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".

Divergence

Divergence of a vector field is defined as the vector operation that 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}

Divergence Theorem Formula

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 V_j bounded by the surface S_j of area d\overrightarrow{\rm S} then the surface integral of the vector V over S_j 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 S_i is the inward volume of S_i+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 ΔV_i→0, ∑ (ΔV_i ) 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

Related Articles

• Scalars and Vectors
• Vector Calculus
• Vector Algebra

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 +ye^3x) 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 , xe^3z) 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 , xz⁵e^3y) and -1 ≤ x ≤ 3, 1 ≤ y ≤ 3 and -2 ≤ z ≤ 2.