Making the rigorous link between the conceptual interpretation of curl and the formula EDIT: I would like to clarify for potential new readers what the gist of this question is, and how it differs from other questions about curl on this site. Namely, I'm confused about:


*

*The specifics of the physical interpretation of curl (is it infinitesimal torque, average infinitesimal torque, average infinitesimal angular acceleration, etc.)

*How this specific physical interpretation translates into a formula (ideally, the integral below as seen on Wikipedia)

*How this formula becomes the usual definition of curl.


I'm not confused about what curl is on a general level, I'm confused about why the formula for curl coincides with the physical interpretation, on an intuitive and rigorous basis.

I'm having a bit of difficulty making the connection between the intuitive idea of the curl of a vector field (as the torque experienced on an infinitesimally small paddle wheel placed at a point in the vector field), and the formula for it (I'll stick to 2-D first off, which is how the MIT OpenCourseware lectures define it at first):
$$
\nabla\times F = N_x-M_y
$$
where $F(x,y)=(M(x,y),N(x,y))$. (Yes this is a scalar whereas curl is normally a vector, but I interpret this definition of curl to be the signed magnitude of the torque, since the direction is perpendicular to the $xy$ plane for all 2-D fields).
I can accept that this formula aligns with the idea of "infinitesimal torque", but I'd like to see it in action in a rigorous setting. So, I asked myself, what would the torque be at a point in a vector field?
If we were to place a non-infinitesimal paddle wheel (a circle of radius $r$) centered at $(x_1, y_1)$, it would be experiencing forces at each point in its area. If $\vec{r}$ is the position vector from the center of the wheel to where the force is exerted, then the torque is $\vec{r}\cdot\vec{F}$. The net torque is just the sum of all the torques, so the net torque experienced by the paddle wheel should be:
$$
\tau=\mathop{\iint}_{(x-x_1)^2+(y-y_1)^2\le r^2}(x-x_1,y-y_1)\cdot F(x,y)\ \text{d}A
$$
If we want to find the infinitesimal torque, we would supposedly take the limit as $r\to 0$. However, this doesn't make sense, as this would imply that the infinitesimal torque is zero (since the integral goes to zero). What's missing?
I assume we needed some sort of limiting factor to divide by in the limit. I know from physics that $\tau = \alpha I$ where $I$ is the moment of inertia (for our 2-D paddle wheel with unit density, $I=\frac\pi2r^4$). So maybe curl is actually the infinitesimal angular acceleration? If so, then we have
$$
\nabla\times F = \lim_{r\to0}\frac{2}{\pi r^4}\mathop{\iint}_{(x-x_1)^2+(y-y_1)^2\le r^2}(x-x_1,y-y_1)\cdot F(x,y)\ \text{d}A
$$
This looks similar but not quite what the formula on Wikipedia is:
$$
(\nabla\times F)\cdot\mathbf{\vec{n}} = \|\nabla\times F\| = \lim{A\to 0}\frac1{|A|}\oint_C\mathbf{F}\cdot d\mathbf{r}
$$
Why, here, is the area $|A|$ the limiting factor? That doesn't make much sense to me. Is curl then the average infinitesimal torque, averaged over an infinitesimal area (since $|A|\to 0$)? Wikipedia also says that the curl is the "circulation density", rather than simply the torque. So is my initial interpretation of the curl wrong?
Furthermore, what change would I have to make to my interpretation to go from my integral to the integral on wikipedia (since the integrands are different and I'm not sure what to do about that), and then how do you go from the integral to the initial formula for curl?
Any and all help would be appreciated :)
 A: $
\newcommand\dd{\mathrm d}
\newcommand\R{\mathbb R}
\newcommand\diff{\mathbf D}$Let $L_x(v)$ be a field of functions linear in $v$. I will define the derivative $L_x(\nabla)$ of $L$ by
$$
  L_x(\nabla) = \sum_{i=1}^n\frac{\partial L_x(e_i)}{\partial x_i}
$$
where $\{e_i\}_{i=1}^n$ is the standard basis and $x = \sum_ix_ie_i$ is expressed in Cartesian coordinates. We can generalize this formula to any coordinate basis and show that it is independent of the coordinates chosen, and even do a coordinate/basis-free treatment, but in the interest of not drowning in detail I will not go into that here.
We begin by showing that
$$
  L_x(\nabla) = \lim_{|R|\to0}\frac1{|R|}\int_{\partial R}L_y(\hat n(y))\,\dd S(y)
\tag{IntDeriv}
$$
where $R$ is a region of space contaning $x$ with non-zero volume $|R|$, $\partial R$ is its boundary, $\dd S(y)$ is the corresponding surface area measure with variable of integration $y$, and $\hat n(y)$ is the outward-pointing unit normal of $\partial R$ at $y$. I will not rigorously define this limit. However, the specific case of the divergence ($L_x(v) = v\cdot f(x)$ for $f : \R^n \to \R^n$) is more well known,
$$
  \nabla\cdot f(x) = \lim_{|R|\to0}\frac1{|R|}\int_{\partial R}\hat n(y)\cdot f(y)\,\dd S(y),
\tag{$*$}
$$
and this proves the general case as follows. If $L_x$ is a linear form $\R^n \to \R$, then there is some vector $l_x$ such that $L_x(v) = v\cdot l_x$ and $L_x(\nabla) = \nabla\cdot l_x$; applying ($*$) proves (IntDeriv) in this case. If $L_x : \R^n \to \R^m$, then for any $w \in \R^m$ the function $v \mapsto w\cdot L_x(v)$ is a linear form and
$$
  w\cdot L_x(\nabla) = \lim_{|R|\to0}\frac1{|R|}\int_{\partial R}w\cdot L_y(\hat n(y))\,\dd S(y).
$$
Set $w = e_i$, multiply each side by $e_i$, and sum over $i$; this proves (IntDeriv).

To interpret (IntDeriv) geometrically, consider $R$ as an infinitesimal ball of radius $\epsilon$. The surface area $|\partial R|$ and volume of $R$ in $\R^n$ are related by $|\partial R| = n|R|/\epsilon$ so may write
$$
  L_x(\nabla) \approx \frac n\epsilon\frac1{|\partial R|}\int_{\partial R}L_y(\hat n(y))\,\dd S(y).
$$
$y = x + \epsilon\hat n(y)$, so if $\diff L_x(v) : \R^n \to \R^m$ is the total differential of $x \mapsto L_x(v)$ then
$$
  L_y(\hat n) = L_{x+\epsilon\hat n}(\hat n) \approx L_x(\hat n) + \epsilon\diff L_x(\hat n)(\hat n)
$$
where we've suppressed the $y$-dependence of $\hat n$ for notational clarity. Since $L_x(-\hat n) = -L_x(\hat n)$, the first term cancels upon integration yielding
$$
    L_x(\nabla) \approx \frac n{|\partial R|}\int_{\partial R}\diff L_x(\hat n)(\hat n)\,\dd S.
$$
$\dd S = \epsilon^{n-1}\dd\Omega$ where $\dd\Omega$ is the solid-angle measure, and $|\partial R| = \epsilon^{n-1}S_{n-1}$ where $|S_{n-1}|$ is the area of a unit $(n-1)$-sphere $S_{n-1}$ (i.e. surface area of a unit $n$-ball), so we may finally write
$$
      L_x(\nabla) \approx \frac n{|S_{n-1}|}\int_{S_{n-1}}\diff L_x(y)(y)\,\dd\Omega(y).
\tag{IntDeriv'}
$$
(As a small aside, it turns out that the RHS is the integral representation of the metric contraction of $\diff L_x$.) Careful consideration will show that we may switch $\approx$ to true equality here.
$\diff L_x$ captures all the variation of $L$ at $x$, and in particular $\diff L_x(y)(y)$ is the variation of $L_x(y)$ as $x$ varies in the $y$ direciton. We are lead to the following interpretation of $L_x(\nabla)$:

*

*$L_x(\nabla)$ is the average variation in all directions of $L$ at the point $x$.

Let us apply this to the curl $L_x(\nabla) = \nabla\times f(x)$. Let $h$ be a unit vector; we can see that
$$
  \diff L_x(h)(h) \approx h\times\Delta f/\epsilon,\quad
  \Delta f = f(x + \epsilon h) - f(x).
$$
(This is not the same $\epsilon$ as before, just an arbitrary infinitesimal parameter). The cross product is best interpreted as giving the normal to a plane
together with an associated magnitude.
This magnitude
$$
  |h\times\Delta f/\epsilon| = \frac{|h||\Delta f|}\epsilon\sin\theta
$$
with $\theta$ the angle between $h$ and $\Delta f$ measures how separated they are;
in other words, how much the change of $f$ deviates, or curls away, from the direction of change.
All together, we have the following:
as $x$ varies to $x + \epsilon h$,
$f$ curls away from this direction of change $h$ in the plane (normal to) $h\times\Delta f/\epsilon$.
By (IntDeriv'), the curl $\nabla\times f(x)$ is the (normal to) the average such plane over all directions $h$
weighted by how harshly $f$ curls away from $h$.
A: In general,  you're on your own.   But here is a neat application.  It can be expressed in English as:  "there are no magnetic monopoles"...
For such an object would have nonzero divergence. ..  But a straightforward computation shows divergence of the curl is always zero (note:  magnetic fields are expressed as a certain curl).  If memory serves,  we have :  $$\nabla \cdot \nabla × (x,y,z) =O $$. 
