 Geodesic image on a compact two-pole surface of revolution is not dense

Let a compact smooth surface of revolution have exactly two poles, and let $c=f(v{)}^2\dot{u}=f(v)\cos \theta$ be the Clairaut first integral for a surface of revolution of a complete unit-speed geodesic. If $c\ne 0$, then $f(v)\ge |c|$, so the geodesic avoids open neighbourhoods of both poles. If $c=0$, then $u$ is constant away from the poles and the geodesic lies in a meridian of a surface of revolution, whose complement contains a nonempty open set. Thus no complete geodesic has dense image.