A conservation law in fluid dynamics Let's consider a water flow in two dimensional space, assume the velocity field is $\vec{v}(t,x,y)$, we can define: 
$$\rho=\hat{z}\cdot(\partial_x\vec{v}\times \partial_y\vec{v})$$
$$j^x=\hat{z}\cdot(\partial_y\vec{v}\times \partial_t\vec{v})$$
$$j^y=\hat{z}\cdot(\partial_t\vec{v}\times \partial_x\vec{v})$$
we can prove: 
$$\partial_t \rho+\partial_xj^x+\partial_yj^y=0$$
as long as the velocity field is smooth. My question is:
1) What's the physics corresponding to this conservation law?
2) Is there any important consequence due to this conservation law?
 A: Note that
$$\partial_t \rho = \hat z \cdot (\partial_t\partial_x \vec v \times \partial_y \vec v + \partial_x \vec v \times \partial_t\partial_y \vec v)$$
$$\partial_x j^x = \hat z \cdot (\partial_x \partial_y \vec v \times \partial _t \vec v + \partial_y \vec v \times \partial_x\partial_t \vec v)$$
$$\partial_y j^y = \hat z\cdot (\partial_y\partial_t \vec v \times \partial _x \vec v + \partial_t \vec v + \partial_y\partial_x \vec v)$$
Because you can switch the order of partial derivatives and because the cross product is antisymmetric, the first term on line 1 cancels the second term on line 2; the second term on line 1 cancels with the first term on line 3; and the first term on line 2 cancels with the second term on line 3.
This would be true for any smooth function $\vec v$ (whether it is vector-valued or not) so I cannot see how it would encode any meaningful physics.

As an aside, using the symbols $\rho$ and $j$ for this is probably a bad idea.  My initial thought was that you were trying to cast this into the form of a conservation law, but this is misleading.  Conservation laws have physical content - they are either taken to be true by assumption, or they arise from more fundamental considerations (e.g. declaring mass conservation to be a necessary component of fluid mechanics vs. deriving the continuity equation as the zeroth moment of the Boltzmann transport equation).  
In this case, however, your equation is actually an identity.  The fact that it is true does not rely on the Navier-Stokes equations or on any other physics, in the same way as the vector identity  $\vec v \times \vec v=0$.
