Gauss-Ostrogradsky Theorem

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

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


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

where $\mathbf n$ is the normal to $\partial U$.


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


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

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

Then :

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


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


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

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


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

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


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

We turn now to examine $\mathbf n$:

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


\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle -M\) on $A_1$
\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle M\) on $A_2$
\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle -N\) on $A_3$
\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle N\) on $A_4$
\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle -P\) on $A_5$
\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle P\) on $A_6$

We also have:

\(\displaystyle \d S\) \(=\) \(\displaystyle \d y \rd z\) on $A_1$ and $A_2$
\(\displaystyle \d S\) \(=\) \(\displaystyle \d x \rd z\) on $A_3$ and $A_4$
\(\displaystyle \d S\) \(=\) \(\displaystyle \d x \rd y\) 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.


\(\displaystyle \iint_{A_2} \mathbf F \cdot \mathbf n \rd S\) \(=\) \(\displaystyle \iint_{A_2} M \rd y \rd z\)

and in general:

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

Hence the result:

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


Also known as

The Gauss-Ostrogradsky Theorem is also known as:

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

Also see

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 $1828$.