I am asking this question on SE Physics because it's about an intuitive derivation, not a rigorously made proof of a theorem.
Reading Mathematical Methods for Physics and Engineering by Riley, et al. (actually the book does not matter) I found the following intuitive definition of the curl:
The notation might be a bit confusing - here $\vec{a}$ represents a vector field, $d\vec{S}$ is a surface differential with orientation.
Yes, I understand this. I can also do an intuitive proof on my own, reaching the conclusion with the following expression:
$$dxdydz \ (\nabla \times \vec{a}) = d\vec{S} \times \vec{a}$$
which is pretty much the same as the statement.
But another problem rises - the author states another intuitive definition of the curl:
I tried to derive this by applying the dot product with $\vec{dA}=\hat{n} \ dA$ to the above expression, where $\hat{n}$ is the normal vector to a specific point of the surface in three dimension.
It doesn't work. The problem is that I don't know how to deal with an arbitrary part of the surface, which obviously has an arbitrary orientation.
How do I show this? I noticed that it is, if integrated, equivalent to the Stokes' theorem in three dimension:
$$\int_{\partial D} \vec{a} \cdot d\vec{r} = \iint_D (\nabla \times \vec{a}) \cdot d\vec{S}$$
So the derivation of this formula should have greater importance than the above one.