I'm having a hard time understanding where exactly the formula for computing area using Green's theorem comes from. It is typically:
\begin{equation}
\int_C x\,dy = \int_C -y\,dx = Area
\end{equation}
Or the equation with:
\begin{equation}
\frac{1}{2}\int_C x\,dy -ydx= Area
\end{equation}
But I don't understand and can't seem to find anywhere where this is proven or from where it is derived. I do know that when you do a little algebraic manipulation on the first equation, you can get:
\begin{equation}
\int_C x\,dy + y\,dx = Area
\end{equation}
And then using Green's theorem, I seem to get the partial derivative of x with respect to x and the partial derivative of y with respect to y to subtract each other, which gives me Area = 0. This is where I'm stuck. Any and all help is appreciated.
Thanks.
Best Answer
First, Green's theorem states that $$ \int_C P dx+Q dy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA $$
where $C$ is positively oriented a simple closed curve in the plane, $D$ the region bounded by $C$, and $P$ and $Q$ having continuous partial derivatives in an open region containing $D$. Since the area of $D$ is simply $\iint_D 1dA$, you want to choose $P$ and $Q$ such that $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$ and then apply Green's Theorem to derive a formula for the area of $D$.
A possible choice is $P=0$, $Q=x$, which gives the formula $$ A=\int_C x\,dy. $$ Or you could take $P=-y$, $Q=0$, giving $$ A=\int_C -y\,dx $$ or finally take $P=-y/2$, $Q=x/2$, to yield $$ A=\frac{1}{2}\int_C x\,dy -ydx. $$ You could derive this third one by adding the first two and dividing as well. These are the three formulas which you provided in your question.