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.


  • $\begingroup$ I don't think your third equation is consistent. If $\int_C xdy = \int_C -ydx$, then by adding $\int_C ydx$ to both sides, we get $\int_C xdy + ydx = 0$. $\endgroup$ – Michael Albanese Sep 19 '12 at 7:30

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.

  • $\begingroup$ Thanks so much. I can't believe I didn't see that. It seems so obvious now... $\endgroup$ – M. Choi Sep 19 '12 at 8:17

As yunone' noted completely: $$\oint_C Pdx+Qdy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$$ This can be proved that the left hand side closed integral is independent of the path $C$ joining any two given points in the riegi $\mathcal{R}$ iff $$\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$$. It is also supposed that the partial differential above are continues on $\mathcal{R}$. In this case; $Pdx+Qdy$ is an exact differential, i.e. $\exists \phi(x,y); Pdx+Qdy=d\phi$. Now if the end points of the curve $C$ are $(x_1,y_1),(x_2,y_2)$ we have: $$\int_{(x_1,y_1)}^{(x_2,y_2)} Pdx+Qdy=\int_{(x_1,y_1)}^{(x_2,y_2)} d\phi=\phi(x_1,y_1)-\phi(x_2,y_2)=0$$ since $C$ is closed and two end points are coincided.


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