Triple integral changing order of integration(impossible textbook question) Related to the question here:

can anyone see this post, please help or atleast comment, I can't go on in the textbook the next three problems have the same issue!! I don't understand the upvotes and no answers

$$ \int_0^1 \int_\sqrt{z}^1 \int_0^{2-y-z} f(x,y,z) \; \mathrm{d}x\; \mathrm{d}y\; \mathrm{d}z$$
I want to change that order to $\mathrm{d}z\; \mathrm{d}x\; \mathrm{d}y$
${}$
Now I have:
$$R = \{(x,y,z) \in \mathbb R^3 \mid (0 \le z \le 1) \wedge (z \le y^2 \le 1) \wedge (0 \le x+y+z \le 2)\}.$$
But because of the order, I have both $y \leq y^2$ and $ z \leq 2-y-x$, so I have set up my attempt as $R = \{{(x,y,z)|0\leq y \leq 1,0 \leq x\leq 2-y,0\leq z \leq y^2}\}$, but this is missing a bounding, and I don't know what to do. Please help me!
Note: Just the bounds for the region as the function is not known.
Breaking into two regions:
I don't have any software to confirm my two regions as encompassing the total region.
1: $0 \leq y \leq 1, 0 \leq x \leq 2-y, 0 \leq z \leq y^2$
and 2: $0 \leq y \leq 1, 0 \leq x \leq 2-y, 0 \leq z \leq 2-x-y$
Is this question impossible, no-one here can work it out , so I assume the textbook did it as a trick.

Why is this question impossible to set up as triple integrals?


Image courtesy of @heropup
 A: Your domain $R\subset{\mathbb R}^3$ is defined by the following three conditions:
$$({\rm a})\quad 0\leq z\leq1,\qquad({\rm b})\quad \sqrt{z}\leq y\leq 1,\qquad({\rm c})\quad 0\leq x\leq 2-y-z\ .$$
The second condition is equivalent with
$$({\rm b}')\qquad 0\leq y\leq 1\quad\wedge\quad z\leq y^2\ .$$

It follows that $R$ is a subset of the simplex $$S:=\bigl\{(x,y,z)\bigm| x\geq0,\ y\geq0,\ z\geq0, \ x+y+z\leq2\bigr\}$$
shown in the above figure; see also heropup's figure shown as an appendix to the question. Simple extra conditions are $y\leq1$ and $z\leq1$, but most notorious is $z\leq y^2$, indicating that admissible points $(x,y,z)$ have to lie below the parabolic cylinder $z=y^2$. This cylinder intersects the oblique facet of $S$ in the curve
$$\gamma:\quad y\mapsto (x,y,z):=(2-y-y^2, y, y^2)\qquad(0\leq y\leq1)\ ,$$
and this curve projects vertically down onto the curve
$$\gamma':\quad x\mapsto(x,y):=(2-y-y^2,y)\qquad(0\leq y\leq1)\ ,$$
resp.
$$\gamma':\quad x\mapsto(x,y):=\left(x,{\sqrt{9-4x}-1\over2}\right)\qquad(0\leq x\leq2)\ ,$$
shown in blue in the figure.
The vertical stalk erected on points $(x,y)$ to the left of $\gamma'$ ends on the cylinder $z=y^2$, and the stalk erected on points to the right of $\gamma'$ ends on the facet $z=2-x-y$ of $S$. It follows that we have to foresee three triple integrals when the integration order has to be $\ \ldots dz\>dy\>dx$, namely
$$\eqalign{&\int_0^2\int_0^{(\sqrt{9-4x}-1)/2}\int_0^{y^2}\ldots dz\>dy\>dx \quad +\cr
&\int_0^1\int_{(\sqrt{9-4x}-1)/2}^1\int_0^{2-y-x}\ldots dz\>dy\>dx \quad+\cr
&\int_1^2\int_{(\sqrt{9-4x}-1)/2}^{2-x}\int_0^{2-y-x}\ldots dz\>dy\>dx\quad.\cr} $$
