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

Set $H(x,y,z)=x^3+xy+yz-1$. The gradient $\nabla H=(3x^2+y,x+z,y)$ is a normal vector to its level set. Since $y\ge 1/2$ on the given patch, division by its last component is valid and gives the unique normalized fieldIts dot product with $\mathbf{k}$ is one, as required; it is not generally a unit vector.

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.

Take the dot product of the given field with the upward vector area element. The surface relation simplifies its integrand:The projected domain is exactly the rectangle $1/2\le x,y\le 1$, since $y>0$ determines a unique $z=(1-x^3)/y-x$. Therefore the surface integral is

The defining function $x^3+xy-1$ is independent of $z$, so a normal vector is $(3x^2+y,x,0)$. It never vanishes here because $x>0$. Every normal is horizontal, whereas $\mathbf{G}=(0,0,x)$ is vertical and tangent to the surface. Their dot product is zero at every point, givingThis is independent of orientation. Although the surface extends without a bound on $x$, the integrand vanishes identically, so no convergence problem arises for this flux.