Why does a stream of falling water get narrower at the bottom? At first glance, I thought that the flux of the velocity field, $\mathbf u$, should be constant. The velocity of the water increases due to gravity, hence the imaginary area the velocity field is passing through has to decrease.
$$
\frac{d}{dt} \int_{C}{ \mathbf u  \cdot d \mathbf A } = 0
$$
$$
|\mathbf u(t)| |\mathbf A(t)|=k\\
u(t)=u_0 +gt\\
A(t) = \frac{k}{u_0 +gt} \\
A(0) = A_0 \rightarrow k = A_0 u_0
$$
Finally arriving at:
$$
A(t) = \frac{A_0 u_0}{u_0 + gt} \\
A(t) \propto \frac{1}{t}
$$
Then, supposing the faucet has a circular hole, we can assume this area, $A(t)$, is from a circle.
$$
\frac{A_0 u_0}{u_0 + gt} = \pi r^2 \\
\frac{A_0 u_0}{g \pi r^2} - \frac{u_0}{g} = t
$$
Now, plotting each time slice of the stream area gives the following (time is in the z-axes):

This is pretty similar to what you would see in real life. Each slice (now in 3D space) of the stream of water as you go down the z-axis (because  the further down you go in the stream, the more time has passed) the narrower it gets.
Is my line of reasoning correct? Is there any incorrect assumptions that I made? Is this result coherent or even correct?
 A: 
Is my line of reasoning correct? Is there any incorrect assumptions that I made? Is this resoult coherent or even correct?

Seems reasonable to me, at least to zeroth order.
You are basically using:
$$ 
v_1 A_1 = v_2 A_2\; \qquad(1)
$$
and
$$
v_2 = v_1 + gt\;. \qquad (2)
$$
If you want to get rid of the Eq. (2) and the time dependence in favor of, say, distance ($d$) below the faucet, you could integrate Eq. (2) again and solve for time.
Alternatively, you could ignore Eq. (2) and instead use Bernoulli's equation in the form:
$$
P+\rho g h_1 + \frac{1}{2}\rho v_1^2 = P+\rho g h_2 + \frac{1}{2}\rho v_2^2
$$
So, you have:
$$
v_2 = \sqrt{v_1^2 + 2gd}\;,
$$
where $d$ is the distance below the faucet.
So you get the $d$ dependence of the area as:
$$
A(d) = \frac{v_1 A_1}{\sqrt{v_1^2 + 2gd}}
$$
Or, the $d$ dependence of the radius as:
$$
r(d) = \sqrt{\frac{v_1 A_1}{\pi\sqrt{v_1^2 + 2gd}}}
$$
A: There will be an additional contraction term due to the fact that the water entering the stream sideways carries with it a velocity component in the x and y directions, but I do not know how to calculate it.
