= Solution
Parametrize the <surface of revolution> by $X(s,\theta)=(f(s)\cos\theta,f(s)\sin\theta,g(s))$. The coordinate tangent vectors have squared lengths $f'^2+g'^2=1$ and $f^2$, and inner product zero. Thus the <first fundamental form of an arc-length surface of revolution> is \b[$I=ds^2+f(s)^2d\theta^2$].
For an arc-length parameter $t$, the <geodesic equations> are
$$
\boxed{\ddot s-ff'\dot\theta^2=0,\qquad \ddot\theta+2\frac{f'}f\dot s\dot\theta=0,\qquad \dot s^2+f^2\dot\theta^2=1.}
$$
These follow either from the metric's <Christoffel symbols> or from the <Euler-Lagrange equation> for $\tfrac12(\dot s^2+f^2\dot\theta^2)$. On a parallel $s=s_0$, unit speed requires $\dot\theta=\pm1/f(s_0)$; the first <geodesic equation> then holds exactly when $f'(s_0)=0$. Thus <geodesic parallels of a surface of revolution> are precisely the critical points of the radius function.
To obtain any prescribed positive integer $n$, set
$$
H(s)=e^{-s^2}\prod_{j=1}^n(s-j),\qquad f(s)=R+\varepsilon\int_0^sH(t)\,dt,\qquad g(s)=\int_0^s\sqrt{1-f'(t)^2}\,dt.
$$
The function $H$ is bounded and integrable. Choose $\varepsilon>0$ so $\varepsilon\sup|H|<1$, and choose $R>\varepsilon\int_{\mathbb R}|H|$. Then $f>0$, $g$ is smooth, and $f'^2+g'^2=1$. The derivative $f'=\varepsilon H$ vanishes at exactly $1,\ldots,n$. Also $g'>0$, so different meridian parameters give different heights. \b[The resulting surface has exactly $n$ geodesic parallels.]
For explicit sketches, use the following two radius functions, again with $g'=\sqrt{1-f'^2}$:
$$
\begin{aligned}
f_1(s)&=2-0.6e^{-s^2},\\
f_2(s)&=2+0.6\left(-\frac12se^{-s^2}-\frac{\sqrt\pi}{4}\operatorname{erf}(s)\right).
\end{aligned}
$$
Here $f_1'=1.2se^{-s^2}$ has only the zero $0$, while $f_2'=0.6(s^2-1)e^{-s^2}$ has exactly the zeros $-1,1$. Both radii stay positive and both derivatives have absolute value below one. The highlighted rings are the geodesic parallels; the surfaces continue beyond the displayed window.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-4-geodesic-parallels.png]
{title=Surfaces of revolution with exactly one and exactly two geodesic parallels}
If every parallel is a <geodesic>, then $f'=0$ everywhere, so $f=R$ is a positive constant. The arc-length condition gives $g'^2=1$; smoothness forces a constant sign, hence $g(s)=\pm s+c$. \b[The surface is a circular cylinder of radius $R$.] It is a <circular cylinder> because the meridian is a straight line parallel to the axis.
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