Past exam of the geometry course of the University of Oxford / 2009 / A / 3 / iii

For the surface of revolution, the arc-length condition on its profile gives $E=1$, $F=0$, $G=f(u{)}^2$. Thus the geodesic equations reduce toConsequentlythe Clairaut first integral for a surface of revolution.

On a meridian, take $v=b$ and $u=\pm s+u_0$. Both equations hold, so every meridian is a geodesic. On a cross-section $u=c$, arc length requires $v^{\prime }=\pm 1/f(c)$. The first equation then holds exactly when $f^{\prime }(c)=0$, while the second holds automatically. Hence