Double integral with transformations.

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

Re: Double integral with transformations.

Sorry for putting anyone out. I think I found the problem. Simple math errors have brought me undone.

First off I calculated the transformed points wrong. They should of been (-4,0), (-4,8), (4,0) and (4,8) which gives a nice square.

Also the integrand should of been $\displaystyle 3v - 5u$.

So,

$\displaystyle \int_{-4}^{4}\int_0^8 (3v-5u) \mid4\mid dv du = 3072$

Thanks to anyone you read this. Please help with my other post about the TRIPLE INTEGRAL.