Intuitive Definition of Curl and Stokes' Theorem 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.
 A: Your idea is correct: you have to exploit Stokes' formula for a decreasing sequence of surfaces $\Sigma_n$ tending to a point $p$. Here the surfaces of the sequence should be viewed as restrictions of an initially given surface $\Sigma$ containing $p$.
When the surfaces tend to $p$, the averaged value with respect to the area of the line integral of $\vec{a}$ around the boundary of the surfaces,
$$\frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}$$
tends to the $\vec{n}_p$ component of the curl of $\vec{a}$ at $p$.
Indeed, taking advantage of Stokes' theorem,  if $\Sigma_n$ is very concentrated around $p$,
$$\frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}
= \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \nabla \times \vec{a}(q) \cdot \vec{n}_q \:dS(q)
\sim
  \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \nabla \times \vec{a}(p) \cdot \vec{n}_p \:dS(q)$$
$$= \nabla \times \vec{a}(p) \cdot \vec{n}_p   \frac{1}{S(\Sigma_n)}\int_{\Sigma_n}1 \:dS(q) =  \nabla \times \vec{a}(p) \cdot \vec{n}_p
$$
I write below a formal statement with a proof.
Theorem.
Let $\Sigma$ be a $C^1$ surface in $\mathbb{R}^3$ whose boundary $\partial \Sigma$ is a regular  closed curve (without self-intersections). Consider a sequence of similar surfaces $\Sigma_n\subset \Sigma$, restrictions of $\Sigma$, such that $\Sigma_n \to p \in \Sigma$ for $n\to +\infty$
(in other words, $\Sigma=  \vec{x}(D_{r}(p_0))\subset \mathbb{R}^3$ for some $C^1$  parametrization from the disk $D_{r}(p_0) \subset \mathbb{R}^2$ of radius $r>0$ such that $\vec{x}(p_0) =p$ and  $\Sigma_n=  \vec{x}(D_{r_n}(p_0))$ with $r_n\to 0$ as $n\to +\infty$. Obviously I am also assuming that $\partial \Sigma_n = \vec{x}(\partial B_{r_n}(p_0))$.)
If $\vec{a}: O \to \mathbb{R}^3$, where $O\supset \Sigma$
is an open set, is $C^1$ then
$$\vec{n}_p \cdot \nabla \times\vec{a}(p) = \lim_{n\to +\infty} \frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}$$
where $\vec{n}_p$ is the unit vector normal to $\Sigma$ at $p$ and the orientation of $\partial \Sigma_n$ is chosen positively with respect to $\vec{n}$, finally
$$S(\Sigma_n) = \int_{\Sigma_n} 1 dS$$
is the area of $\Sigma_n$.
PROOF. From Stokes' theorem,
$$ \frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}=
\frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \nabla \times \vec{a} \cdot \vec{n} \:dS =
\nabla \times \vec{a}(p) \cdot \vec{n}_p + \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \left(\nabla \times \vec{a} \cdot \vec{n} -  \nabla \times \vec{a}(p) \cdot \vec{n}_p\right)\:dS \:.$$
Now observe that
$$\left|  \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \left(\nabla \times \vec{a} \cdot \vec{n} -  \nabla \times \vec{a}(p) \cdot \vec{n}_p\right)\:dS \right|
\leq \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \left|\nabla \times \vec{a} \cdot \vec{n} -  \nabla \times \vec{a}(p) \cdot \vec{n}_p\right|\:dS $$ $$\leq  \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} 
\sup_{q \in \Sigma_n}\left|\nabla \times \vec{a}(q) \cdot \vec{n}_q -  \nabla \times \vec{a}(p) \cdot \vec{n}_p\right|\:dS
= \sup_{q \in \Sigma_n}\left|\nabla \times \vec{a}(q) \cdot \vec{n}_q -  \nabla \times \vec{a}(p) \cdot \vec{n}_p\right|\:.
$$
Since $\Sigma \ni q \mapsto \nabla \times \vec{a}(q) \cdot \vec{n}_q -  \nabla \times \vec{a}(p) \cdot \vec{n}_p$ is continuous (from the hypotheses) and $\Sigma_n \to p$, then
$$\sup_{q \in \Sigma_n}\left|\nabla \times \vec{a}(q) \cdot \vec{n}_q -  \nabla \times \vec{a}(p) \cdot \vec{n}_p\right| \to 0$$
as $n\to +\infty$ and the thesis follows. QED
