Divergence and Curl in Mathematics (Definition and Examples) (2024)

In Mathematics, divergence is a differential operator, which is applied to the 3D vector-valued function. Similarly, the curl is a vector operator which defines the infinitesimal circulation of a vector field in the 3D Euclidean space. In this article, let us have a look at the divergence and curl of a vector field, and its examples in detail.

Table of Contents:

• Divergence and Curl Definition
• Divergence of a Vector Field
• Curl of a Vector Field
• Examples
• Practice Questions
• FAQs

Divergence and Curl Definition

In Mathematics, divergence and curl are the two essential operations on the vector field. Both are important in calculus as it helps to develop the higher-dimensional of the fundamental theorem of calculus. Generally, divergence explains how the field behaves towards or away from a point. Similarly, curl is used to measure the rotational extent of the field about a particular point.

Divergence of Vector Field

The divergence of a vector field is a scalar field. The divergence is generally denoted by “div”. The divergence of a vector field can be calculated by taking the scalar product of the vector operator applied to the vector field. I.e., ∇ . F(x, y).

If F(x, y) is a vector field in the two dimensions, then its divergence is given by:

\(\begin{array}{l}\triangledown . F(x, y)=\left ( \frac{\partial i}{\partial x} + \frac{\partial j}{\partial y} \right ).(F_{1(x,y)}i +F_{2(x,y)}j )\end{array} \)

\(\begin{array}{l}\triangledown . F(x, y)= \frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}\end{array} \)

The divergence of a vector field can be extended to three dimensions and it is given as follows:

I.e., F(x, y, z) = F₁i + F₂j + F₃k

\(\begin{array}{l}\triangledown . F(x, y, z)= \frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}\end{array} \)

Curl of a Vector Field

The curl of a vector field is again a vector field. The curl of a vector field is obtained by taking the vector product of the vector operator applied to the vector field F(x, y, z).

I.e., Curl F(x, y, z) = ∇ × F(x, y, z)

It can also be written as:

\(\begin{array}{l}\triangledown\times F(x, y, z)= \left ( \frac{\partial F_{3}}{\partial y}- \frac{\partial F_{2}}{\partial z}\right )i – \left ( \frac{\partial F_{3}}{\partial x}- \frac{\partial F_{1}}{\partial z}\right )j+\left ( \frac{\partial F_{2}}{\partial x}- \frac{\partial F_{1}}{\partial y}\right )k\end{array} \)

\(\begin{array}{l}\triangledown\times F(x, y, z)= \begin{vmatrix}i & j & k \\\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\F_{1} & F_{2} & F_{3} \\\end{vmatrix} \end{array} \)

Divergence and Curl Examples

Example 1:

Determine the divergence of a vector field in two dimensions: F(x, y) = 6x²i + 4yj.

Solution:

Given: F(x, y) = 6x²i + 4yj.

We know that,

\(\begin{array}{l}\triangledown . F(x, y)= \frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y}\end{array} \)

\(\begin{array}{l}\triangledown . F(x, y)= \frac{\partial (6x^{2})}{\partial x}+\frac{\partial (4y)}{\partial y}\end{array} \)

\(\begin{array}{l}\triangledown . F(x, y)= 12x + 4\end{array} \)

Example 2:

Find the divergence of a vector field in three dimensions: F(x, y, z) = x²i + 2zj – yk.

Solution:

Given: F(x, y, z) = x²i + 2zj – yk

As we know,

\(\begin{array}{l}\triangledown . F(x, y, z)= \frac{\partial F_{1}}{\partial x}+\frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}\end{array} \)

\(\begin{array}{l}\triangledown .F(x, y, z)= \frac{\partial (x^{2})}{\partial x}+\frac{\partial(2z)}{\partial y} + \frac{\partial (-y)}{\partial z}\end{array} \)

\(\begin{array}{l}\triangledown .F(x, y, z)= 2x + 0 + 0 = 2x.\end{array} \)

Example 3:

Find the curl of the vector field F(x, y, z) = y³i + xyj – zk.

Solution:

Given: F(x, y, z) = y³i + xyj – zk.

Here, F₁ = y³, F₂ =xy, F₃ = -z

We know that,

\(\begin{array}{l}\triangledown\times F(x, y, z)= \left ( \frac{\partial F_{3}}{\partial y}- \frac{\partial F_{2}}{\partial z}\right )i – \left ( \frac{\partial F_{3}}{\partial x}- \frac{\partial F_{1}}{\partial z}\right )j+\left ( \frac{\partial F_{2}}{\partial x}- \frac{\partial F_{1}}{\partial y}\right )k\end{array} \)

\(\begin{array}{l}\triangledown\times F(x, y, z)= \left ( \frac{\partial (-z)}{\partial y}- \frac{\partial (xy)}{\partial z}\right )i – \left ( \frac{\partial (-z)}{\partial x}- \frac{\partial (y^{3})}{\partial z}\right )j+\left ( \frac{\partial (xy)}{\partial x}- \frac{\partial (y^{3})}{\partial y}\right )k\end{array} \)

= 0i – 0j +(y-3y²) k

= (y-3y²)k.

Therefore, the curl of the vector is in the direction of “k”.

Practice Questions on Divergence and Curl

Solve the following problems.

1. Compute the divergence of the vector field F(x, y) = y³i + xyj.
2. Calculate the divergence of the vector field F(x, y, z) = x²i +2zj -yk.
3. Find the curl of the vector field F(x, y, z) = y³i +xyj -zk.