You want to integrate over:
$z\le y-2x$
$0\le z$
$0\le x$
$y\le 2$
If you sketch a picture, you'll see that this is a tetrahedron with vertices $(0,0,0)$, $(0,2,0)$, $(1,0,2)$, $(1,2,0)$.
Obviously, you have 6 possible permutations of variables $x$, $y$, $z$.
Let's have a look at the range of these variables. We have $y\ge 2x+z \ge 0$, hence $0\le y \le 2.$
We also have $x\le \frac{y-z}2 \le \frac{2-0}2=1$, thus $0\le x\le 1.$
From $z\le y-2x\le 2-0 = 2$ we have $0\le z\le 2$.
All you have to do now is to choose some order. E.g. we can start with $x$. Then express possible range for $y$ using $x$. And then find in what range $z$ will be, if $x$ and $y$ are given. In this way we get: $\begin{align} 0 &\le x \le 1\\ 2x &\le y \le 1\\ 0 &\le z \le y-2x \end{align}$ For example, we have used $y\ge 2x+z \ge 2x$ in the second line. (Since $z\ge 0$.)
If we choose different ordering $x$, $z$, $y$ we get $z\le y-2x \le 2-2x$, i.e. $\begin{align} 0 &\le x \le 1\\ 0 &\le z \le 2-2x\\ 2x+z &\le y \le 2 \end{align}$
We can try to start with $y$: $\begin{align} 0 &\le y \le 2\\ 0 &\le z \le y\\ 0 &\le x \le \frac{y-z}2 \end{align}$
I guess you'll be able to do the remaining three possibilities yourself.