Let $E$ be a region in $\mathbb{R}^2$ with a smooth and non self-intersecting boundary $\partial E$ oriented in the counterclockwise direction, then from green theorem, we know that $$Area(E)=\frac{1}{2}\int_{\partial E} x\ dy-y\ dx$$ What is the analogue for Green theorem for volume of $E$ in 3 dimension? Also, a proof will be nice?
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$\begingroup$ an application of the divergence theorem (Gauss-Ostrogradsky) $\endgroup$– janmarqzJan 1, 2015 at 17:02
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$\begingroup$ example found here at MSE: math.stackexchange.com/questions/797915/… $\endgroup$– janmarqzJan 1, 2015 at 17:26
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1 Answer
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The analogue becomes almost obvious if you think of $\frac{1}{2}\int_{\partial E} x\ dy-y\ dx$ not as the line integral of $\frac12 (-y,x)$ along the boundary, but rather as the flux of $\frac12(x,y)$ across the boundary. Which is what it is, since $(dy,-dx)$ represents the exterior normal. The significance of $\frac12(x,y)$ is that it's a field with divergence equal to $1$, hence the integral of divergence over a region is equal to its area.
In any number $n$ dimensions, the same thing holds (by the divergence theorem): the volume is equal to the flux of any field with constant divergence $1$ across the boundary. A nice symmetric field to use is $\frac1n (x_1,\dots,x_n)$, which is $\frac13(x,y,z)$ in three dimensions. As janmarqz pointed out, this question has an example of this field used. But any other field $v(x) =Ax$ where $A$ is a matrix of trace $1$ would work too.
