= Vector area element of an implicit surface
{title2=$d\mathbf S=(\nabla f/f_z)\,dx\,dy$}
For a regular implicit surface $f(x,y,z)=0$ with $f_z\ne0$, the <implicit function theorem> gives a local graph $z=h(x,y)$. Its upward <vector area element> is $(-h_x,-h_y,1)dx\,dy$. Differentiating $f(x,y,h(x,y))=0$ yields $h_x=-f_x/f_z$, $h_y=-f_y/f_z$, proving the formula. Division by $f_z$ makes the final component positive independently of the sign of the original gradient. At a point where $f_z=0$, another projection or parametrization is needed.
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