# what is the problem with this variable transformation?

$$\iint\limits_D (x − y)^2 \sin^2(x + y) \, dx \, dy$$ where $D$ is a parallelogram with vertices at $(π, 0), (2π, π), (π, 2π)$ and $(0, π)$.

We can change the variables as $$x=\frac{u-v}{2} \text{ and } y=\frac{u+v}{2}.$$ We can also do $$x=\frac{u+v}{2} \text{ and } y=\frac{u-v}{2}.$$ In both cases the limits of $u$ and $v$ are the same but the jacobian is the negative of one another (row is switched making it negative), thus integral turns out to be negative.

This is very confusing for me, because I am not able to understand why. Thanks for your help.

Note: In a few youtube videos I have seen that the absolute value of Jacobian is to be used, in the case no problem arises, but in the university I am studying in they have not specified to use the absolute value.

• what should i do when jacobian depends on the variables itself, in this case it comes out constant , if it is of the form $f(u,v)$ then what should we do? – avz2611 Oct 6 '14 at 15:15
• You take the absolute value as well. When the Jacobian varies with each point you should consider the regions on the considered domain where it is positive and negative. Consider as an example the Jacobian $J(x,y) = x-y$ and suppose the region is the unit square $0 \leq x \leq 1, 0 \leq y \leq 1$. You will consider the absolute value of $J(x,y)$ and check where $x-y \geq 0$ so $J(x,y) = x-y$ and where $x-y < 0$ so $J(x,y) = y-x$. – Mark Fantini Oct 6 '14 at 15:18