Prove that $2\text{Area}(\omega) \leq \text{length}(\gamma)\,\text{distance}(\gamma)$ So basically, I am given the following to prove:

Let $+\gamma$ be a positively oriented smooth Jordan arc, and let $\omega$ denote the interior of $+\gamma$. Recall that if $F = (F_1, F_2):D \to \mathbb{R}^2$ is a continuously differentiable vector field in an open set $D$ containing $\omega \cup(+\gamma)$, then
$$\iint_\omega \left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) dxdy = \oint_{+\gamma}F \cdot \overrightarrow{ds} $$
where the right hand-side is the line-integral of $F$ along the path $+\gamma$.
By suitably choosing $F$, prove that $$ \DeclareMathOperator{\Area}{Area} \DeclareMathOperator{\diameter}{diameter} \DeclareMathOperator{\length}{length} 2\Area(\omega) \leq \diameter(+\gamma) \length(+\gamma)$$
where
$\diameter(+\gamma) = \sup\{|z(s)-z(t)| : s,t \in [a,b]\}$ and $\length(+\gamma) = \int_{a}^{b} |z'(t)| dt$.

The only thing I know so far is that I need to find an $F$ such that  $\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} =1$ because then $$\iint_{\omega}  dxdy = \Area(\omega).$$ However, I do not know where to proceed from there!
 A: I assume that $z$ is $\gamma$ in the definition of $\operatorname{diameter}(+\gamma)$.  Without loss of generality, we can assume that $\gamma(a)=(0,0)$ (otherwise, we can translate $\gamma$ until the point $\gamma(a)$ hits the origin).
I would take $\vec{F}=(F_1,F_2)$ with $F_1=-y$ and $F_2=x$,
so that
$$\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=2.$$
By Green's theorem, we have
$$2\operatorname{Area}(\omega)=\iint_\omega \left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\ dx\ dy=\oint_{+\gamma}\vec{F}\cdot \vec{dr}.$$
Hence,
$$2\operatorname{Area}(\omega)\leq \oint_{+\gamma}\left\vert\vec{F}\cdot \vec{dr}\right\vert\leq \oint_{+\gamma}\left\Vert\vec{F}\right\Vert\ dr.$$
Because $\left\Vert\vec{F}\right\Vert\leq \operatorname{diameter}(+\gamma)$, we conclude that
$$2\operatorname{Area}(\omega)\leq \oint_{+\gamma}\operatorname{diameter}(+\gamma)\ dr=\operatorname{diameter}(+\gamma)\oint_{+\gamma}dr=\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma).$$

This inequality is a quite weak.  If we define $\operatorname{radius}(+\gamma)$ to be 
$$\inf\Big\{r>0:\exists p\in\mathbb{R}^2,\ \forall u\in[a,b],\ \big\vert\gamma(u)-p\big\vert< r\Big\}\,,$$
then we have
$$2\operatorname{Area}(\omega)\leq \operatorname{radius}(+\gamma)\operatorname{length}(+\gamma).$$
The equality holds iff $\gamma$ traces a circle (once).  You can use Jung's theorem to show that
$$2\operatorname{Area}(\omega)\leq \frac{1}{\sqrt{3}}\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma),$$ which is stronger than the required result, but is still weak.  So, it is an interesting question to find the infimum $\lambda_{\min}$ of all $\lambda>0$ such that 
$$\operatorname{Area}(\omega)\leq \lambda\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma).$$
We already know that $\lambda_{\min} \leq \frac1{2\sqrt{3}}$.
A: Choose $F(x,y) = (-y+y(a),x-x(a))$ and define a $1$-form $\alpha := (x-x(a))\mathrm{d}y+(y(a)-y)\mathrm{d}x = F_2(x,y)\mathrm{d}y+F_1(x,y)\mathrm{d}x$. Then
$$\mathrm{d}\alpha = 2\mathrm{d}x \wedge \mathrm{d}y$$ 
and so
$$\int_{\omega} \mathrm{d}\alpha= \int_{\omega}\left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) \mathrm{d}x\mathrm{d}y= 2\int_{\omega} \mathrm{d}x \wedge \mathrm{d}y = 2\mathrm{Area}(\omega). $$
By Stokes theorem (which reduces exactly to the curl theorem you are recalling in this case), you also get
\begin{align}
\int_{\omega} \mathrm{d}\alpha &= \int_{\partial \omega} \alpha = \int_{\gamma+} \alpha\\
&= \int_a^b (x(t)-x(a))y'(t)+(y(a)-y(t))x'(t)\,\mathrm{d}t = \int_a^b \langle F(x,y),(x'(t),y'(t))\rangle\,\mathrm{d}t,
\end{align}
where $\langle \cdot, \cdot \rangle$ denotes the dot product. By Cauchy-Schwarz inequality you get
\begin{align}
\int_a^b \langle F(x,y),(x'(t),y'(t))\rangle\,\mathrm{d}t & \leq \int_a^b \sqrt{(x(t)-x(a))^2+(y(t)-y(a))^2}\sqrt{x'(t)^2+y'(t)^2}\, \mathrm{d}t \\ & 
\leq \mathrm{diameter}(\omega)\int_a^b \lvert z'(t)\rvert \, \mathrm{d}t\\
&\leq \mathrm{diameter}(\omega)\mathrm{length}(\gamma+).
\end{align}
