Graph density and outward orientation

The tangent columns are $(e_i,{\mathrm{\partial }}_{i}h)$, so $D{X}^{T}DX=I+Dh(Dh{)}^T$ and F1 gives density $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$. The vector $(-Dh,1)$ has zero dot product with each tangent column, length $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$, and positive last component. It points out of $z<h(y)$, as moving in its direction increases $z-h(y)$ to first order by $1+\mathrm{\mid }Dh{\mathrm{\mid }}^2>0$. Dividing by its length and multiplying by the density proves $\nu dS=(-Dh,1)dy$.

For affine h, $Dh=a$, so $dS=\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}dy$ and $\nu =(-a,1)\mathrm{/}\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}$. In the stated instance $a=(2,-1)$, giving $D{X}^{T}DX=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)$, determinant $10-4=6$, and $\nu =(-2,1,1)\mathrm{/}\sqrt{6}$. Integration over the unit square gives area $\sqrt{6}\cdot 1=\sqrt{6}$ and flux $\int (e_3\cdot \nu )\sqrt{6}dy=\int 1dy=1$. For the upper unit hemisphere, $h(y)=\sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}$ on $\mathrm{\mid }y\mathrm{\mid }<1$. Here $Dh=-y\mathrm{/}h$, so $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}=1\mathrm{/}h$ and $\nu =(y,h)$; its last component is positive and it is the radial outward unit vector. The equator is outside this graph. Rotated full-sphere graph charts cover it in an atlas of the full sphere, but no graph chart contained in the closed upper hemisphere covers an equator point.