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    Green's theorem

    T-4M73O

    $\oint_{\bd\Omega}f\,dx+g\,dy=\iint_\Omega(g_x-f_y)\,dA$; fields with $g_x-f_y=1$ give the area of $\Omega$ as a boundary integral.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be a bounded open set whose boundary \(\bd\Omega\) is a finite union of piecewise smooth closed curves, oriented positively (with \(\Omega\) on the left), and let \(f,g\in C^1(\overline\Omega)\) be real-valued. Then \[ \int_{\bd\Omega}f\dx+g\dy=\iint_\Omega\Big(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\Big)\dA. \] In vector form, for the vector field \(F=(f,g)\) with scalar curl \(\curl F=g_x-f_y\), \[ \int_{\bd\Omega}F\cdot\dr=\iint_\Omega\curl F\dA. \]

    Taking \((f,g)=(0,x)\), \((-y,0)\), and \(\big(-\frac y2,\frac x2\big)\), each with \(g_x-f_y=1\), gives the area of \(\Omega\): \[ \mu(\Omega)=\oint_{\bd\Omega}x\dy=-\oint_{\bd\Omega}y\dx=\frac12\oint_{\bd\Omega}(x\dy-y\dx). \]

    For \(f\) holomorphic and injective on a neighborhood of \(\overline\Omega\), the real Jacobian determinant of \(f\) is \(\abs{f'(z)}^2\), so the area of the image is \(\mu(f(\Omega))=\iint_\Omega\abs{f'(z)}^2\dA\).

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