Green's Theorem for Finding Area of a Region by Using Perimeter Values

Question

My textbook says that Green’s theorem can be used to find the area of a region by using only values on its perimeter:

$$\text{Area} = \iint_D \,dA = \oint x \,dy = - \oint y \,dx = \frac{1}{2} \oint x \,dy - y\,dx$$

I understand that Area $= \iint_D dA$, but where does the rest come from?

How do we get $ \iint_D dA = \oint x \, dy$?

And $\oint x \, dy = - \oint y \, dx$?

And $- \oint y \, dx = \dfrac{1}{2} \oint x \,dy - y\,dx$?

I would greatly appreciate it if people could please take the time to explain this.

Answer 1

Green's theorem says for sufficiently nice functions $M(x,y)$ and $N(x,y)$ and a sufficiently nice region $D$ that $$ \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial N}{\partial y}\right) \, dA = \oint_{\partial D} N\,dx + M\,dy.$$ So, we can apply this to $M=x$ and $N=0$ to get $$ \iint_D dA = \oint_{\partial D} x\,dy.$$

The other formulae can be gotten by applying it to $M=0$ and $N=-y$ and $M=x/2$ and $N=-y/2.$

Answer 2

The symbol $\displaystyle \oint$ means you're traversing the curve in a counterclockwise direction. Say you're going around something that looks approximately circular. Draw a vertical line through it, i.e. a line parallel to the $y$-axis, and that line hits the curve twice. As you move along the "top" part of the curve going counterclockwise, $x$ is decreasing, so $dx$ is negative. At any fixed $x$ value, the change in $x$ is $dx$ and the distance from the $x$-axis is $y$, and $dx$ is negative, so $-\int y\,dx$ is how much area is added by doing the infinitely small step of length $dx.$ Then as you move along the bottom part of the curve, $dx$ is positive, so $y\,dx$ is the area between that curve and the $x$-axis, but this time you want to subtract that area, so you add $-y\,dx.$ Thus you have $$ \text{area} = -\oint y\,dx. $$ By reversing the roles of horizontal and vertical you see how you get $\displaystyle \oint x\,dy.$