 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 $D=\{x,y\ge 0:x+y\le 4\}$. Its oriented area element isThus the surface integral isThe divergence theorem checks this result: $\nabla \cdot \mathbf{F}=1$, the tetrahedron's volume is $4\cdot 4\cdot 2/6=16/3$, and each coordinate-plane face has zero flux. The opposite orientation gives $-16/3$; a signed flux requires an orientation.