The area of $D$ that is bounded by the closed curve $C$ given parametrically by $x=x(t)$ and $y=y(t)$ is $\frac{1}{2}\int_Cx\,dy-y\,dx.\tag{1}$
It seems obvious (this naivete may may be incorrect) that (1) pops out of the equation for the area of $D$, $\int\int_D1dxdy$, when applying Green's theorem to it:
$\int\int_D1\,dx\,dy=\int\int_D\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)\,dx\,dy=\int_CF_y\,dy+F_x\,dx$
$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=1.\tag{2}$
But isn't $F_y=\frac{1}{2}x$ and $F_x=-\frac{1}{2}y$ just one of the infinite solutions to equation (2), giving a different equation (1)?
This would imply, for example, that $\frac{1}{2}\int_Cx\,dy-y\,dx=\int_C 2x\,dy+y\,dx$ ($F_y=2x$ and $F_x=y$), which looks false (is it?).