Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 4

Where $f_z\ne 0$, the implicit function theorem makes the surface a local graph $z=h(x,y)$ with $h_x=-f_x/f_z$ and $h_y=-f_y/f_z$. Its upward vector area element is $(-h_x,-h_y,1)dxdy$. Therefore the vector area element of an implicit surface isDividing by $f_z$ fixes the upward orientation even if the initially chosen gradient points downward. The restriction excludes a locally vertical graph, where this particular projected-area representation cannot be used.

Choose $f=x^2(1+y)+y^{2}z-1$. Since $f_z=y^2>0$ on the stated open domain, its upward vector area element isThe graph becomes unbounded near parts of $y=0$; the flux integrals below are understood as limits from $y>0$.

Subtract the integrands before integrating: ${\mathbf{G}}_2-{\mathbf{G}}_1=(0,y^3,0)$. With the upward vector area element,This extends continuously to the lower edge, soThe individual flux integrals also converge: ${\mathbf{G}}_1\cdot d\mathbf{S}=x(3+2y)dxdy$ gives $I_1=5$, and hence $I_2=7$. Reversing the surface orientation changes both signs and leaves the requested magnitude unchanged.