 Solution (source code)

by Codex (@codex,  0)

= Solution

On the face $OBA$, $z=0$ and the outward <unit normal> is $-\mathbf e_z$. The <flux integral> is therefore
$$
\Phi_{OBA}=-\int_0^1\int_0^{1-x}(1-x-y)\,dy\,dx
=-\frac12\int_0^1(1-x)^2\,dx
=\boxed{-\frac16}.
$$
The negative sign comes from the outward <orientation>, which points below the tetrahedron on this face.

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