 Solution

Take the normal vector to point outward from the box. The divergence theorem changes the closed-surface flux into a volume integral. The divergence isThe first term integrates to zero in $x$, since ${\int }_{-1}^2\pi \cos (\pi x)dx=\sin 2\pi -\sin (-\pi )=0$. ThereforeHenceA direct face check gives zero from the two $x$ faces, $45/2$ from the pair of $y$ faces, and $45$ from the pair of $z$ faces. This also checks the outward-orientation signs.