Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 15F / b

The generating circle has $f^{\prime 2}+g^{\prime 2}=1$, and its revolution is a torus. From part (a), $K=\cos u/(2+\cos u)$ and the area element of a surface is $dA=fdudv=(2+\cos u)dudv$. ThereforeThe positive Gaussian curvature on the outer side exactly cancels the negative curvature on the inner side. Equivalently, the Gauss-Bonnet theorem gives ${\int }_{S}KdA=2\pi \chi (T^2)=0$ because the Euler characteristic of the torus is zero.