The divergence theory states that
$ \iiint\limits_T \nabla \cdot F \; dxdydz = \iint\limits_S F \cdot n \; dA $
where $ F=[F_1,F_2,F_3]$ is a vector field.
But in some homework solutions I see without a proof that they also claim that
$ \iint\limits_S F \cdot n\; dA = \iint\limits_S F_1 dydz + F_2 dxdz + F_3 dxy $
How do you prove this claim ?
In fact, $F \cdot n \, dA = F_1\, dy \wedge dz + F_2\, dx \wedge dz + F_3 \, dx \wedge dy \, = i_{F}\omega_{T}$, where $\omega_{T}$ is the volume form on $T$ (basically, in any oriented coordinate system on $T$, $\omega_T = dx \, dy \, dz = dV$), and $i$ denotes the interior product. A way to see this is to pick oriented coordinates $(x_1, \dots, x_3) \in \{x \in \mathbb{R}^3 : x_1 \leq 0\}$ for $T$ for which $S = \partial T$ is given by $\{x_1 = 0\}$, and $e_1$ is the outward unit normal to $S = \partial T$, i.e. $n = e_1$. Then $$dV = \sqrt{g}\,dx_1 \wedge dx_2 \wedge dx_3$$ $$dA = \sqrt{g}\,dx_2 \wedge dx_3,$$ $$F \cdot n \, dA = F_1 \sqrt{g} \, dx_2 \wedge dx_3,$$ $$i_{F}\omega_{T} = i_{F}dV = \sqrt{g}F_1 \, dx_2 \wedge dx_3.$$
If you aren't familiar with this algebraic machinery, do the computations in some coordinate system using things you are familiar with (graph coordinates are an excellent choice!).