Proof of Gauss Divergence Theorem
To prove Gauss Divergence theorem, consider a surface S with volume V. It has a vector field P in it. Let the total volume of solid consists of different small volumes of parallelepipeds.
Since the solid is divided into small elementary volumes, let’s take one of them Vj bounded by the surface Sj of area then the surface integral of the vector V over Sj is represented as
The total volume is divided into parts volume I, II, III …. The outward volume of Si is the inward volume of Si+1. So, these elementary volumes cancel each other, and we get the surface integral derived by surface S.
——(1)
Now, we will multiply and divide equation (1) by ΔVi,
Again, we consider that the volume of surface S is divided into infinite parts i.e., ΔVi → 0
——–(2)
Since,
Putting above value in equation 2
As ΔVi→0, ∑ (ΔVi ) results in the volume integral V
Divergence Theorem
Divergence Theorem is one of the important theorems in Calculus. The divergence theorem relates the surface integral of the vector function to its divergence volume integral over a closed surface.
In this article, we will dive into the depth of the Divergence theorem including the divergence theorem statement, divergence theorem formula, Gauss Divergence theorem statement, Gauss Divergence theorem formula, and Gauss Divergence Theorem proof.
We will also go through some points on Gauss’s Divergence theorem vs Green’s theorem, solve some examples, and answer some FAQs related to the divergence theorem.
Table of Content
- What is Divergence Theorem?
- Divergence Theorem Formula
- Gauss Divergence Theorem
- Proof of Gauss Divergence Theorem
- Gauss’s Divergence Theorem vs. Green’s Theorem
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