral: $$\iint(2x-3y)\,dA$$ I found the jachobian to be -1/5 and the limits of integration to be 1<=u<=2 2<=v<=5 so i set up the integral like this: $$\frac{-1}{5}\int_{2}^{5}\int_{2}^{1} vdv$$ and I get -21/5 which doesn't seem right(a negitive number??),what am I doing wrong?
 P: 291 I'm getting -21/10, but that's just because you forgot a 1/2 in the integration. I am wondering, though, why your bounds of integration on u are going from 2 to 1 instead of 1 to 2, especially since you don't evalute the integral as such. What's wrong with a negative number? The integral in u-v obviously isn't going to be negative, and the Jacobian is in fact negative, so obviously the answer should be negative. From the x-y integral, we're integrating 2x-3y. If y is sufficiently large compared to x, the integral will be negative, as in this case. --J
 P: 40 Oh sorry the 2 to 1 is just a typo , and lol yea it would be negitive (have been doing to many volume problems lately) thanks for the help

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