Using transformation to evaluate double integral 
Given the transformation $T(x, y) = (x - y, x + y)$, evaluate the double integral
  $\iint_R (x^2+y^2) dA$, where $R$ is the rectangle in the $xy$-plane with vertices $A(1, 1)$, $B(2, 2)$, $C(-1, 5)$ and $D(-2, 4)$.

Computing the Jacobian of the transformation, I get $\det[DT(x,y)] = 2$. Change the coordinates of the vertices respectively, we get 


*

*$A(1,1)$ becomes $(0,2)$

*$B(2,2)$ becomes $(0,4)$

*$C(-1,5)$ becomes $(-6,4)$

*$D(-2,4)$ becomes $(-6,2)$
Solving the double integral, I did:
$$
\int_2^4 \int_{-6}^0 2\left[(x-y)^2-(x+y)^2\right]dxdy
$$
which gave me an answer of 864, however, I should be getting $-54$ as given in the answer sheet, may I know what went wrong?
 A: HINT
One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
\begin{split}
A &= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[\left(\frac{u+v}{2}\right)^2 + \left(\frac{v-u}{2}\right)^2\right] 2\ du\ dv \\
  &= \frac12
     \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[(u+v)^2 + (v-u)^2\right] du\ dv \\
  &= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[u^2+v^2\right] du\ dv
\end{split}
$$
UPDATE
$$
\begin{split}
A &= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[u^2+v^2\right] du\ dv \\
  &= 2 \int_{u=-6}^{u=0} u^2 du
   + 6\int_{v=2}^{v=4} v^2 dv \\
  &= \left.\frac23 u^3 \right|_{u=-6}^{u=0}
   + \left.2 v^3 \right|_{v=2}^{v=4} \\
  &= -\frac23 (-6)^3 + 2\left(4^3-2^3\right)\\
  &= 36\times 4+2(64 - 8)\\
  &= 256.
\end{split}
$$
The arithmetic agrees with Wolfram Alpha.
As for your hint, you are (originally) integrating 
$$
\iint_A \left(x^2+y^2\right)dA,
$$
where the integrand is always non-negative, so the integral should be positive as well.
