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Multiple Integrals with parallelogram

  • #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
 

Answers and Replies

  • #2
Sounds good. Did you also transform the "dA"?
 
  • #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?
 
  • #4
changing it du dudv is not quite correct.

What you do is employ the transformation

[tex]
\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)
[/tex]
This transformation matrix has determinant 2 so you should account for this dilatation of the area element dA:

dxdy = 2dudv
 
Last edited:
  • #5
so the answer would be 640 instead of 320
 

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