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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?).

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You shouldn't write of "infinite solutions" if you mean "infinitely many solutions". "Infinite solutions" means "solutions, each one of which, by itself, is infinite". –  Michael Hardy Dec 28 '12 at 17:20

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That sounds right. But your last equation is not false, because $\int_C xdy+ydx$ is always zero, so you can add or subtract any multiple of it to the integral without changing its value.

This is again because $\int_C xdy$ and $\int_C (-y)dx$ are both expressions for the same area of $D$, so their difference must vanish.

The "canonical" $\frac12 \int_C xdy-ydx$ can be understood as the limit of successive polygonal approximations to $D$, computed by the shoelace formula. $\int_C xdy$ and $\int_C (-y)dx$ as the area between the curve and one of the coordinate axes, with sign depending on whether the curve is moving up or down (so the parts of the areas outside the closed curve cancel out).

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