= 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 parametrization
$$
X(\theta,t)=(\phi(t)\cos\theta,\phi(t)\sin\theta,t).
$$
It is locally one-to-one and has <tangent vectors>
$$
X_\theta=(-\phi\sin\theta,\phi\cos\theta,0),
\qquad
X_t=(\phi'\cos\theta,\phi'\sin\theta,1).
$$
Their cross product has magnitude
$$
|X_\theta\times X_t|=\phi(t)\sqrt{1+\phi'(t)^2}>0,
$$
so 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 therefore
$$
2\pi\int_{a_0}^{b_0}\phi(t)\sqrt{1+\phi'(t)^2}\,dt.
$$
By hypothesis this equals $2\pi r(b_0-a_0)$ for every subinterval. Since the integrand is continuous,
$$
\phi(t)\sqrt{1+\phi'(t)^2}=r,
$$
and squaring gives
$$
r^2=\phi(t)^2+\phi(t)^2\phi'(t)^2.
$$
If $0<\phi(t)<r$, then $\phi'$ never vanishes, so its sign $\sigma\in\{1,-1\}$ is constant. The last equation gives
$$
\phi'=\sigma\frac{\sqrt{r^2-\phi^2}}{\phi},
$$
and hence
$$
\frac d{dt}\sqrt{r^2-\phi(t)^2}=-\sigma.
$$
Thus $\sqrt{r^2-\phi(t)^2}=-\sigma(t-t_0)$ for a constant $t_0$, and
$$
(t-t_0)^2+\phi(t)^2=r^2.
$$
The graph lies on a circle of radius $r$, exactly as described by <constant strip-area density of a surface of revolution>.
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