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Math Help - change of variables, triangular region

  1. #1
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    change of variables, triangular region

    so i have the double integral (x-3y)da, and R is a triangular region with vertices (0,0) (1,2) (2,1). transformations are x=2u+v and y=u+2v

    So i drew out the triangle and found equations for the 3 lines.
    I got:
    y=.5x
    y=2x
    y=-x+3.

    then i plugged in my transformations for each of these three equations.
    the first equation left me with v=0
    The second left me with u=0
    the third with u=1-v

    So i drew out this new triangle shape.
    I want to make sure my bounds are correct, if i'm integrating wrong, i can figure that out.
    so integrating with respect to u first, my bounds should be
    0<u<1-v
    0<v<1
    right?

    then plug in the transformations for x-3y, and integrate away.
    i came out with an answer of -6. but it's supposed to be -3. again, if i integrated wrong, i can figure that out. i've done it a few times and am still not getting -3, so i'm wondering if my bounds are wrong.
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  2. #2
    MHF Contributor
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    Hi

    I am getting -3
    Could you show your work ?
    Maybe your problem comes from the Jacobian matrix
    Jacobian matrix and determinant - Wikipedia, the free encyclopedia
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  3. #3
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    yeah, i got that the matrix came out to be 3. is that correct?

    alright, so i must be integrating wrong and i still can't figure it out

    so bounds of
    0<u<1-v
    0<v<1

    and plugging in the transformations of x=2u +v and y = u +2v into x-3y, i get
    2u+v-3u-6v which is -u -5v.

    so integrate that with respect to u, i get
    -.5u^2 - 5vu.
    use the top bound of 1-v, i get
    -.5(1-2v+v^2) - (5v-5v^2)
    so i multiplied everything by 2 to get rid of the .5. (and i remember there is a 3 outside the integral from the jacobian matrix)

    so i would get 6 outside the integral (bounds of 0 and 1)
    and then inside is 1-2v+v^2 - 10v-10v^2
    simplify to
    1-12v-9v^2.
    integrate for v
    v-6v^2-3v^3. plug in top bound of 1

    1-6-3 = -8. and times the 6 from outside, -48. so not even close... i don't know what i am doing wrong
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  4. #4
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    Quote Originally Posted by isuckatcalc View Post
    yeah, i got that the matrix came out to be 3. is that correct?

    alright, so i must be integrating wrong and i still can't figure it out

    so bounds of
    0<u<1-v
    0<v<1

    and plugging in the transformations of x=2u +v and y = u +2v into x-3y, i get
    2u+v-3u-6v which is -u -5v.

    so integrate that with respect to u, i get
    -.5u^2 - 5vu.
    use the top bound of 1-v, i get
    -.5(1-2v+v^2) - (5v-5v^2)
    so i multiplied everything by 2 to get rid of the .5. (and i remember there is a 3 outside the integral from the jacobian matrix)
    It's OK up to this point

    -.5(1-2v+v^2) - (5v-5v^2) = -.5-4v+4.5v^2 that you integrate from 0 to 1 and multiply the result by 3
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