I want to write $z^2 \ge x^2 + y^2$, $x^2 +y^2 +z^2 \le 1$ and $z \ge 0$ in the form $$a \le z \le b, \quad c(z) \le y \le d(z), \quad f(y,z) \le x \le g(y,z)$$
or
$$a \le z \le b, \quad c(z) \le x \le d(z), \quad f(x,z) \le x \le g(x,z)$$
I get $0 \le z \le \dfrac{1}{\sqrt{2}}$, $-\sqrt{\frac{1}{2}-y^2} \le x \le \sqrt{\frac{1}{2}-y^2}, -\dfrac{1}{\sqrt{2}} \le y \le \dfrac{1}{\sqrt{2}}, $ which is wrong.
Any ideas?
It is not possible to answer that question as was first written.
The first form of your question implies that $z$ and $y$ have limits that are independent of each other, while the second form says the same about $z$ and $x$. But that is not the case. Here is a graph of the region.
The region is above the red cone and below the blue hemisphere. As you can see, the limits on both $x$ and $y$ do depend on the value of $z$, both in the cone part of the region and in the upper "cap" part.
Putting it another way, the limits on $x$ (or similarly on $y$) are
$$0 \le |x| \le \min(z,\sqrt{1-z^2})$$
which clearly depends on $z$.
The question would make more sense as
$$a \le z \le b, \quad c(z) \le y \le d(z), \quad f(y,z) \le x \le g(y,z)$$
or
$$a \le z \le b, \quad c(z) \le x \le d(z), \quad f(x,z) \le x \le g(x,z)$$
Using the first style, I get the answer
How to get those: For $z$, you have $z\ge 0$, and also $$z\le 1-x^2-y^2 \le 1$$
Once $z$ is set, for $y$ we get $$y^2 \le z^2-x^2 \le z^2 \implies |y|\le |z|=z$$ $$y^2 \le 1-z^2-x^2 \le 1-z^2 \implies |y|\le \sqrt{1-z^2}$$ Combining those two and using $|r|\le s \implies -s\le r\le s$ gives the limits on $y$.
Once $z$ and $y$ are set, for $x$ we get $$x^2 \le z^2-y^2 \implies |x|\le \sqrt{z^2-y^2}$$ $$x^2 \le 1-z^2-y^2 \implies |x|\le \sqrt{1-z^2-y^2}$$ Combining those two gives the limits on $x$.