We are given the region $D= {\{(x,y) | 1 \leq x-y \leq 2, x \leq 0, y\leq 0\} \subseteq \mathbb R^2}$
I drew this region on a piece of paper, it resembles an infinite parallelogram on the third quadrant. $x \in (-\infty,0]$, $y\in [x-2,x-1]$.
We are given the transformation $v=x-y$, $u=x+y$. We are asked to show that the transformation turned the parallelogram into a trapezoid, confined by $v=1, v=2, u=v,u=-v$
I don't completely understand this.
I understand why $v$ goes from $1$ to $2$.
we were given $1 \leq x-y \leq 2$ and $v=x-y$ and so $1 \leq v \leq 2$. that's easy enough.
I don't understand why the $u$ part is true.
I mean, if $1 \leq x-y \leq 2$, then surely $1+2y \leq x+y = u \leq 2+2y$, so I would assume $1+2y \leq u \leq 2+2y$ but that is not very helpful at all...
You have been lied to :-)
"Parallelogram" remains "parallelogram", defined by $1\le v\le 2, u \le -v$.
The original domain is defined by $x\le 0, 1\le x-y\le 2$.
This is immediately translated to $u+v\le 0, 1\le v\le 2$ because $2x=u+v$.
ps. Just to make sure: $u=x+y, v=x-y$.