Greens theorem on the circle $x^2 + y^2 = 16$.

Question

Use Greens theorem to calculate the area enclosed by the circle $x^2 + y^2 = 16$.

I'm confused on which part is $P$ and which part is $Q$ to use in the following equation

$$\iint\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right){\rm d}A$$

Answer 1

Hint: You want

$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$$

so the integral is

$$\iint_{x^{2}+y^{2}\leq 16}{\rm d}A$$

Can you find $P$ and $Q$ that satisfy this? Notice that there is more than one choice.

Answer 2

It suffices to take $Q =0$ and $P =-y$ then $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$ and by Green formula we get, $$ \iint_{x^2+y^2\le4^2} 1dxdy =\iint_{x^2+y^2\le4^2}\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dxdy= -\oint_{x^2+y^2=4^2} ydx =\\- 16\int_0^{2\pi} \sin td(\cos t)= 16\int_0^{2\pi} \sin^2 t dt= 16\int_0^{2\pi} \frac{1-\cos (2t)}{2}dt= \color{red}{ 16\pi}$$ Where we used the parametrization, $$x= 4\cos t~~~and ~~y =4\sin t$$