Gauss's Theorem

From ProofWiki
(Redirected from Divergence Theorem)
Jump to: navigation, search

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.



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.



Let:

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

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


Then:

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


We turn now to examine $\mathbf n$:

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


Hence:

\(\displaystyle \mathbf F \cdot \mathbf n\) \(=\) \(\displaystyle -M\)          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:

\(\displaystyle \mathrm d S\) \(=\) \(\displaystyle \mathrm d y \ \mathrm d z\)          on $A_1$ and $A_2$          
\(\displaystyle \mathrm d S\) \(=\) \(\displaystyle \mathrm d x \ \mathrm d z\)          on $A_3$ and $A_4$          
\(\displaystyle \mathrm d S\) \(=\) \(\displaystyle \mathrm d x \ \mathrm d y\)          on $A_5$ and $A_6$          

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:

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

and in general:

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


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.