# Green's Theorem confusion

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?

It looks like you should use the vector field $$\vec F = \langle -y,x,0 \rangle$$, not $$\vec F = \langle x,y,0 \rangle$$.
• You stated you wanted to use the second form of Green's theorem. If $$\int_C x dy - y dx = \int_C P dx + Q dy$$ you need $P = -y$ and $Q = x$. – Umberto P. Apr 18 at 12:55
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.)
• If I use your F=(x/2,y/2) then the area of a circle calculated after parametrisation will be 0 when integrating $F.dr$ unless I made a mistake in calculating that integral. – JustWandering Apr 18 at 12:55