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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 $(\ast )$.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.