Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 2G / Solution

Set $D=1+u^2+v^2$. The two tangent vectors to the Enneper surface areTheir cross product is $D(-2u,2v,1-u^2-v^2)$. Since $4u^2+4v^2+(1-u^2-v^2{)}^2=D^2$, its length is $D^2>0$, proving regularity and giving the unit normalFor the first fundamental form, direct dot products give $E=G=D^2$ and $F=0$. The second derivatives are ${\sigma }_{uu}=(-2u,2v,2)$, ${\sigma }_{uv}=(2v,2u,0)$, and ${\sigma }_{vv}=(2u,-2v,-2)$. Taking their dot products with $\mathbf{n}$ gives the second fundamental form coefficients $e=2$, $f=0$, $g=-2$. Therefore the Gaussian curvature isHere curvature means Gaussian curvature; the opposite signs of the second-form diagonal entries also give zero mean curvature, as expected for this minimal surface.