differentiation - What is the physical significance of curl $nablatimesboldsymbol{V}$?

What is the physical significance of curl $$\nabla\times\boldsymbol{V}~?$$ I mean I read 'curl V represents the rotation of the vector $V$. My question what is it about the term $\nabla\times\boldsymbol{V}$ that it represents the rotation of the vector?

Answer

Curl can be equated with the closed line integral in the limit that the encircled area $\Delta S$ goes to zero. However, we would have to do this in three components because curl is a vector. $$(\nabla \times \vec{v})_x = \lim_{\Delta S \rightarrow 0} \frac{1}{\Delta S} \oint \vec{v}\cdot d\vec{l}$$ in the $yz$ plane and so on.

But what does it mean? Well it is easy to show that $$\nabla \times \vec{v} = 2 \vec{ \omega}$$ As follows: $$(\nabla \times \vec{v})_x = \partial_y v_z - \partial_z v_y = \partial_y (\vec{\omega} \times \vec{r})_z - \partial_z (\vec{\omega} \times \vec{r})_y$$ $$(\nabla \times \vec{v})_x = \partial_y (\omega_x y - \omega_y x) - \partial_z ( \omega_z x - \omega_x z) = 2 \omega_x$$

and ditto for the other components $$(\nabla \times \vec{v})_y = \partial_z v_x - \partial_x v_z = 2\omega_y$$ $$(\nabla \times \vec{v})_z = \partial_x v_y- \partial_y v_x = 2\omega_z$$ i.e. the curl of a velocity field equals twice the angular velocity at that point. In other words it is angular velocity within a fluid flow that creates curl! You can imagine constructing a ``curl meter'' out of a little (infinitesimally small) paddle wheel which could be inserted into the fluid flow. If the paddle wheel turns then there is curl.