Use the Divergence Theorem to compute the surface integral?

Question

Use the Divergence Theorem to compute the surface integral where Q is the cube -1<=x<=1, -1<=y<=1, -1<=z<=1, F=<4y^2, 3z-cosx, z^3-x>. (Answer: 8)

The divergence of the vector field is 3z^2, which is what I found.

How to set up the integral and solve for it?

Answer 1

By the Divergence Theorem,

the surface integral

= ∫[-1,1] dx∫[-1,1] dy ∫[-1,1] 3z^2 dz

= 2*2*z^3 from -1 to 1

= 2^3

= 8 <==Answer

Answer 2

Div F = 3z^2

 

∫∫ F ds  =   ∫∫∫ div F dV

∫∫ F ds = ∫∫∫ div F dx dy dz

           

            = ∫∫∫  dx dy 3z^2 dz

            =3∫∫∫dx dy z^2 dz.   The limits of integration for each integral is  -1 to 1.

            = 8