# Difference between revisions of "Dictionary:Divergence theorem"

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<center> <math> \int_D \mathbf{\nabla} \cdot \mathbf{Q} \; dV = \int_{\partial D} \mathbf{\hat{n}} \cdot \mathbf{Q} \; dS. </math> </center> | <center> <math> \int_D \mathbf{\nabla} \cdot \mathbf{Q} \; dV = \int_{\partial D} \mathbf{\hat{n}} \cdot \mathbf{Q} \; dS. </math> </center> | ||

− | Here, <math> Q</math> is a vector field and the integrand of the (hyper-)volume integral is an exact divergence of that vector field. The quantity <math> \mathbf{\hat{n}}=({\bar{n}}_1, {\bar{n}}_2, ... \bar{{n}}_m),</math> is the outward-pointing | + | Here, <math> Q</math> is a vector field and the integrand of the (hyper-)volume integral is an exact divergence of that vector field. The quantity <math> \mathbf{\hat{n}}=({\bar{n}}_1, {\bar{n}}_2, ... \bar{{n}}_m),</math> is the outward-pointing unit |

normal vector to the (hyper-) surface <math> \partial D. </math> Each <math> {\bar{n}}_k </math> is the respective direction cosine | normal vector to the (hyper-) surface <math> \partial D. </math> Each <math> {\bar{n}}_k </math> is the respective direction cosine | ||

obtained by forming the inner product of the unit normal to the boundary <math> \partial D </math> with the respective <math> \hat{x}_k </math> coordinate axis unit vector. | obtained by forming the inner product of the unit normal to the boundary <math> \partial D </math> with the respective <math> \hat{x}_k </math> coordinate axis unit vector. |

## Latest revision as of 17:07, 17 August 2021

The flux through a surface (or the integral of the vector flux density **g** over a closed surface) equals the divergence of the flux density integrated over the volume contained by the surface:

Another way of writing this is

where is the area element on the bounding surface of volume , with volume element .

Commonly called **Gauss's theorem** or the ** Gauss-Ostrogradski theorem **.

## Generalization to arbitrary dimensions

The divergence theorem is a general mathematical result that may be applied in arbitrary dimensions.

If we have an integral over the volume in bounded by in then the following relation holds

Here, is a vector field and the integrand of the (hyper-)volume integral is an exact divergence of that vector field. The quantity is the outward-pointing unit normal vector to the (hyper-) surface Each is the respective direction cosine obtained by forming the inner product of the unit normal to the boundary with the respective coordinate axis unit vector.

Because we are in arbitrary dimensions and .

### Proof^{[1]}

Consider the case when and is a three-dimensional volume bounded by the two-dimensional boundary . The volume integral may be written in three terms, one for each coordinate direction

If we concentrate on the volume integral and apply the fundamental theorem of calculus we obtain

```
```

Here, the functions and define points on of greater
and lesser respectively.

Now, if we consider the surface integral

and concentrate on the term

On the part of that is of greater the values of and . On the part of the surface that of lesser values of the values of and . Here on the respective portion of the surface.

Thus, the surface integral for the is the same as the volume integral of the term:

Here

Similar results may be found for the and , proving the divergence theorem for 3 dimensions.

This same method generalizes for the case of dimensions, proving the theorem for the arbitrary case.

## References

- ↑ Greenspan, H. P., Benney, D. J., & Turner, J. E. (1986). Calculus: an introduction to applied mathematics. McGraw-Hill Ryerson Ltd.