Multiple Integrals with parallelogram

[ghostrider989]

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

 

[Pere Callahan]

Sounds good. Did you also transform the "dA"?

 

[ghostrider989]

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?

 

[Pere Callahan]

changing it du dudv is not quite correct.

What you do is employ the transformation

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

dxdy = 2dudv

 

[ghostrider989]

so the answer would be 640 instead of 320