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.
What is Divergence Theorem?
Divergence Theorem states that,
"Surface integral of the normal of the vector point function P over a closed surface is equal to the volume integral of the divergence of P under the closed surface".
What is Divergence?
Divergence of a vector field is defined as the vector operation which results in the scalar field calculating the rate of change of flux. The divergence is denoted by the symbol ∇ or div(vector).
For a vector field P:
Divergence of P(x, y) in 2-D, P = P1i + P2j is given by:
∇P(x, y) = \frac{\partial P_1}{\partial x} + \frac{\partial P_2}{\partial y}
Divergence of P(x, y) in 3-D, P = P1i + P2j + P3k is given by:
∇P(x, y, z) = \frac{\partial P_1}{\partial x} + \frac{\partial P_2}{\partial y}+\frac{\partial P_3}{\partial z}
Learn more about, Divergence and Curl
The formula for the Divergence Theorem is given by:
\oiint\limits_S\overrightarrow{\rm P}.\overrightarrow{\rm n} .d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dv
Gauss Divergence Theorem
Gauss Divergence theorem gives us the relation between the surface integral of the vector to the volume of the vector in a closed surface. Below we will learn about the Gauss Divergence Theorem in detail.
Statement of Gauss Divergence Theorem
The Gauss Divergence Theorem states that:
"Surface integral of the vector field P over the closed surface is equal to the volume integral of the divergence of the vector over the closed surface".
It can be mathematically represented as:
\oiint\limits_S\overrightarrow{\rm P}.d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dV
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 d\overrightarrow{\rm S}
then the surface integral of the vector V over Sj is represented as \oiint\limits_S\overrightarrow{\rm P}.d\overrightarrow{\rm S_j}
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.
\sum\oiint\limits_{Sj}\overrightarrow{\rm P}.d\overrightarrow{\rm S_j}=\oiint\limits_{S}\overrightarrow{\rm P}.d\overrightarrow{\rm S}
------(1)
Now, we will multiply and divide equation (1) by ΔVi,
\oiint\limits_S\overrightarrow{\rm P}.d\overrightarrow{\rm S}=\sum\frac{1}{\Delta V_i}(\oiint\limits_{Si}\overrightarrow{\rm P}.d\overrightarrow{\rm S})\Delta V_i
Again, we consider that the volume of surface S is divided into infinite parts i.e., ΔVi → 0
\oiint\limits_S\overrightarrow{\rm P}.d\overrightarrow{\rm S}=lim_{\Delta V_i\rightarrow0}\sum\frac{1}{\Delta V_i}(\oiint\limits_{Si}\overrightarrow{\rm P}.d\overrightarrow{\rm S})\Delta V_i
--------(2)
Since,
lim_{\Delta V_i\rightarrow0}\big[\frac{1}{\Delta V_i}(\oiint\limits_{Si}\overrightarrow{\rm P}.d\overrightarrow{\rm S})\bigg] = (\overrightarrow{\rm V}.\overrightarrow{\rm P})
Putting above value in equation 2
\oiint\limits_S\overrightarrow{\rm P}.d\overrightarrow{\rm S}=\sum(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})\Delta V_i
As ΔVi→0, ∑ (ΔVi ) results in the volume integral V
\oiint\limits_S\overrightarrow{\rm P}.d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dV
Gauss's Divergence Theorem vs. Green's Theorem
Gauss Divergence Theorem and Green's Theorem both the theorems have their specific advantages.
|
|
Gauss Divergence Theorem deals with 3-D solid bounded by closed curve
| Green's Theorem deals with the 2-D figures bounded by simple closed curve
|
Gauss divergence theorem uses the volume integralin its results.
| Green's theorem uses the surface integral in its results.
|
\oiint\limits_S\overrightarrow{\rm P}.\overrightarrow{\rm n} .d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dv
| \oint\limits_C(Ldx~+~Mdy)~=~\iint\limits_D(\delta M/\delta x~-~\delta L/\delta y)dxdy
|
Divergence Theorem - Conclusion
In the above article we have discussed about the Divergence Theorem and Gauss Divergence theorem. After discussing these theorems, we can conclude that the surface integral of the vector field under a closed surface is equal to the volume integral of the divergence of the vector field under the closed region.
Read More,
Divergence Theorem Example
Example 1: Compute \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, F = (4x + y, y2 - cos x2 z, xz +ye3x) and 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 and 0 ≤ z ≤ 2.
Solution:
According to Divergence Theorem
\oiint\limits_S\overrightarrow{\rm P}.\overrightarrow{\rm n} .d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dv
The intervals given are: 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 and 0 ≤ z ≤ 2
First, we find div(F)
div(F) = (4 + 2y + x)
Now, putting the values of div(F) and intervals in the formula
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iiint\limits_{000}^{132}
(4 + 2y + x) dz dy dx
Integrate the above expression w.r.t x, y and z respectively.
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iint\limits_{00}^{13}
(8 + 4y + 2x) dy dx
Putting all the limits and integrating
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
42 +3
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
45
Example 2: Compute \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, F = (x2 + 4y, 4y - tan z, z +y) and 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1.
Solution:
According to Divergence Theorem
\oiint\limits_S\overrightarrow{\rm P}.\overrightarrow{\rm n} .d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dv
The intervals given are: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1
First, we find div(F)
div(F) = (2x + 4 + 1) = 2x + 5
Now, putting the values of div(F) and intervals in the formula
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iiint\limits_{000}^{111}
(2x + 5) dz dy dx
Integrate the above expression w.r.t x, y and z respectively.
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iint\limits_{00}^{11}
(2x + 5) dy dx
Putting all the limits and integrating
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
12 + 5(1)
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
6
Example 3: Compute \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, F = (x + y + z, y2, x3 + z3) and 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 and 0 ≤ z ≤ 2.
Solution:
According to Divergence Theorem
\oiint\limits_S\overrightarrow{\rm P}.\overrightarrow{\rm n} .d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dv
The intervals given are: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 and 0 ≤ z ≤ 2
First, we find div(F)
div(F) = (1 + 2y + 3z2)
Now, putting the values of div(F) and intervals in the formula
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iiint\limits_{000}^{122}
(1 + 2y + 3z2) dz dy dx
Integrate the above expression w.r.t x, y and z respectively.
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iint\limits_{00}^{12}
(2 + 4y + 8) dy dx
Putting all the limits and integrating
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
28(1)
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
28
Example 4: Compute \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, F = (2x + 3y + 4z, 2y2, 5x3 + z) and 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 3.
Solution:
According to Divergence Theorem
\oiint\limits_S\overrightarrow{\rm P}.\overrightarrow{\rm n} .d\overrightarrow{\rm S}= \oiiint\limits_V(\overrightarrow{\rm \nabla}.\overrightarrow{\rm P})dv
The intervals given are: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 3
First, we find div(F)
div(F) = (2 + 4y + 1) = 3 + 4y
Now, putting the values of div(F) and intervals in the formula
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iiint\limits_{000}^{113}
(3 + 4y) dz dy dx
Integrate the above expression w.r.t x, y and z respectively.
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}= \iint\limits_{00}^{11}
(9 + 12y) dy dx
Putting all the limits and integrating
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
15(1)
\oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}=
15
Practice Problems on Gauss Divergence Theorem
P1. By using the Divergence theorem, evaluate \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, \overrightarrow{\rm F}
= sin (πx) i + zy3 j + (z2 + 4x) k and S is the surface of the box with -1 ≤ x ≤ 2, 0 ≤ y ≤1 and 1 ≤ z ≤ 4. All the six sides are included in S.
P2. By using the Divergence theorem, evaluate \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, \overrightarrow{\rm F}
= yx2 i + (x y - 3x5) j + (x - 6y) k and S is the surface of the sphere with radius of sphere is 5, y ≤ 0 and z ≤ 0. All the three surfaces of solid are included in S.
P3. By using Gauss Divergence Theorem, compute \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, F = (x4 + 4y, 5y2 - cot x2 , xe3z) and 0 ≤ x ≤ 2, 1 ≤ y ≤ 5 and -1 ≤ z ≤ 1.
P4. By using Gauss Divergence Theorem, compute \oiint\limits_S\overrightarrow{\rm F}.d\overrightarrow{\rm S}
where, F = (x2 + y2 + z2, 5y3 - z4 , xz5e3y) and -1 ≤ x ≤ 3, 1 ≤ y ≤ 3 and -2 ≤ z ≤ 2.
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