A particular geodesics problem I am trying to find the geodesics on the surface defined by $M:=\{(x,y,z)\in\mathbb{R}^3|x^2 + y^2 = f(z)\}$, for $f:\mathbb{R}\rightarrow \mathbb{R}_+$.
I have parametrized the surface by
$$
\vec{r} = \sqrt{f(z)}\cos\theta\hat{i} + \sqrt{f(z)}\sin \theta \hat{j} + z\hat{k}
$$
and calculated the metric components to be:
$$
g_{zz} = \frac{f'(z)^2}{4f(z)} + 1$$
and
$$
g_{\theta\theta} = f(z).
$$
Calculating the Christoffel symbols really yields a mess and thus a difficult set of coupled differential equations for the geodesics. I am only asked to prove that the intersection lines of $M$ with the planes $ax + by = 0$ are geodesics, but I have no clue how to do this besides calculating the geodesics explicitly. Does anyone have a hint on how to solve this?
 A: Hints to make solving for geodesics much easier:


*

*Assume that geodesic is affinely parameterized, namely $x^i(\lambda) = (z(\lambda), \theta(\lambda))$ with unit speed.  This translates into the following first order differential equation: $g_{ij}\dot x^i\dot x^j = 1$ where overdots represent derivative with respect to affine parameter $\lambda$.

*Notice that the metric components are not explicitly dependent on $\theta$.  This means that the metric has a $\theta$-translation symmetry, so $\partial_\theta$ is a killing vector.  This gives a conservation equation $g_{ij}\dot x^i(\partial_\theta)^j = g_{\theta\theta}\dot\theta = \mathrm{const.}$.
Solving the resulting set of first order differential equations for $z(\lambda)$ and $\theta(\lambda)$.
Here's another hint: the planes $ax+by=0$ are just vertical planes perpendicular to the $x$-$y$ plane.  It follows that their intersections with the given surface correspond to curves for which $\dot\theta = 0$.  If you look at the equations obtained from the two hints above, it should be clear that there are geodesics with this property.
