Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 7 / a / ii / Solution

Orthogonal projection of a small surface patch onto the $xy$ plane multiplies its scalar area by $\cos \alpha =\mathbf{n}\cdot \mathbf{k}>0$. Hence $dxdy=\cos \alpha dS$ and $d\mathbf{S}=\mathbf{n}dxdy/\cos \alpha$. Since $\mathbf{n}=\mathbf{u}/|\mathbf{u}|$ and $\mathbf{u}\cdot \mathbf{k}=1$, one has $\cos \alpha =1/|\mathbf{u}|$. The vector area element of an implicit surface is thusThe positive final component implements the specified upward orientation.