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For Green's theorem in complex plane , with the help of Cauchy-Riemman equations , we can modify it to become $\int\int_B{f^{'}(z)}dxdy=\frac{i}{2} \int_{\partial B}{f(z)}d\overline{z}$ where $d\overline{z}=dx-idy$. Can we do the same for divergence theorem ? In other words, can we change the divergence theorem involves only function instead of vector field ?

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What do you mean? $f$ is a field. You haven't changed anything to not involve a field with this manipulation of Green's theorem. –  Muphrid Dec 22 '12 at 6:04
    
Divergence theorem and Green's theorem are the same thing in the plane. Apply Green's to the field rotated by 90 degrees, and you get divergence. –  user53153 Dec 24 '12 at 3:09
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I don't know what exactly you mean, but here's a variant of the divergence theorem:

Let $\Omega$ be a bounded $C^1$-domain in $\mathbb R^n$. Let $\nu: \partial \Omega \to \mathbb R^n$ denote the outward pointing unit normal to $\partial \Omega$. Let $\varphi$ be a $C^1$-function. Denote by $e_i$ (for $i=1, \dots, n$) the usual basis vectors of $\mathbb R^n$. Applying the divergence theorem to the vector field $\varphi e_i$ yields $$\int_\Omega \partial_i\varphi = \int_{\partial \Omega} \varphi \langle \nu, e_i\rangle.$$ Multiplying this with $e_i$ and summing over $i=1, \dots,n$, we obtain that $$\int_\Omega \nabla \varphi = \int_{\partial \Omega} \varphi \, \nu.$$ This is a form of the divergence theorem for functions.

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