Gauss's 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 $\mathbf F: \R^3 \to \R^3$ be a smooth vector function defined on a neighborhood of $U$.
Then:
- $\displaystyle \iiint \limits_U \left({\nabla \cdot \mathbf F} \right) \mathrm d V = \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \ \mathrm d S$
where $\mathbf n$ is the normal to $\partial U$.
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.
Link to a result demonstrating the above, or explain it better here
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Let:
- $R = \left\{ {\left({x, y, z}\right): a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le c_2}\right\}$
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.
A diagram here is de rigueur
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Let:
- $\mathbf F = M \mathbf i + N \mathbf j + P \mathbf k$
where $M, N, P: \R^3 \to \R$.
Then:
| \(\displaystyle \iiint_R \nabla \cdot \mathbf F \ \mathrm d V\) | \(=\) | \(\displaystyle \iiint_R \left({ \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z} }\right) \ \mathrm d x \ \mathrm d y \ \mathrm d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \iiint_R \frac{\partial M}{\partial x} \ \mathrm d x \ \mathrm d y \ \mathrm d z + \iiint_M \frac{\partial N}{\partial y} \ \mathrm d x \ \mathrm d y \ \mathrm d z + \iiint_M \frac{\partial P}{\partial z} \ \mathrm d x \ \mathrm d y \ \mathrm d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \int_{c_1}^{c_2} \int_{b_1}^{b_2} \left({M \left({a_2, y, z}\right) - M \left({a_1, y, z}\right)}\right) \ \mathrm d y \ \mathrm d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \int_{c_1}^{c_2} \int_{a_1}^{a_2} \left({N \left({x, b_2, z}\right) - N \left({x, b_1, z}\right)}\right) \ \mathrm d x \ \mathrm d z\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \int_{b_1}^{b_2} \int_{a_1}^{a_2} \left({P \left({x, y, c_2}\right) - P \left({x, y, c_1}\right)}\right) \ \mathrm d x \ \mathrm d y\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \iint_{A_2} M \ \mathrm d y \ \mathrm d z - \iint_{A_1} M \ \mathrm d y \ \mathrm d z\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \iint_{A_4} N \ \mathrm d x \ \mathrm d z - \iint_{A_3} N \ \mathrm d x \ \mathrm d z\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \iint_{A_6} P \ \mathrm d x \ \mathrm d y - \iint_{A_5} P \ \mathrm d x \ \mathrm d y\) | $\quad$ | $\quad$ |
We turn now to examine $\mathbf n$:
| \(\displaystyle \mathbf n\) | \(=\) | \(\displaystyle \left({-1, 0, 0}\right)\) | $\quad$ on $A_1$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf n\) | \(=\) | \(\displaystyle \left({1, 0, 0}\right)\) | $\quad$ on $A_2$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf n\) | \(=\) | \(\displaystyle \left({0, -1, 0}\right)\) | $\quad$ on $A_3$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf n\) | \(=\) | \(\displaystyle \left({0, 1, 0}\right)\) | $\quad$ on $A_4$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf n\) | \(=\) | \(\displaystyle \left({0, 0, -1}\right)\) | $\quad$ on $A_5$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf n\) | \(=\) | \(\displaystyle \left({0, 0, 1}\right)\) | $\quad$ on $A_6$ | $\quad$ |
Hence:
| \(\displaystyle \mathbf F \cdot \mathbf n\) | \(=\) | \(\displaystyle -M\) | $\quad$ on $A_1$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf F \cdot \mathbf n\) | \(=\) | \(\displaystyle M\) | $\quad$ on $A_2$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf F \cdot \mathbf n\) | \(=\) | \(\displaystyle -N\) | $\quad$ on $A_3$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf F \cdot \mathbf n\) | \(=\) | \(\displaystyle N\) | $\quad$ on $A_4$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf F \cdot \mathbf n\) | \(=\) | \(\displaystyle -P\) | $\quad$ on $A_5$ | $\quad$ | |||||||||
| \(\displaystyle \mathbf F \cdot \mathbf n\) | \(=\) | \(\displaystyle P\) | $\quad$ on $A_6$ | $\quad$ |
We also have:
| \(\displaystyle \mathrm d S\) | \(=\) | \(\displaystyle \mathrm d y \ \mathrm d z\) | $\quad$ on $A_1$ and $A_2$ | $\quad$ | |||||||||
| \(\displaystyle \mathrm d S\) | \(=\) | \(\displaystyle \mathrm d x \ \mathrm d z\) | $\quad$ on $A_3$ and $A_4$ | $\quad$ | |||||||||
| \(\displaystyle \mathrm d S\) | \(=\) | \(\displaystyle \mathrm d x \ \mathrm d y\) | $\quad$ on $A_5$ and $A_6$ | $\quad$ |
where $\mathrm d S$ is the area element.
This is true because each side is perfectly flat, and constant with respect to one coordinate.
Hence:
| \(\displaystyle \iint_{A_2} \mathbf F \cdot \mathbf n \ \mathrm d S\) | \(=\) | \(\displaystyle \iint_{A_2} M \ \mathrm d y \ \mathrm d z\) | $\quad$ | $\quad$ |
and in general:
| \(\displaystyle \sum_{i \mathop = 1}^6 \iint_{A_i} \mathbf F \cdot \mathbf n \ \mathrm d S\) | \(=\) | \(\displaystyle \iint_{A_2} M \ \mathrm d y \ \mathrm d z - \iint_{A_1} M \ \mathrm d y \ \mathrm d z + \iint_{A_4} N \ \mathrm d x \ \mathrm d z - \iint_{A_3} N \ \mathrm d x \ \mathrm d z + \iint_{A_6} P \ \mathrm d x \ \mathrm d y - \iint_{A_5} P \ \mathrm d x \ \mathrm d y\) | $\quad$ | $\quad$ |
Hence the result:
- $\displaystyle \iiint_R \nabla \cdot \mathbf F \ \mathrm d V = \iint_{\partial R} \mathbf F \cdot \mathbf n \ \mathrm d S$
$\blacksquare$
Also known as
This result is also known as the divergence theorem.
Source of Name
This entry was named for Carl Friedrich Gauss.
Historical Note
Gauss's Theorem was proved by Carl Friedrich Gauss during the course of his investigations into electromagnetism.
