TITLE: Derivation of the Curl formula in cartesian coordinates.
QUESTION [5 upvotes]: By calculating the circulation per area of a vector field 
$$F(x,y,z) = F_x(x,y,z)\vec{x} + F_y(x,y,z)\vec{y} + F_z(x,y,z)\vec{z}$$
in a small rectangle around $(x_0, y_0, z_0)$ on the $xy$ plane, it can be shown the limit as the sides of the rectangle approach zero is
$$\left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y}\right)$$ 
The same calculation however is not that straightforward if the rectangle does not lie in the $xy$, $yz$, or $xz$ planes. Now if $\vec{n}$ is the normal of the plane, I thought that by performing a change of basis such that $\vec{n} \rightarrow \vec{z'} $ and by following the previous calculations we could show that the limit of the circulation per area is 
$$ \left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x_0, y_0, z_0)}{\partial y'}\right) $$
This is also the inner product of the curl of the vector field and the normal $\vec{n}$
As such the two should be equal:
$$\left(\frac{\partial F_{y'}(x'_0, y'_0, z'_0)}{\partial x'} - \frac{\partial F_{x'}(x'_0, y'_0, z'_0)}{\partial y'}\right) = \\
\left[\left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial y} - \frac{\partial F_y(x_0, y_0, z_0)}{\partial z} \right)\vec{x} +
\left(\frac{\partial F_z(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial z} \right)\vec{y} + 
\left(\frac{\partial F_y(x_0, y_0, z_0)}{\partial x} - \frac{\partial F_x(x_0, y_0, z_0)}{\partial y} \right)\vec{z}\right] \cdot \vec{n} $$
I've been trying to prove the above equality for some time without success, specifically I am not sure how to handle the transformations correctly. Any help with this is much appreciated!

REPLY [6 votes]: A simpler approach is via integral theorems.
As stated in the question, the special cases for a rectangle in the $xy$ , $yz$ , $zx$ planes are well understood.
According to Green's theorem :
$$
\begin{cases}
\iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy
= \oint_{xy} \left( F_x\, dx + F_y\, dy \right) \\
\iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz
= \oint_{xy} \left( F_y\, dy + F_z\, dz \right) \\
\iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx
= \oint_{xy} \left( F_z\, dz + F_x\, dx \right) \end{cases}
$$
But instead of rectangles, we take half rectangles, or better: the triangles $OAB$ , $OBC$ , $OAC$ respectively:

Thanks to Green's theorem we can replace area integrals by line-integrals; mind that they are counter-clockwise.
Then it is clear that, irrespective of any further content:
$$
\oint_{OAB} + \oint_{OBC} + \oint_{OAC} + \oint_{ABC} = 0
$$
Assuming that the operator rot(ation) is not defined yet in general, this means that we now have an expression for it:
$$
2 \iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA = \\
- \iint_{xy} \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) dx\, dy
- \iint_{yz} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) dy\, dz
- \iint_{zx} \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) dz\, dx
$$
Continuing with infinitesimal volumes / areas and flipping normals on the right hand side, so that they become
the components of the normal at the left hand side:
$$
\vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, \Delta A = \\
\left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right)\cdot n_x\, \Delta A +
\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\cdot n_y\, \Delta A +
\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\cdot n_z\, \Delta A 
$$
Leaving out the infinitesimal area $\,\Delta A\,$ gives us the same answer as found by the OP themselves.
A somewhat neater approach is to calculate mean values and let the area of the (red) triangle go to zero:
$$
\vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n} = \lim_{ABC \to 0}
\frac{\iint_{ABC} \vec{\operatorname{rot}}(\vec{F}) \cdot \vec{n}\, dA}{\iint_{ABC} dA}
$$
Note. I've encountered essentially the same method at several places elsewhere in physics
(I think it's with stress and strain). Aanyway, a related subject is :
What does shear mean?