I need help in determine the limits of the double integrals where I had to use transformations.

The double integral is $\displaystyle \int\int_R (4x + 8y) dA$ where R is a parallelogram with vertices (-1,3), (1,-3), (3,-1) and (1,5).

The transformations were given as: $\displaystyle x = \frac{1}{4}(u+v)$, $\displaystyle y=\frac{1}{4} (v-3u)$.

Using the transformation equations of $\displaystyle u=x-y$ and $\displaystyle v=3x+y$ (solving simultaneoulsy from above), the points in the transformed plane $\displaystyle T^{-1}$ become (-4,-2), (-4,-6), (4,10) and (4,6).

I calculated the Jacobian as $\displaystyle \mid 4 \mid$ and the integrand becomes $\displaystyle 3v-6u$.

So,

$\displaystyle \int\int_R (4x+8y) dA = \int\int_r (3v-6u) \mid 4 \mid dv du$ where r is the region in the transformed plane.

$\displaystyle = 12 \int_{-4}^{4} \int_*^* (v-2u) dv du$

Could someone please help to me with what the * * limits of the inner integral would be and correct any errors I have made in getting this far?

Thanks in advance,

Tammy