If we rotate the graph of $z=r$ in the $(r,z)$ plane where $r \geq 0$. We get the same graph of the function,
$$z=r$$
In the $(x,y,z)$ plane where $r$ is the distance from the $z$ axis, i.e $r=\sqrt{x^2+y^2}$.
The same with $z=r^2$, rotating it about the $z$ axis gives $z=r^2$.
Why does this work? Can we show this is the case for all functions $z=f(r)$?
Yes. Call the inverse function g(z) ( like $ z= r^2 \rightarrow r = \sqrt z = g(z) $. You are rotating $g$ around $z-$ axis, works in all such bijective situations.
The components are $(x,z), ( y,z), (\sqrt{x^2+y^2},z). $