# Gauss divergence theorem for a scalar

I have to integrate $$\int_S^\ x~dy~dz + y~dz~dx + z~dx~dy\$$ where '$S$' is the cube of unit length.

I have the solution of the problem which converts this surface integral into a volumetric integral using the gauss divergence theorem, in this way $$\nabla\cdot F = 1 + 1 + 1 = 3$$ $$\int_V\ \nabla \cdot F~dV = \int_V3~dV = 3\cdot 1 = 3$$

From what I have been doing till now is that I used to have a vector on which I applied the gauss divergence theorem where $\hat{i}$,$\hat{j}$ and $\hat{k}$ were defined.

How did they apply $\nabla \cdot F$ to a scalar? It makes no sense to me.

Thank You

$\newcommand{\Vec}[1]{\mathbf{#1}}$The integral $$\iint_{S} x\, dy\, dz + y\, dz\, dx + z\, d\, dy$$ is a vector surface integral, giving the flux of the radial field $F(x, y, z) = x\Vec{i} + y\Vec{j} + z\Vec{k}$ over the surface of the unit cube. This explains the Gauss' theorem calculation you sketch.
Note generally that if $S$ is an oriented surface with continuous unit normal field $\Vec{n}$, and if $F$ is a continuous vector field on $S$, then $$\iint_{S} F \cdot d\Vec{S} = \iint_{S} (F \cdot \Vec{n})\, dS$$ (once expressed using a parametrization) is the integral of a scalar function, the flux density $F \cdot \Vec{n}$, over $S$.
• A differential $2$-form $\omega = F_{1}\, dy\, dz + F_{2} \, dz\, dx + F_{3}\, dx\, dy$ corresponds to the vector field $F = (F_{1}, F_{2}, F_{3})$ in the sense that if $X$ and $Y$ are vectors at some point $p$, then $\omega(p)(X, Y) = F(p) \cdot (X \times Y)$. – Andrew D. Hwang Jan 14 '17 at 14:33