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Surface Integrals of Vector Fields, Flux Integral

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Let $ S$ be a surface in $ \mathbb{R}^3$ parametrized by

$\displaystyle \sigma: A \longrightarrow \mathbb{R}^3 \ \ (t,u) \mapsto \sigma (t,u) .$

Assume that $ \sigma $ is regular, i.e. $ \sigma $ is differentiable and $ \sigma_t $ and $ \sigma_u $ are linearly independent. Further let $ \Phi $ be a continuous differentiable vector field defined on an open set containing $ \sigma (A) .$

Then the surface integral of $ \Phi $ is defined as

$\displaystyle \iint_S \Phi \cdot n \ d\sigma := \iint_A \Phi(\sigma(t,u)) \cdot (\sigma_t \times \sigma_u) \ dt du . $

Here $ n$ denotes the unit normal into the direction of $ \sigma_t \times \sigma_u .$ The integral depends on the direction of the normal and in this sense from the parametrization. For the other unit normal $ \tilde{n}$ of the surface one gets

$\displaystyle \iint_S \Phi \cdot \tilde{n} \ d\sigma = \iint_S \Phi \cdot n \ d\sigma \ .$

If $ \sigma (t,u)$ and $ \tilde{\sigma} (t,u)$ are parametrizations of $ S$ such that the normal vectors $ \sigma_t \times \sigma_u $ and $ \tilde{\sigma}_t \times \tilde{\sigma}_u $ have the same direction then the corresponding surface integrals are equal (this justifies the used notation).

Physical interpretation: The surface integral gives the amount of fluid passing through the surface per unit time.

This explains why a surface integral of a vector field is also called the flux of the vector field through the surface or why surface integrals are simply called flux integrals.

\includegraphics[width=.6\linewidth]{a_flussintegral_bild}

The conditions on the smoothness of $ \Phi $ und $ \sigma $ may be weakened using suitable limit considerations.

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  automatically generated 7/ 5/2005