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 ΔVi – 0
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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