Area using Green Theorem

Question

Please can anyone tell me how to prove that the area of a region enclosed by a simple closed curve $C$ is

$$\frac 1 2 \int \limits _C x \Bbb dy - y \Bbb dx$$

using Green theorem? Thanks in advance.

Answer 1

Well, Green's theorem says that if $C = \partial D$ with the positive orientation, then

$$\int \limits _C P(x,y) \ \Bbb d x + Q(x,y) \ \Bbb d y = \iint \limits _D \left( \frac {\partial Q} {\partial x} - \frac {\partial P} {\partial y} \right) \Bbb d x \Bbb d y .$$

Choosing $P = - \dfrac y 2$ and $Q = \dfrac x 2$ in Green's theorem gives

$$\int \limits _C - \frac y 2 \Bbb d x + \frac x 2 \Bbb d y = \iint \limits _D \left( \frac 1 2 - \Big( -\frac 1 2 \Big) \right) \Bbb d x \Bbb d y = \iint \limits _D 1 \ \Bbb d x \Bbb d y$$

which is precisely the area of $D$.