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Letbe the Clairaut first integral for a surface of revolution from part a(ii). If $c\ne 0$, then $|\cos \theta |\le 1$ givesfor every $s$. Since $f$ tends to zero at both poles, the geodesic avoids nonempty open neighbourhoods of the poles.

If $c=0$, then $u^{\prime }(s)=0$ whenever the geodesic is away from a pole. It consequently lies in one meridian of a surface of revolution; after crossing a pole, the angular coordinate changes by $\pi$ but remains in the same plane through the rotation axis. A small surface neighbourhood of any point outside that meridian is disjoint from the geodesic. In either case there is a nonempty open set $U$ withThis proves that the geodesic image on a compact two-pole surface of revolution is not dense.