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The curl of $\mathbf{F}$ at some point $(x_0,y_0)$ is defined as $$\lim_{A\to0}\frac{1}{|A|}\oint_\gamma\mathbf{F}\cdot ds$$ I was curious to see what happens when we expand this, in the simple two dimensional case, and where I simply let $\gamma$ enclose a circle, $A$, so rewriting: $$\lim_{r\to0}\frac{1}{\pi r^2}\int_0^{2\pi}\mathbf{F}(\gamma(t))\cdot\gamma'(t)\,dt$$ $$\gamma(t)=\begin{pmatrix}x_0+r\cos t\\y_0+r\sin t\end{pmatrix},\mathbf{F}=\begin{pmatrix}\mathbf{F}_1\\\mathbf{F}_2\end{pmatrix}$$ And I am stuck after the following steps, which may or may not have a confused misapplication of integration by parts! I am not sure how this works in multivariable calculus. Note also that any appearance of $x,y$ in the following refers to the $x$ and $y$ coords returned by $\gamma$, which $\mathbf{F}$ takes in: $$\begin{align*}&\lim_{r\to0}\frac{1}{\pi r^2}\left(\int_0^{2\pi}\mathbf{F}_2(\gamma(t))\cdot r\cos t\,dt-\int_0^{2\pi}\mathbf{F}_1(\gamma(t))\cdot r\sin t\,dt\right) \\&=\lim_{r\to0}\frac{1}{\pi r}\left(\int_0^{2\pi}\mathbf{F}_2(\gamma(t))\cdot\cos t\,dt-\int_0^{2\pi}\mathbf{F}_1(\gamma(t))\cdot\sin t\,dt\right) \\&=\lim_{r\to0}-\frac{1}{\pi r}\left(\int_0^{2\pi}\sin t\left(-r\sin t\frac{\partial\mathbf{F}_2}{\partial x}\frac{dx}{dt}+ r\cos t\frac{\partial\mathbf{F}_2}{\partial y}\frac{dy}{dt}\right)\,dt+\\\int_0^{2\pi}\cos t\left(-r\sin t\frac{\partial\mathbf{F}_1}{\partial x}\frac{dx}{dt}+r\cos t\frac{\partial\mathbf{F}_1}{\partial y}\frac{dy}{dt}\right)\,dt\right) \\&=\lim_{r\to0}-\frac{1}{\pi}\left(\int_0^{2\pi}\sin t\left(r\sin^2 t\frac{\partial\mathbf{F}_2}{\partial x}+r\cos^2 t\frac{\partial\mathbf{F}_2}{\partial y}\right)\,dt+ \\\int_0^{2\pi}\cos t\left(r\sin^2 t\frac{\partial\mathbf{F}_1}{\partial x}+r\cos^2 t\frac{\partial\mathbf{F}_1}{\partial y}\right)\,dt\right)\end{align*}$$ And it is at this point that I have no idea how to proceed, since continuing to integrate by parts would yield nothing, I think. And in a limit letting $r\to0$ would zero out the entire expression, so I am sure I have made a mistake somewhere. Going from the second line to the third, I noticed that in an integration by parts, the term outside the integral would have either $\sin,\cos$ which would zero out at the evaluation points $0,2\pi$. I also presumed that the one-dimensional integration by parts process of integrating one, differentiating the other, would correlate to me needing to find the directional derivative of $\mathbf{F}_1,\mathbf{F}_2$ w.r.t $t$, which also would need a chain rule expansion. I thought this would require $\nabla_\mathbf{F}\cdot\gamma'(t)$. This may also be the source of my mistake, but I haven't a clue.

FShrike
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    Nitpick: Either it's $\oint_\gamma\mathbf{F}\cdot,ds$, or it's $\int_0^{2\pi}\mathbf{F}(\gamma(t))\cdot\gamma'(t),dt$. Never $\oint_\gamma\mathbf{F}(\gamma(t))\cdot\gamma'(t),dt$. – Arthur Jul 01 '21 at 11:16
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    If you're trying to derive the formula for curl in cartesian coordinates, then using circles is probably not a good idea; using squares centered at the point $(x_0,y_0)$ is better. I've actually written up an answer a while back for formula of curl in 3D, hopefully reading that will allow you to specialize to your case. (Also, what you have in your first equal sign is not the equal to the curl; it is only one of the components of the curl.) – peek-a-boo Jul 01 '21 at 11:31
  • @peek-a-boo how else? Would it be easier and still accurate to consider $A$ as a square around $(x_0,y_0)$? – FShrike Jul 01 '21 at 11:35
  • @peek-a-boo plus, your answer is highly formal and I need a little assistance: what is the meaning of the braces <$F,\tau$>, <$H,n$>? Is this the vector notation <$x,y$> or something else? – FShrike Jul 01 '21 at 11:37
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    yup, squares are good. Then $\gamma$ is a union of the 4 straight lines, so you'll have 4 integrals, which you can group together into two pairs of 2 terms, involving differences of $F_1$ and $F_2$ at various points. Then, you apply the integral and differential mean-value theorem. THe braces $\langle \rangle$ refer to the dot product $\cdot$ (so $\int\langle F,\tau\rangle,dl$ is the same as your $\int\mathbf{F}\cdot d\mathbf{s}$ etc; I didn't use boldface in that answer for the vector fields because it was too much typing for me). – peek-a-boo Jul 01 '21 at 11:38
  • @peek-a-boo thank you – FShrike Jul 01 '21 at 11:45
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    oh and having thought about it somemore, you should ignore my paranthetical remark in my first comment. In two-dimensions only, it makes sense to think of the curl as a scalar as in your first expression. – peek-a-boo Jul 01 '21 at 12:38

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