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

  1. 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?
     
  2. jcsd
  3. 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
     
  4. 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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