Apparently, the curl of a vector field is a function that outputs the "rotationality" of the vector field at some point, as a function of that point's coordinates. I want to go from this qualitative understanding to a formula, though in a way that is understandable with my basic mathematical knowledge. This video from 3b1b gives some insights, and interpreting them literally, I get the formula below. This is my literal interpretation:
To get the curl of a vector field $\mathbf{F}$at the point $(x,y)$, take some small step vector $\mathbf{s}$ away from $(x,y)$, thus arriving at some new point $(c_1,c_2)$. Look now at the vector at $(c_1,c_2)$; that is, the vector $\mathbf{F}(c_1,c_2)$. Take the difference between this vector and the one situated at the point of interest, $(x,y)$: this difference is $\mathbf{F}(c_1,c_2) - \mathbf{F}(x,y) = \mathbf{d}$. Now, do this many times, with every possible step vector extending radially from $(x,y)$, making the points $(c_i,c_j)$ the different points on the circumference of some circle with radius $r$.
Now, take the cross product of the step vector and the difference vector, getting some new vector. Then, take the average of all of these cross products. This average will be close to equalling the curl at the point $(x,y)$ if $r$ is small. To make it equal to the curl, you must let $r$ approach $0$.
I assume a literal interpretation is false, because 3b1b was aiming at inuitiveness and not rigor, and since a cross product would give a vector perpendicular to the plane of the step and difference vectors. AFAIK, the curl gives vectors in the plane, not perpendicular to it. Anyways, trying to capture my interpretation above in a formula, I have made this:
$$ \begin{align} \nabla \times \mathbf{F}(x,y) = &\lim_{r \rightarrow 0} \biggr( \frac{1}{2r} \int_{-r}^r [s, \sqrt{r^2 - s^2}]\times \mathbf{d} \ ds \biggr) + \\ &\lim_{r \rightarrow 0} \biggr( \frac{1}{2r} \int_{-r}^r [s, -\sqrt{r^2 - s^2}]\times \mathbf{d} \ ds\biggr) \end{align} $$
The limit function takes the radius (the magnitude of the step vectors) to zero. The quantity $s$ is the integrand variable, representing the step along the $x$-direction, thus locking the step in the $y$ direction to $\pm \sqrt{r^2 - s^2}$. Due to the $\pm$, I split the equation up for the sake of notational clarity. The $[s, \pm \sqrt{r^2 - s^2}]$ is the step vector. The fraction is there to make the integral expressions into averages. The $\mathbf{d}$ is the difference vector (defined in the section above), which I wonder if I could turn to $\mathbf{F}'(x,y)$. If so, it would make the expression a bit simpler.
Since the cross product is distributive over addition, I think my equation must be wrong, since:
$$\begin{align} \text{integrand} &= [s, +\sqrt{r^2 - s^2}]\times \mathbf{d} + [s, -\sqrt{r^2 - s^2}]\times \mathbf{d} \\ &= ([s, +\sqrt{r^2 - s^2}] + [s, -\sqrt{r^2 - s^2}])\times \mathbf{d} \\ &= [2s, 0]\times \mathbf{d} \end{align}$$
My question: How do I make my formula correct?
The question that motivates my above question is merely "how to connect the equation for the curl with the idea the curl represents". This question has already been answered here. I am afraid the answer is a bit too advanced for me (but I will try), so I therefore ask this question as a way to perhaps answer that deeper question. This question is separate from the linked-to question because it asks about creating a formula for the curl from a certain, specific interpretation; that of taking the limit of the cross-products of the step and difference vectors, as the radius approaches zero. The correct way to find a formula from this interprettion is what my question requests (assuming any such way exists).
