Use Gauss’s theorem to evaluate ∬S F⋅dS, where F=zex2 i+3y j+(2−yz7) k and S is the union of the five “upper” faces of the unit cube [0, 1]×[0, 1]×[0, 1]. That is, the z=0 face is not part of S. (Hint: Note that S is not closed, so to apply Gauss’s theorem you will have to close it up.)