 Solution

A subset $S\subset {\mathbb{R}}^3$ is a smooth surface if every point has a neighbourhood in $S$ parametrized by a map $X:U\to S$, where $U\subset {\mathbb{R}}^2$ is open, $X$ is a homeomorphism onto that neighbourhood, $X$ is smooth, and $DX$ has rank two everywhere. These are the embedded surface parametrization conditions.

For the given set use local angular intervals in the parametrizationIt is locally one-to-one and has tangent vectorsTheir cross product has magnitudeso the derivative has rank two. The angular charts cover $\Sigma$, proving that it is a smooth surface.

The area between heights $a_0$ and $b_0$ is thereforeBy hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,and squaring gives

If $0<\varphi (t)<r$, then ${\varphi }^{\prime }$ never vanishes, so its sign $\sigma \in \{1,-1\}$ is constant. The last equation givesand henceThus $\sqrt{r^2-\varphi (t{)}^2}=-\sigma (t-t_0)$ for a constant $t_0$, andThe graph lies on a circle of radius $r$, exactly as described by constant strip-area density of a surface of revolution.