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The surface of revolutionis geodesically incomplete: a meridian reaches the missing origin in finite length. Its Gaussian curvature, computed using the curvatures of a parametrized surface of revolution, iswhich tends to infinity as $r\downarrow 0$. Every finite-time endpoint of a constant-speed geodesic must approach the origin, the only finite point in the ambient closure missing from $S$, so this surface satisfies the curvature-blowup criterion.

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 $(\ast )$.

WriteThe regularity of the profile makes $q>0$. The coordinate tangent vectors areso the first fundamental form of a surface of revolution has coefficientsChoose the unit normalTaking scalar products with the second derivatives of $\varphi$ gives the second fundamental form coefficientsThus the coordinate directions are principal directions, with principal curvaturesThe curvatures of a parametrized surface of revolution are thereforeReversing the orientation of a surface reverses the sign of $H$ but leaves $K$ unchanged.