Question: Consider $\Bbb D=\{|x|+|y|\le2\}$; I'm trying to solve:
$$\int\_{\Bbb D}(x^2 - y^2)\, dx\, dy$$

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My goal is to solve that through a **change of variables**.  

I was thinking to something like: $\begin{cases} |x|=u^2 \\ |y|=v^2 \end{cases}$. In such a case I would had the new domain as a circle; but the transformation is not invertible and I should study case by case: $\{x+y\le 2,x\ge0,y\ge0\},\{x-y\le2,x\ge0,y\le0\},\{-x+y\le2,x\le0, y\ge0\},\{-x-y\le2,x\le0,y\le0\}$.  

Now, another transformation that would fit the problem is: $\begin{cases} x+y=u \\ x-y=v \end{cases}$, but I wasn't able to write any single domain in a comfortable way.

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Is there another suitable change of variables?  

(otherwise) an someone handle one of the transformation I just proposed?

Answer: 
The integral is $0$ because the domain is symmethric relative to coordinate change $(x,y)\to (y,x)$ therefore
$$\int\_{\Bbb D}x^2\, \mathrm{d}x\, \mathrm{d}y=
\int\_{\Bbb D}y^2\, \mathrm{d}x\, \mathrm{d}y.$$
I highly recommend your second change of variables. For this choice of $u,v$, we find that $$
\mathbb D = \{x,y: -1 \leq u(x,y) \leq 1, -1 \leq v(x,y) \leq 1\},
$$
and we find that
$$
x^2 - y^2 = u(x,y)v(x,y).
$$