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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.

For an example satisfying $(\ast )$, consider the incomplete surface z equals r to three halvesA meridian of a surface of revolution reaches the missing origin in the finite lengthso $S_{\ast }$ is geodesically incomplete. The curvatures of a parametrized surface of revolution giveBy constant speed of an affinely parametrized geodesic, a geodesic with a finite endpoint has finite length and converges in ${\mathbb{R}}^3$. The finite-dimensional continuation criterion rules out an endpoint in a compact subset of $S_{\ast }$, and infinity is infinitely far away. It must approach the origin, so $(\ast )$ holds.

The answer to the second question is no. The punctured circular coneis incomplete because a generator reaches the missing vertex in finite time. It is inextendible: a proper connected smooth extension would have to add the origin, but the tangent planes approaching the origin depend on $\theta$ and therefore cannot be the continuous tangent planes of a smooth surface. Yetby the Gaussian curvature of a cone away from its vertex. Thus this incomplete inextendible surface does not satisfy $(\ast )$.

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.