Deriving 2-D curl from the formal definition, the hard way

Question

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.