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I'm attempting to use Green's Theorem to express the area of a region in the complex plane in terms of a contour integral, but I'm a little confused as to how this works. I have a simple closed curve $\gamma$ with interior $D$, and I believe I'm supposed to get $$\mathrm{Area}(D)=\frac{1}{2i} \oint_\gamma \overline{z} \,dz.$$ Can anyone help me justify this?

Fabian
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Rick
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2 Answers2

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$$\int_\gamma \bar{z}dz = \int_\gamma (x - iy)(dx + idy) = \int_\gamma (xdx + ydy) + i \int_\gamma( xdy - ydx)$$now hit this with stokes:$$ = \int_D d(xdx + ydy) + i\int_D d(xdy - ydx) = i \int_D 2 dx \wedge dy$$

Fabian
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uncookedfalcon
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  • from $\int_D{d(xdx+ydy)} +i\int_D{d(xdy-ydx)} = i\int_D{2dxdy}$ , how you get this equality ? – hong wai Dec 22 '12 at 14:44
  • Using the definition of derivative of a differential form, you get $dx dx + dy dy = 0$ for the first form, and $dx dy - dy dx = 2 dx dy$ for the second form – Mark Jul 20 '14 at 19:48
  • Because the first integral is zero, you can also multiply it by $i$ and add it on to the second one to justify $ i \int 2 x dx = 2 i \text{Area(D)}$ i.e. $\int x dx = \text{Area(D)}$ – Mark Jul 20 '14 at 20:24
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You can basically use Greens theorem twice: It's defined by

\begin{align} \oint_{\mathcal{C}}\left(L\mathrm{d}x+M\mathrm{d}y\right)=\iint_{D}\mathrm{d}x\mathrm{d}y\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right) \end{align}

where $D$ is the area bounded by the closed contour $\mathcal{C}$.

For the term $\oint_{\mathcal{C}}\left(x\mathrm{d}x+y\mathrm{d}y\right)$ we identify $L = x$ and $M=y$, then using Greens theorem, we see that it vanishes and for the second term $i\oint_{\mathcal{C}}\left(x\mathrm{d}y-y\mathrm{d}x\right)$ we obtain

\begin{align} i\int\oint_{\mathcal{C}}\left(x\mathrm{d}y-y\mathrm{d}x\right) = 2i \iint_{D}\mathrm{d}x\mathrm{d}y \end{align}

which confirm your result.

blackhole123
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