Is there a relationship between curl and area? The cross product of two vectors is a new vector which lies on a new direction perpendicular to the plane of the multiplicand vectors. Its magnitude is the area of the parallelogram formed between these multiplicand vectors. 
On the other hand, the definitions of curl operation are all related with rotations. Since this operation also involves a cross product $(\nabla\times\vec A)$, I wonder if there is a relationship between curl operation and some kind of (may be infinitesimal) area? 
 A: You are correct, the area of parallelogram spanned by vectors is the magnitude of the cross product $$A = || a|| \, ||b|| \sin \theta = |\vec{a} \times \vec{b}|$$  As a vector $ \vec{a}\times \vec{b}$ points in a direction normal to the plane of $\vec{a}$ and $\vec{b}$. Does it then make sense that $\nabla \times \vec{a}$ is the area of the "parallelogram" spanned by $\nabla = (\partial_x, \partial_y, \partial_z)$ and $\vec{a}$?
Curl can be defined as the limit of a line integral over an infinitesimally small circle:
$$ (\nabla \times \vec{F})\cdot \hat{n} = \lim_{|A| \to 0} \frac{1}{|A|} \oint_C \vec{F} \cdot d\vec{r} $$
The macroscale version of is the Kelvin-Stokes theorem that relates the curl across a circle to the total vector field around a loop:
$$ \int_S (\nabla \times \vec{F}) \cdot dS = \oint_{\partial S} \vec{F} \cdot d \vec{r}$$
Another way to think of curl is to use exterior algebra.  In which case, $\nabla \times F$ is technically a 2-vector:
\begin{eqnarray} d(\ast F) &=& d(F_x dx + F_y dy + F_z dz) \\
&=& (\partial_x F_y - \partial_y F_x) dx \wedge dy 
+ (\partial_y F_z - \partial_z F_z) dy \wedge dz
+ (\partial_z F_x - \partial_x F_y) dz \wedge dx\end{eqnarray}
This 2-form can represent a kind of area.  


Philosophically it could be that same data that defines an area (a plane), also defines defines the axis of rotation normal to it in 3D space.
