Green's Theorem confusion

Question

I have two equivalent forms of Green's theorem, namely $$ \int\int_D \frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}dxdy = \int_C pdx + qdy $$ $$ \int\int_D \frac{\partial p}{\partial x}+\frac{\partial q}{\partial y}dxdy = \int_C pdy - qdx $$ Moreover I know that the area of a simple closed curve is $A=\int_C xdy - ydx$. I tried to verify this result using the second version of Green's Theorem by plugging $$F=(x,y)$$ However, though if I used this idea to evaluate the area of a circle by firstly parametrize it as $r=(a\cos\theta,a\sin\theta)$ Then calculate $\int_{0}^{2\pi} F.dr$ will just gives $0$. I know I can use the first version to get the correct, but why it fails if I try to use the second one?

Answer 1

It looks like you should use the vector field $\vec F = \langle -y,x,0 \rangle$, not $\vec F = \langle x,y,0 \rangle$.

Answer 2

The area formula you are referring to is $$A:={\rm area}(D)={1\over2}\int_{\partial D}(x\>dy-y\>dx)\ .$$ Here the RHS can be written as $\int_{\partial D}(p\>dy-q\>dx)$ with $$p(x,y)={1\over2}x,\qquad q(x,y)={1\over2}y\ .$$ It follows that $${\partial p\over\partial x}+{\partial q\over\partial y}={1\over2}+{1\over2}=1\ ,$$ so that your second formula gives $${1\over2}\int_{\partial D}(x\>dy-y\>dx)=\int_D 1\>{\rm d}(x,y)\ ,$$ as desired.

(The function $F$ of three variables you have introduced has no place here.)