I'm starting a course on Vector Calculus, and I got to the intuitive relation between the divergence and the density of the fluid, namely that we can see the divergence as the opposite of the change of density per unit of time, or:
$$div(F)=-\frac{d\rho}{dt}$$
where $F$ is the vector field, and $\rho$ is the density of the fluid, both functions of position and time.
From this, we deduce that given the divergence, we can compute the density by integrating the divergence with respect to the time. Now, if the field $F$ does not depend on the position, e.g., $F(x,y,t)=(t,t+1)$, then the divergence is $0$, which by the above formula would imply the density is also $0$.
This seems rather counterintuitive to me, as the fluid does seem to be increasing in quantity as time passes, yet the density is $0$? Care to help me understand this?
$\rho$ isn't $0$, $\frac{d\rho}{dt}=0$. That means that density is constant and not necessarily $0$.