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Integral Theorem of Gauss

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Let $ U \subseteq \mathbb{R}^3$ be a region which is the interior of a closed surface $ S$. Assume that $ S$ is smooth except for a finite number of smooth curves. Denote by $ n$ the unit normal pointing outward. Let $ \Phi $ be a continuous differentiable vector field defined on an open set containing $ U$ and $ S .$ Then

$\displaystyle \iiint\limits_{U} \operatorname{div} \Phi \,dU = \iint\limits_{S} \Phi \cdot n d\sigma \, , $

i.e. the triple integral of the divergence of $ \Phi $ coincides with the flux of $ \Phi $ through $ S .$ Note that it is geometrically clear what is meant by outward for the unit normal.

A version of Gauss' theorem with respect to a more general boundary is as follows.

Let $ U \subseteq \mathbb{R}^3$ be a bounded open set whose boundary consists of a finite number of surfaces $ S_1, \ldots , S_m$. Supoose that these surfaces are oriented such that their unit normal vectors $ n_i$ point outward. Let $ \Phi $ be a continuous differentiable vector field defined on an open set containing $ U$ and its boundary. Then

$\displaystyle \iiint\limits_{U} \operatorname{div} \Phi \,dU = \iint\limits_{S... ...i \cdot n_1 d\sigma + \ldots + \iint\limits_{S_m} \Phi \cdot n_m d\sigma \, . $

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  automatically generated 7/ 5/2005