Editing Divergence theorem (section)

==Informal derivation==
The divergence theorem follows from the fact that if a volume {{mvar|V}} is partitioned into separate parts, the [[flux]] out of the original volume is equal to the algebraic sum of the flux out of each component volume.<ref name="Benford">{{cite web
  | last = Benford
  | first = Frank A.
  | title = Notes on Vector Calculus
  | work = Course materials for Math 105: Multivariable Calculus
  | publisher = Prof. Steven Miller's webpage, Williams College
  | date = May 2007
  | url = https://web.williams.edu/Mathematics/sjmiller/public_html/105Sp11/handouts/Benford_NotesVectorcalculus.pdf
  | access-date = 14 March 2022}}</ref><ref name="Purcell">{{cite book
  | last = Purcell
  | first = Edward M.
  |author2=David J. Morin
  | title = Electricity and Magnetism
  | publisher = Cambridge Univ. Press
  | date = 2013
  | pages = 56–58
  | url = https://books.google.com/books?id=A2rS5vlSFq0C&pg=PA56
  | isbn = 978-1-107-01402-2}}</ref>  This is true despite the fact that the new subvolumes have surfaces that were not part of the original volume's surface, because these surfaces are just partitions between two of the subvolumes and the flux through them just passes from one volume to the other and so cancels out when the flux out of the subvolumes is summed.

[[File:Divergence theorem 1 - split volume.png|thumb|upright=2|A volume divided into two subvolumes. At right the two subvolumes are separated to show the flux out of the different surfaces.]]
See the diagram.  A closed, bounded volume {{mvar|V}} is divided into two volumes {{math|''V''<sub>1</sub>}} and {{math|''V''<sub>2</sub>}} by a surface {{math|''S''<sub>3</sub>}} ''<span style="color:green;">(green)</span>''.  The flux {{math|Φ(''V''<sub>i</sub>)}} out of each component region {{math|''V''<sub>i</sub>}} is equal to the sum of the flux through its two faces, so the sum of the flux out of the two parts is

:<math>\Phi(V_\text{1}) + \Phi(V_\text{2}) =  \Phi_\text{1} + \Phi_\text{31} + \Phi_\text{2} + \Phi_\text{32}</math>

where {{math|Φ<sub>1</sub>}} and {{math|Φ<sub>2</sub>}} are the flux out of surfaces {{math|''S''<sub>1</sub>}} and {{math|''S''<sub>2</sub>}},  {{math|Φ<sub>31</sub>}} is the flux through {{math|''S''<sub>3</sub>}} out of volume 1, and {{math|Φ<sub>32</sub>}} is the flux through {{math|''S''<sub>3</sub>}} out of volume 2.  The point is that surface {{math|''S''<sub>3</sub>}} is part of the surface of both volumes.  The "outward" direction of the [[normal vector]] <math>\mathbf{\hat n}</math> is opposite for each volume, so the flux out of one through {{math|''S''<sub>3</sub>}} is equal to the negative of the flux out of the other so these two fluxes cancel in the sum.

:<math>\Phi_\text{31} = \iint_{S_3} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S = -\iint_{S_3} \mathbf{F} \cdot (-\mathbf{\hat n}) \; \mathrm{d}S = -\Phi_\text{32}</math>

Therefore:
:<math>\Phi(V_\text{1}) + \Phi(V_\text{2}) =  \Phi_\text{1} + \Phi_\text{2}</math>
Since the union of surfaces {{math|''S''<sub>1</sub>}} and {{math|''S''<sub>2</sub>}} is {{mvar|S}}
:<math>\Phi(V_\text{1}) + \Phi(V_\text{2}) = \Phi(V)</math>
{{clear}}

[[File:Divergence theorem 2 - volume partition.png|thumb|upright=2|The volume can be divided into any number of subvolumes and the flux out of ''V'' is equal to the sum of the flux out of each subvolume, because the flux through the <span style="color:green;">green</span> surfaces cancels out in the sum. In (b) the volumes are shown separated slightly, illustrating that each green partition is part of the boundary of two adjacent volumes]]

This principle applies to a volume divided into any number of parts, as shown in the diagram.<ref name="Purcell" />  Since the integral over each internal partition ''<span style="color:green;">(green surfaces)</span>'' appears with opposite signs in the flux of the two adjacent volumes they cancel out, and the only contribution to the flux is the integral over the external surfaces ''<span style="color:grey;">(grey)</span>''.  Since the external surfaces of all the component volumes equal the original surface.
:<math>\Phi(V) = \sum_{V_\text{i}\subset V} \Phi(V_\text{i})</math>
{{clear}}

[[File:Divergence theorem 3 - infinitesimals.png|thumb|upright=1|As the volume is subdivided into smaller parts, the ratio of the flux <math>\Phi(V_\text{i})</math> out of each volume to the volume <math>|V_\text{i}|</math> approaches <math>\operatorname{div} \mathbf{F}</math>]]
The flux {{math|Φ}} out of each volume is the surface integral of the vector field {{math|'''F'''('''x''')}} over the surface

:<math>\iint_{S(V)} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S = \sum_{V_\text{i}\subset V} \iint_{S(V_\text{i})} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S</math>

The goal is to divide the original volume into infinitely many infinitesimal volumes.  As the volume is divided into smaller and smaller parts, the surface integral on the right, the flux out of each subvolume, approaches zero because the surface area {{math|''S''(''V''<sub>i</sub>)}} approaches zero. However, from the definition of [[divergence]], the ratio of flux to volume, <math>\frac{\Phi(V_\text{i})}{|V_\text{i}|} = \frac{1}{|V_\text{i}|} \iint_{S(V_\text{i})} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S</math>, the part in parentheses below, does not in general vanish but approaches the [[divergence]] {{math|div '''F'''}} as the volume approaches zero.<ref name="Purcell" />

:<math>\iint_{S(V)} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S = \sum_{V_\text{i} \subset V} \left(\frac{1}{|V_\text{i}|} \iint_{S(V_\text{i})} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S\right) |V_\text{i}|</math>

As long as the vector field {{math|'''F'''('''x''')}} has continuous derivatives, the sum above holds even in the [[limit (mathematics)|limit]] when the volume is divided into infinitely small increments

:<math>\iint_{S(V)} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S = \lim_{|V_\text{i}|\to 0}\sum_{V_\text{i}\subset V} \left(\frac{1}{|V_\text{i}|}\iint_{S(V_\text{i})} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S\right) |V_\text{i}|</math>

As <math>|V_\text{i}|</math> approaches zero volume, it becomes the infinitesimal {{math|''dV''}}, the part in parentheses becomes the divergence, and the sum becomes a [[volume integral]] over {{mvar|V}}
{{Equation box 1 |indent =: |cellpadding = 0 |border = 1 |border colour = black |background colour = transparent
|equation = <math>\;\iint_{S(V)} \mathbf{F}\cdot\mathbf{\hat n}\; \mathrm{d}S = \iiint_{V} \operatorname{div} \mathbf{F}\;\mathrm{d}V\;</math>
}}
Since this derivation is coordinate free, it shows that the divergence does not depend on the coordinates used.