Gauss-Ostrogradsky Theorem/Intuitive Illustration

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.

Let $\mathbf V: \R^3 \to \R^3$ be a smooth vector field defined on a neighborhood of $U$.


Then:

$\ds \iiint \limits_U \paren {\nabla \cdot \mathbf V} \rd v = \iint \limits_{\partial U} \mathbf V \cdot \mathbf n \rd S$

where $\mathbf n$ is the unit normal to $\partial U$, directed outwards.


Proof

Consider a closed surface $S$ within a body of fluid which at an arbitrary point $P$ is in motion with velocity $\mathbf V$.

The total rate of flow of fluid passing through $S$ can be found in $2$ different ways:

$(1): \quad$ By calculating $\mathbf V \cdot \rd \mathbf S$, which is the product of a small element of $S$ and the component of velocity which is perpendicular to it, and then adding all these contributions
$(2): \quad$ By investigating the divergence of an element of volume, which is the excess of the sources of fluid over its sinks per unit volume and integrating $\operatorname {div} \mathbf V \rd v$ throughout the volume enclosed by $S$.

These $2$ results are physically equivalent, because the excess fluid leaving $S$ over that entering must be because of the fluid injected into $B$ by the aggregate of sources and sinks.


Also known as

The Gauss-Ostrogradsky Theorem is also known as:

the Divergence Theorem
Gauss's Theorem
Gauss's Divergence Theorem or Gauss's Theorem of Divergence
Ostrogradsky's Theorem
the Ostrogradsky-Gauss Theorem.


Also see


Source of Name

This entry was named for Carl Friedrich Gauss and Mikhail Vasilyevich Ostrogradsky.


Sources