Flux to Volume

According to the Divergence Theorem :

$\int \int \int_E (\nabla \cdot \vec{F}) dV = \int \int_S (\vec{F} \cdot \vec{n}) dS$

We are told that the left side is equal to the volume, and we are given an explicit way to evaluate the right side (the net flux):

$L^3 = -1 + 2 - 3 + 4 - 5 + 6 = 3 \\ \implies L = \sqrt[3]{3}$

We are given that $\iiint_E \nabla \cdot \mathbf{F} \ dV$ is the volume of the cube $E.$ For any region $\mathcal{S}$ in $\mathbb{R}^3$ , the volume of $\mathcal{S}$ can be written as $\iiint_{\mathcal{S}} dV$ . As such, we see that $\nabla \cdot \mathbf{F} = \text{div} \ \mathbf{F} = 1$ .

The net flux of $\mathbf{F}$ over $E$ is $\displaystyle \sum_i \text{flux } S_i$ . Since $\dfrac{\text{ flux(} \mathbf{F} \text{)} }{\text{Volume (} E \text{)}} = \text{div } \mathbf{F}$ , we see that the volume, $V$ , of the cube is $\displaystyle \dfrac{\text{flux } \mathbf{F}}{\text{div } \mathbf{F} } = \dfrac{\displaystyle \sum_{i=1}^6 (-1)^i i}{\nabla \cdot \mathbf{F}}$ .

Therefore $V$ is simply $\dfrac{-1+2-3+4-5+6}{1} = 3$ . Thus, the edge length of the cube is $\sqrt[3]{3}$ , and therefore $a+b = \boxed{6}$ .