Gauss's divergence theorem for a scalar field I know Gauss's divergence theorem for a vector field:
$$\iint{\vec{F}\cdot\hat{n}}\space{dS}=\iiint\nabla\cdot\vec{F}\space{dV}$$
But how do you apply this to a scalar field? For example, if you wanted to find the surface integral of $z^2$ over a unit cube:
$$\iint_{S}z^2{dS}$$
where $S$ is the surface of unit cube, how would you approach this using Gauss's divergence theorem?
 A: You can always write $ z^2 = z^2 \hat{n}.\hat{n} $ where $\hat{n} $ is the unit normal then using divergence theorem your expression becomes 
$$ \iiint_{|z|\leq 1} \nabla.(z^2\hat{n})dV $$
A: Indeed we can use divergence theorem if we could find an $F$ such that:
$$
\iint_S g\,dS = \iint_S F\cdot n\,dS.\tag{1}
$$
Though the example you gave is not very illustrative it seems. Say the unit cube is $D = [0,1]\times[0,1]\times [0,1]$. The faces perpendicular to the $x$-axis have unit outward normals $(\pm 1,0,0)$, if we want $F\cdot (1,0,0) = z^2$ on the face $\{x = 1\}\cap D$, while $F\cdot (-1,0,0) = z^2$ on the face $\{x = 0\}\cap D$ as well? 
Here we can't find a well-defined $F$ such that $F\cdot n = z^2$ on all six faces of the unit cube, and we have to evaluate directly:
$$
\mathrm{Top} + \mathrm{Bottom} =  \iint_{[0,1]\times[0,1]\times \{1\}} 1 \,dxdy + \iint_{[0,1]\times[0,1]\times \{-1\}} 0 \,dxdy,
\\
\mathrm{Right} + \mathrm{Left} =  \iint_{[0,1]\times\{1\}\times [0,1]} z^2 \,dzdx + \iint_{[0,1]\times \{-1\}\times[0,1]} z^2 \,dzdx,
\\
\mathrm{Front} + \mathrm{Back} =  \iint_{\{1\}\times[0,1]\times [0,1]} z^2 \,dydz + \iint_{\{-1\}\times[0,1]\times [0,1]} z^2 \,dydz.
$$

Based on my teaching experience, the problem to illustrate (1) is actually taking $S$ to be the unit 2-sphere:
$$
S = S^2 = \{(x,y,z): x^2 + y^2 + z^2 = 1\},
$$
so that the unit outward normal is $n=(x,y,z)$:
$$
\iint_S z^2 \,dS = \iint_S (0,0,z)\cdot (x,y,z) \,dS = \iiint_{B(0,1)} \nabla\cdot (0,0,z)\,dV = |B(0,1)|.
$$
