Constant strip-area density of a surface of revolution
= Constant strip-area density of a surface of revolution
For
$$
X(\theta,z)=(\phi(z)\cos\theta,\phi(z)\sin\theta,z),
$$
the area density after integrating over $\theta$ is $2\pi\phi\sqrt{1+\phi'^2}$. If every height interval has area $2\pi r$ times its length, continuity forces
$$
\phi^2(1+\phi'^2)=r^2.
$$
Where $0<\phi<r$, the sign of $\phi'$ is constant and $\sqrt{r^2-\phi^2}$ has derivative $\pm1$. Hence the profile lies on a circle of radius $r$.