What Do We Understand By Gauss’s Divergence Theorem?

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You must have read about Quantum Theory or Matrix Theory or Probability Theory in your math class or in some sci-fi movie. But have you ever heard of Divergence Theory? Well, here we are today to learn more about this theory. Divergence Theorem is a theorem that talks about the flux of a vector field through a closed area to the volume enclosed in the divergence of the field. It is a part of vector calculus where the divergence theorem is also called  Gauss’s divergence theorem or Ostrogradsky’s theorem.

State and Prove the Gauss’s Divergence Theorem

The divergence theorem is the one in which the surface integral is related to the volume integral. More precisely, the Divergence theorem relates the flux through the closed surface of a vector field to the divergence in the enclosed volume of the field. It states that the outward flux through a closed surface is equal to the integral volume within the surface of the divergence over the area. 

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The net flow of an area will be received by subtracting the sum of all sources by the sum of every sink. The result describes the flow by a surface of a vector and the behavior of the vector field within it. 

To state Gauss’s divergence theorem in an easier way, let’s break it into parts. The surface integral of a vector field over a sealed area is known as flux through the surface. Therefore, the flux through the surface is equivalent to the volume integral of the divergence over the area inside the surface. Thus, the total sum of all sources of the field in an area gives the net flux out of the area.

This is an essential result for mathematics in engineering and physics. It is one of the most important theorems and is used to solve tough integral problems in calculus. It is used particularly in the field of electrostatic and fluid dynamics.

Divergence Theorem is generally applied in 3 dimensions, but it can be used in any number of dimensions. When you use it in 2 dimensions, it becomes equivalent to Green’s theorem which states that the line integral around any simple closed curve is equal to a double integral over the plane region. When you use it in 1 dimension, it becomes equivalent to integration by parts. 

The Divergence Theorem Proof 

Let us consider a surface denoted by S which encloses a volume denoted by V. 

Suppose vector A is the vector field in the given region. Suppose this volume is made up of a large number of parallelepipeds (1- 6 parallelepipeds) which represent elementary volumes.

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Consider the volume of the jth parallelepiped is \[ \Delta V_{j} \]which is bounded by a surface \[ S_{j} \] of area \[ d \overrightarrow{S_{j}}\] . The surface integral of \[ \overrightarrow{A} \] over the surface \[ S_{j} \]  will be-

\[ \oint \oint\limits_{S} \overrightarrow{A} . d \overrightarrow{S_{j}}\]

Now consider if the whole volume is divided into elementary volumes I, II, and III as shown below. 

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The elementary volume I outward is the elementary volume II inward and the elementary volume II outward is the elementary III inward and so on. 

Thus, the sum of the elementary volumes integrals will cancel each other out and the surface integral arising from the surface S will be left. 

\[ \sum \oint \oint \limits _{S_{j}}  \overrightarrow {A}. d \overrightarrow {S_{j}} = \oint \oint \limits _{S} \overrightarrow {A}. d \overrightarrow {S}  …. (1) \]

Multiplying and dividing the left-hand side of the equation (1) by \[ \Delta V_{i} \], we get

\[ \oint \limits _{S_{j}} \overrightarrow {A} . d \overrightarrow {S} = \sum \frac{1}{\Delta V_{i}} (\oint \oint \limits _{S_{i}} \overrightarrow {A} . d \overrightarrow {S_{i}} ) \Delta V_{i} \]

Now, suppose the volume of surface S is divided into infinite elementary volumes such that \[ \Delta V_{i} \rightarrow 0\].

\[ \oint \oint \limits _{S} \overrightarrow {A} . d \overrightarrow{S} = \lim\limits_{\Delta V_{i} \rightarrow {0}}\sum \frac{1}{\Delta V_{i}} (\oint \oint \limits _{S_{i}} \overrightarrow {A} . d \overrightarrow {S} ) \Delta V_{i} ….. (2) \]

Now,

\[ \lim \limits_ {\Delta V_{i} \rightarrow {0}} (\frac{1}{\Delta V_{i}} (\oint \oint \limits _{S_{i}} \overrightarrow {A} . d \overrightarrow {S})) = (\overrightarrow {V}.\overrightarrow {A} ) \]

Therefore, eq (2) becomes

\[ \oint \oint \overrightarrow {A} . d \overrightarrow {S} = \sum (\overrightarrow{\nabla} . \overrightarrow {A} ) \Delta V_{i} …. (3) \]

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We know that \[ \Delta V_{i}\rightarrow{0} \], thus\[\sum \Delta V_{i} \] will become the integral over volume V.

\[ \oint \oint \overrightarrow {A} .d \overrightarrow {S} = \int\int\int\limits_{v} (\overrightarrow {\nabla} . \overrightarrow {A} ) dV \]

This is the Gauss divergence theorem.

Gauss’s Divergence Theorem History

Lagrange was the first one to discover the Divergence Theorem in 1762. Later on in 1813, it was rediscovered independently by Carl Friedrich Gauss. He also gave the first proof of the general theorem in 1826. Many other mathematicians like Green, Simeon-Denis Poisson, and Frédéric Sarrus also discovered this theorem.

Conclusion

This article entails detailed information on Divergence Theory and its derivation. You can go through it for a comprehensive understanding. Also, download the PDFs to read at your ease. 

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