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Multiple Integrals with parallelogram

  1. Nov 12, 2008 #1
    Use the given transformation to evaluate the given integral, where R is the parallelogram with vertices (-2, 2), (2, -2), (4, 0), and (0, 4).

    ∫∫(2x+8y)dA; x=1/2(u+v) y=1/2(v-u)

    I found the bounds of the parallelogram of -4≤u≤4 and 4≤v≤0

    so i set the equation to be

    ∫∫(2(1/2(u+v))+8(1/2(v-u))dudv with the bounds been written in there

    am i on the right track
  2. jcsd
  3. Nov 12, 2008 #2
    Sounds good. Did you also transform the "dA"?
  4. Nov 12, 2008 #3
    yeah i change it to dudv but the answer i got was 320 and i am not sure that is right afther doing all teh integration and plugging in what am i doing wrong?
  5. Nov 12, 2008 #4
    changing it du dudv is not quite correct.

    What you do is employ the transformation

    \left(\stackrel{x}{y}\right)\mapsto\left(\stackrel{u}{v}\right)=\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array} \right)\left(\stackrel{x}{y}\right)
    This transformation matrix has determinant 2 so you should account for this dilatation of the area element dA:

    dxdy = 2dudv
    Last edited: Nov 12, 2008
  6. Nov 12, 2008 #5
    so the answer would be 640 instead of 320
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