Geodesic curve on rotation surface

Question

Let $S$ be the surface obtained by a rotation around the $z$-axis of $\gamma$ where $$\gamma:(0,+\infty) \rightarrow \mathbb R \quad \gamma(s) = \left(\frac{1}{\cosh(s)},0,s-\tanh(s)\right)$$ Let $f: \mathbb R \rightarrow \mathbb R$ be a $2\pi$-periodic differentiable function and $R_\theta$ the function that rotates around the z-axis of an angle $\theta$. So $\alpha_f(\theta)=R_\theta(\gamma(f(\theta))) $ is a curve on $S$. I need to prove that for any choice of $f$, $\alpha_f$ isn't a geodetic curve on $S$.The only way I came up with to solve this exercise was to use the Gauss-Bonnet theorem to find an absurd (supposing $\alpha_f$ is a geodetic) since I know that the curvature of $S$ is $k(s,\theta) = - \tanh(s) < 0 \ \forall s \in (0,+\infty)$ but I really cannot think of a regular region of S useful for my purpose. Thank you for any hint in advance.