## Gauss Divergence Theorem

According to the **Gauss Divergence Theorem**, *the surface integral of a vector field A over a closed surface is equal to the volume integral of the divergence *of a vector field

**A**over the volume (

**V**) enclosed by the closed surface.

**Proof of Gauss Divergence Theorem**

Consider a surface S which encloses a volume** V**. Let vector **A** be the vector field in the given region. Let this volume is made up of a large number of elementary volumes in the form of parallelopipeds.

Consider jth parallelopiped of volume **Δ** Vj and bounded by a surface **Sj** of area d vector **Sj**. The surface integral of vector **A** over the surface** Sj** is given by

For simplicity, consider the whole volume is divided into elementary volumes I, II, and III as shown in figure 1. The outward of elementary volume I is inward of elementary volume II and the outward of elementary volume II is inward of elementary volume III and so on.

Therefore, the sum of integrals of elementary volumes will cancel each other and we are left with the surface integral which arises from the surface of **S**.

Multiply and divide left hand side of eqn. (1) by **Δ** **Vi** , we get

Now, let us suppose the volume of surface S is divided into infinite elementary volumes so that **Δ**Vi – 0

Now,

Hence eqn. (2) becomes

Since **Δ** Vi – 0, therefore Σ **Δ** **Vi **becomes integral over volume **V**

Which is the **Gauss divergence theorem**.

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