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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
∫∫(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