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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 $(\ast )$.

Completing the square givesThus $V$ is a two-napped cone. On the nappe where $\epsilon y>0$, with $\epsilon \in \{1,-1\}$, useThis has the conical formIts second derivatives obeyThe latter vector is tangent, so for either choice of unit normal,Since the parametrization is regular, $EG-F^2>0$, and thereforeat every point of $V$. This is the Gaussian curvature of a cone away from its vertex.

The punctured circular coneis geodesically incomplete because a generator reaches the missing vertex in finite time. It is an inextendible embedded surface: any proper connected embedded extension would have to add the vertex, but its tangent planes have different limits as $\theta$ varies. Nevertheless its Gaussian curvature of a cone away from its vertex is identically zero.