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

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.