= Solution
Choose the normal pointing away from the origin, equivalently the outward normal of the first-octant tetrahedron. Parametrize the face by $\mathbf r=(x,y,2-x/2-y/2)$ on $D=\{x,y\ge0:x+y\le4\}$. Its oriented area element is
$$
\mathbf r_x\times\mathbf r_y\,dx\,dy=(1/2,1/2,1)\,dx\,dy.
$$
Thus the <surface integral> is
$$
\iint_D\left(-\frac x2+\frac y2+2-\frac x2-\frac y2\right)dx\,dy
=\int_0^4(2-x)(4-x)\,dx
=\boxed{\frac{16}{3}}.
$$
The <divergence theorem> checks this result: $\nabla\cdot\mathbf F=1$, the tetrahedron's <volume> is $4\cdot4\cdot2/6=16/3$, and each coordinate-plane face has zero <flux>. The opposite orientation gives $-16/3$; a signed <flux> requires an orientation.
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