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Question: Is there any formula that bounds the line and double integrals other than the Green one? My guess: No! We know: $$ \int_V \operatorname{div} \vec{F}\, dx\,dy\,dz = \int_{\partial V} \vec{F} \cdot \vec{n} \cdot dS$$ Here: $n$ denotes the unit normal vector of $dS;$ div stands for divergence and defined by the formula through limit, as known. This formula is not the same as the Stokes one, in which one may discern curl. My guess is supported by defining the vector function $$ \vec{F} = (\varphi(x, y) ; \psi(x, y)). $$ So, $$ \operatorname{div} \vec{F} = \vec{\nabla} \cdot \vec{F} = \frac{\partial \varphi}{\partial x} + \frac{\partial \psi}{\partial y}$$

Taking an attempt to get an analogous, we have

$$\int_A \operatorname{div} \vec{F} \,dx\,dy = \int_{\partial A} \vec{F} \,\vec{dl} $$

Taking into consideration that $\vec{dl} = \sum x_k \vec{e_k} $, we get $$\int_A \left(\frac{\partial \varphi}{\partial x} + \frac{\partial \psi}{\partial y}\right) \, dx\,dy = \int_{\partial A} (\varphi \,dx + \psi \,dy) $$ That does not correspond to the Green formula, which implies that this should not work.

Michael Hardy
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sergei ivanov
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  • Green's theorem can be written in terms of the divergence as $$\int_{\partial D} \boldsymbol F \cdot \boldsymbol n , ds = \iint_D \nabla \cdot \boldsymbol F , dS,$$ where $\boldsymbol n$ is the outward unit normal to $\partial D$. – Maxim Sep 21 '20 at 15:15
  • As you may discern, I have written that expression (assuming that surface integrals play the same role as do line ones). Comparing this equation with Green's formula tells us that either there is a solution, a function for which both equations are satisfied ; or that is invalid, meaning that there is only Green theorem – sergei ivanov Oct 04 '20 at 01:39

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