Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 11G (Geometry)

The surface is obtained by revolving the graph $z=f(x)$ about the $x$-axis. It has a central cylindrical section of radius $b$, two smooth transition collars, and unit spherical caps centred at $(\pm 3,0,0)$; its projection onto the $(x,z)$-plane is the region $|z|\le f(x)$.

For a surface of revolution, the Gaussian curvature of a surface of revolution isThus the cylindrical region has $K=0$. On each transition collar, $K<0$ where $f^{\prime \prime }>0$, vanishes at the unique inflection circle, and is positive where $f^{\prime \prime }<0$. The spherical caps have $K>0$.

The area element isConsequentlyAt $x=2-a$, smooth matching to the cylinder gives $f^{\prime }=0$. At $x=2+a$, the spherical formula givesThe total Gaussian curvature of a surface-of-revolution strip is therefore

The two curves produced by $y=0$ are meridians, and meridians of a surface of revolution are geodesics. A boundary circle $x=x_0$ is a geodesic precisely when $f^{\prime }(x_0)=0$, as follows either from the geodesic equations or from its geodesic curvatureThe circle at $x=2-a$ is geodesic because it joins the cylinder smoothly. At the other boundary,which vanishes exactly when $a=1$. Hence the cut pieces are geodesic polygons only for