Rewrite $\int_{0}^1 \int_{0}^{1-y^2} \int_{0}^{1-y} dz \ dx \ dy$ in the five other orders of integration

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I'm struggling to find limits of integration when the outermost integral depends on $x$ and $z$.

For example, when we study the shadow of the region on the $xz$-plane, we find that it's a square. But its projection upwards (in the direction of the positive $y$ axis) depends on two different surfaces (the cylinder $1-y^2$ and the plane $1-y$), and I don't know how to express everything in terms of one variable.

How can I approach these kinds of exercises?

Edit: my textbook says we should verify the following equalities

$$z = 1 - y \iff y = 1- z \\ x = 1 - y^2 \ \ \underset\iff{y\geq0} \ \ y = \sqrt{1-x}$$

Therefore

$$x = z^2 - 2z \\ z=1-\sqrt{1-x}$$

But what are these surfaces bounds of? I'm really confused.

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Simplest (and easiest) way to approach this kind of problems is to look straight at the limits of integration.

The domain can be described by the following inequalities. $$0\leq y \leq 1,\ 0\leq x \leq 1-y^2,\ 0\leq z\leq 1-y $$ Those are equivalent to the following $$0\leq z\leq 1, \ 0\leq x\leq 1,\ y^2 \leq 1-x, \ 0\leq y\leq 1-z $$ Well, the last two inequalities can be written as $0\leq y\leq \min (\sqrt{1-x},\ 1-z)$. This leaves us with $$0\leq z\leq 1,\ 0\leq x\leq 1,\ 0\leq y\leq \min (\sqrt{1-x},\ 1-z) $$ and the integral can be written as $$\int_0^1 \int_0^1 \int_0^{\min (\sqrt{1-x},\ 1-z)} dydxdz. $$