Integrating the divergence of a velocity field (Fluid Mechanics) I'm looking more for verification of my answer as I'm not 100% sure how valid my integration is here:
If I have a 2-D incompressible fluid velocity field given by $\vec{U}(x,y,t)=u(x,t)\vec{x}+v(y,t)\vec{y}$, I know that $\nabla \cdot \vec{U} = 0$, call this eqn (1), by the definition of incompressibility.
In order to find out more about the velocity components $u(x,t)$ and $v(y,t)$, i have been told to integrate eqn (1). Is the following integration w.r.t. space valid?:
\begin{align}
&\int \nabla \cdot \vec{U}  \mathbb{d} r = \int 0 \\
&\int \left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right) \mathbb{d}r = C = \text{const.}  \\
&\int \frac{\partial u}{\partial x}dx+\int \frac{\partial v}{\partial y}\mathbb{d}y = C \\
&u(x,t) + v(y,t) = C(t)
\end{align}
In other words, the magnitude of the sum of the fluid velocity in the $x$ and $y$ directions are constant everywhere in the fluid at any instant in time?
Is there a better way of obtaining more information?
 A: The usual integral for the divergence of the velocity field is over a volume.  Since $u$ does not depend on $y$ and $v$ does not depend on $x$, we have
$$
\begin{align}
\int_V \left(\nabla\cdot \vec{U}\right) \mathrm{d}V & = 
\iint \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{d} y \\
& = \iint \frac{\partial}{\partial x}u(x,t) \mathrm{d}x \mathrm{d}y
+ \iint \frac{\partial}{\partial y}v(y,t) \mathrm{d}x \mathrm{d}y \\
& = \int \left[u(x,t) + c_x\right] \mathrm{d}y + \int \left[v(y,t) + c_y\right] \mathrm{d}x \\
& = y\left[u(x,t) + c_x\right] + x\left[v(y,t) + c_y\right] + d\\
& = 0
\end{align}
$$
where the constants $c_x$, $c_y$, and $d$ will depend on your boundary conditions.  You should also get the same thing if you apply the boundary conditions by taking definite integrals.  Without know what the rest of the problem is, it is difficult to say more about this.
A: If you actually have $\vec{U}(x,y,t)=u(x,t)\vec{x}+v(y,t)\vec{y}$, this is a very special flow, in general you'll have $\vec{U}(x,y,t)=u(x,y,t)\vec{x}+v(x,y,t)\vec{y}$ for a 2D flow.
Thus in general you cannot integrate the continuity equation $\nabla\cdot \vec{U}=0$. 
If you're making the hypothesis above, then what you get is correct, and no, there's no more information just from this equation. However this probably makes a lot easier the resolution of the momentum equation.
