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Take the vector field $F$ to be: $$F(x,\,y,\,z) = \left( f_1(x,y,z), \ f_2(x,y,z), \ f_3(x,y,z)\right).$$

Then, $$\text{div} \ F = \left( \frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} + \frac{\partial f_3}{\partial z} \right).$$ If divergence is the flow through an infinitesimal cube, then couldn't the $x$-component of $F$ be changing, but not contributing to the overall flow through the cube with this definition of flux?

Here's a case that confuses me: $$F(x,\,y,\,z) = (z,\ x,\ y).$$ So, $$\text{div} \ F = 0 + 0 + 0 \text{,} $$ but the vector field is nonzero everywhere besides the origin.

I understand that the $x$-component is not changing as $x$ is varied. But I'm imagining that there is flow through a tiny axis-aligned cube in the $x$-direction (since $z$ is nonzero).

Any tips on where I'm going wrong?

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    If the flow in the $x$ direction does not depend on $x$, then the flux through the left side of the cube will be perfectly balanced by that through the right. – 2017-02-06
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    Divergence is measure the "focus" of the field . The field above is not "focus" on any where . At least that I know. – 2017-02-06

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