Can you explain it clearer that how did S and S1 were calculated differently ? Why are we calculating S1?

10 Divergence Theorem 6. (15 pts) Let S denote the surface of the cube of side length 2, contained in the first quadrant, without the top face (i.e. the cube with vertices (0, 0, 0), (2, 0, 0), (0, 2, 0), (0, 0, 2) (2, 2, 0), (2, 0, 2), (0, 2, 2), (2, 2, Set F(x, y, z) = (3x, 3y, 3z). Calculate SSs F . dS in whatever way is simplest. Use divergence THM solid cube n has boundary _ = S+ S, with outer normal SSSOFdv - JSFd's + SS Fd's SI SSS adv = qvol ( n ) = 9( 2)3 = 72 parametrizati B ( xy ) = Oxx= 2 0sys 2 SS FOS 8 x x by = . dydx = SzS : 6 dydx = 24 - SS Fd's = 12- 24= 48