Let's say i have a closed surface S, for instance the surface of a cube.
and I take the surface integral in a vector field $\vec{F}$ over this closed surface:
$$\oint \limits_S\vec{F} \cdot \hat{n} ~dS$$
I would expect that net flux going into the surface equals flux going out of the surface and the integral should be zero:
$$\oint \limits_S\vec{F} \cdot \hat{n} ~dS = 0$$
However, the example, i'm looking at in the book has this integral having a value that is non-zero.
Its this because the vector field $\vec{F}$ is non-conservative that causes the close surface integral to be non-zero? That is: $\nabla \times \vec{F} = 0$.
The problem is rather lengthy, but I will put it here just so the question has context.
let $\vec{F} = 4zxz\hat{\text{i}} - y^2\hat{\text{j}} + yz\hat{\text{k}}$. Evaluate $\oint \limits_S \vec{F} \cdot \hat{n}~ dS$, where S is cube bounded by: x=0, x=1, y=0, y=1, z=0, z=1.
then its matter of finding the surface integal of each of the sides of the cube.
$\begin{aligned}\oint \limits_S \vec{F} \cdot \hat{n}~ds &= \int \limits_{\text{side 1}} \vec{F} \cdot \hat{n}~ds + \int \limits_{\text{side 2}} \vec{F} \cdot \hat{n}~ds \\ &+ \int \limits_{\text{side 3}} \vec{F} \cdot \hat{n}~ds + \int \limits_{\text{side 4}} \vec{F} \cdot \hat{n}~ds + \int \limits_{\text{side 5}} \vec{F} \cdot \hat{n}~ds + \int \limits_{\text{side 6}} \vec{F} \cdot \hat{n}~ds\end{aligned}$
$\oint \limits_S \vec{F} \cdot \hat{n}~ds = (2) + (0) + (-1) + (0) + (\frac{1}{2}) + (0) = \frac{3}{2}$
solenoidal: a vector field which has zero divergence everywhere is called solenoidal, and any closed surface in the vector field has no a net flux across it, that is: flux in = flux out.