The task is to calculate an integral of the form $$\iint_s x dydz + ydxdz +zdxdy. $$ The text I'm studying says this is just another form of the famed divergence theorem, $$\iint_s F.n \,ds = \iiint_v \operatorname{div} F \,dV,$$ where $F$ is simply $(x, y, z)$, so that the value of the integral is $$\iiint (1+1+1) dV = 3V.$$
The question is why is the original integral equal to the divergence integral? More generally, why is it true that $$\iint_s Pdzdy+ Qdxdz +Rdxdy = \iint_s (P, Q, R)\cdot n \,ds?$$
It is indeed true that $$(F\cdot \hat{n})\,dA = F_x dy\wedge dz + F_y dz\wedge dx + F_z dx\wedge dy.$$ (Be careful about the orientation of each term!)
This follows from the fact that $$\hat{n}_x dA = dy\wedge dz$$ on the surface of $S$, and similarly for the other coordinates; you can prove this by direct calculation on a coordinate chart, but probably the easiest way is to notice that the Hodge stars of both term, $$\hat{n}_x (\hat{n}_x dx + \hat{n}_y dy + \hat{n}_z dz), \qquad dx$$ agree on vectors $\alpha \hat{n}$ orthogonal to $S$.