# Gauss-Ostrogradsky Theorem

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## 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.

## Informal Proof

Let $S$ be the surface of $U$.

By definition, the surface integral of $\mathbf V$ over $S$ is therefore defined as:

$\ds \iint_S \mathbf V \cdot \mathbf n \rd S = \iint_S \mathbf V \cdot \rd \mathbf S$

where $\rd \mathbf S$ is the differential of vector area.

Let $\d v$ be an element of volume within $U$.

In the above diagram, a small cube has been used for convenience.

The total flux emerging from $\d v$ is:

$\operatorname {div} \mathbf V \rd v$

where in this context $\mathbf V$ is the vector quantity at the center of $\d v$.

For the face $abcd$, the positive direction of the $x$ component of $\mathbf V$ and the outward normal are in the same sense, and so the flux is positive.

Consider another small cube contiguous to $\d v$, shown dotted in the above diagram.

This has the same component of $\mathbf V$, but this is acting inward, and so is negative.

Hence the flux through the face of one cube cancels out the flux through the adjacent face of the next cube.

This argument can be applied by imagining a sequence of such cubes until you reach the surface of $U$.

This is the only part of the construction we have just built which contributes to the flux through $S$.

By applying the same treatment to the surfaces of all such small elements of volume, we arrive at a total flux:

$\ds \iint_S \mathbf V \cdot \rd \mathbf S$

But at the same time we have integrated $\operatorname {div} \mathbf V \rd v$ throughout the whole of the volume of $U$.

This also measures the total flux through $S$.

Hence:

 $\ds \iint_S \mathbf V \cdot \rd \mathbf S$ $=$ $\ds \iint_S \mathbf V \cdot \mathbf n \rd S$ $\ds$ $=$ $\ds \iiint \operatorname {div} \mathbf V \rd v$ $\ds$ $=$ $\ds \iiint \nabla \cdot \mathbf V \rd v$

## Formal Proof

It suffices to prove the theorem for rectangular prisms; the Riemann-sum nature of the triple integral then guarantees the theorem for arbitrary regions.

Let:

$R = \set {\tuple {x, y, z}: a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le c_2}$

and let $S = \partial R$, oriented outward.

Then:

$S = A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6$

where:

$A_1, A_2$ are those sides perpendicular to the $x$-axis
$A_3, A_4$ are those sides perpendicular to the $y$ axis

and

$A_5, A_6$ are those sides perpendicular to the $z$-axis

and in all cases the lower subscript indicates a side closer to the origin.

Let:

$\mathbf V = M \mathbf i + N \mathbf j + P \mathbf k$

where $M, N, P: \R^3 \to \R$.

Then:

 $\ds \iiint_R \nabla \cdot \mathbf V \rd v$ $=$ $\ds \iiint_R \paren {\frac {\partial M} {\partial x} + \frac {\partial N} {\partial y} + \frac {\partial P} {\partial z} } \rd x \rd y \rd z$ $\ds$ $=$ $\ds \iiint_R \frac {\partial M} {\partial x} \rd x \rd y \rd z + \iiint_M \frac {\partial N} {\partial y} \rd x \rd y \rd z + \iiint_M \frac {\partial P} {\partial z} \rd x \rd y \rd z$ $\ds$ $=$ $\ds \int_{c_1}^{c_2} \int_{b_1}^{b_2} \paren {\map M {a_2, y, z} - \map M {a_1, y, z} } \rd y \rd z$ $\ds$  $\, \ds + \,$ $\ds \int_{c_1}^{c_2} \int_{a_1}^{a_2} \paren {\map N {x, b_2, z} - \map N {x, b_1, z} } \rd x \rd z$ $\ds$  $\, \ds + \,$ $\ds \int_{b_1}^{b_2} \int_{a_1}^{a_2} \paren {\map P {x, y, c_2} - \map P {x, y, c_1} } \rd x \rd y$ $\ds$ $=$ $\ds \iint_{A_2} M \rd y \rd z - \iint_{A_1} M \rd y \rd z$ $\ds$  $\, \ds + \,$ $\ds \iint_{A_4} N \rd x \rd z - \iint_{A_3} N \rd x \rd z$ $\ds$  $\, \ds + \,$ $\ds \iint_{A_6} P \rd x \rd y - \iint_{A_5} P \rd x \rd y$

We turn now to examine $\mathbf n$:

 $\ds \mathbf n$ $=$ $\ds \tuple {-1, 0, 0}$ on $A_1$ $\ds \mathbf n$ $=$ $\ds \tuple {1, 0, 0}$ on $A_2$ $\ds \mathbf n$ $=$ $\ds \tuple {0, -1, 0}$ on $A_3$ $\ds \mathbf n$ $=$ $\ds \tuple {0, 1, 0}$ on $A_4$ $\ds \mathbf n$ $=$ $\ds \tuple {0, 0, -1}$ on $A_5$ $\ds \mathbf n$ $=$ $\ds \tuple {0, 0, 1}$ on $A_6$

Hence:

 $\ds \mathbf V \cdot \mathbf n$ $=$ $\ds -M$ on $A_1$ $\ds \mathbf V \cdot \mathbf n$ $=$ $\ds M$ on $A_2$ $\ds \mathbf V \cdot \mathbf n$ $=$ $\ds -N$ on $A_3$ $\ds \mathbf V \cdot \mathbf n$ $=$ $\ds N$ on $A_4$ $\ds \mathbf V \cdot \mathbf n$ $=$ $\ds -P$ on $A_5$ $\ds \mathbf V \cdot \mathbf n$ $=$ $\ds P$ on $A_6$

We also have:

 $\ds \d S$ $=$ $\ds \d y \rd z$ on $A_1$ and $A_2$ $\ds \d S$ $=$ $\ds \d x \rd z$ on $A_3$ and $A_4$ $\ds \d S$ $=$ $\ds \d x \rd y$ on $A_5$ and $A_6$

where $\d S$ is the area element.

This is true because each side is perfectly flat, and constant with respect to one coordinate.

Hence:

 $\ds \iint_{A_2} \mathbf V \cdot \mathbf n \rd S$ $=$ $\ds \iint_{A_2} M \rd y \rd z$

and in general:

 $\ds \sum_{i \mathop = 1}^6 \iint_{A_i} \mathbf V \cdot \mathbf n \rd S$ $=$ $\ds \iint_{A_2} M \rd y \rd z - \iint_{A_1} M \rd y \rd z + \iint_{A_4} N \rd x \rd z - \iint_{A_3} N \rd x \rd z + \iint_{A_6} P \rd x \rd y - \iint_{A_5} P \rd x \rd y$

Hence the result:

$\ds \iiint_R \nabla \cdot \mathbf V \rd v = \iint_{\partial R} \mathbf V \cdot \mathbf n \rd S$

$\blacksquare$

## Intuitive Illustration

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.

## Source of Name

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

## Historical Note

The Gauss-Ostrogradsky Theorem was first discovered by Joseph Louis Lagrange in $1762$.

It was the later independently rediscovered by Carl Friedrich Gauss in $1813$ during the course of his investigations into electromagnetism.

Mikhail Vasilyevich Ostrogradsky, who also gave the first proof of the general theorem, rediscovered it in $1826$.

It was also rediscovered by George Green in $1828$, Siméon-Denis Poisson in $1824$ and Pierre Frédéric Sarrus in $1828$.