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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.

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.