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How to evaluate int 2x-3y dA using change of variables

by AeroFunk
Tags: jacobians
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AeroFunk
#1
Dec2-04, 01:33 AM
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:
[tex]
x=1/5(3u+v)[/tex]
[tex]y=1/5(2u-v)[/tex]
to evaluate the integral:
[tex]
\iint(2x-3y)\,dA
[/tex]

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:
[tex]
\frac{-1}{5}\int_{2}^{5}\int_{2}^{1} vdv[/tex]

and I get -21/5 which doesn't seem right(a negitive number??),what am I doing wrong?
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Justin Lazear
#2
Dec2-04, 02:15 AM
P: 290
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
AeroFunk
#3
Dec2-04, 02:31 AM
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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