[Maths Class Notes] on Line Integral Pdf for Exam [2024]

In calculus, a line integral is represented as an integral in which a function is to be integrated along a curve. A line integral is also known as a path integral, curvilinear integral, or curve integral. Line integrals have several applications such as in electromagnetic, line integral is used to estimate the work done on a charged particle travelling along some curve in a force field defined by a vector field. In classical mechanics, line integral is used to compute the word performed on mass m moving in a gravitational field. In this article, we will study a line integral, line integral of a vector field, line integral formulas, etc.

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Line Integral Definition

A line integral is an integral in which a function is integrated along some curve in the coordinate system. The function which is to be integrated can either be represented as a scalar field or vector field. We can integrate both scalar-valued function and vector-valued function along a curve. The value of the vector line integral can be evaluated by summing up all the values of the points on the vector field.

Line Integral of the Vector Field

A line integral (also known as path integral) is an integral of some function along with a curve. One can also incorporate a scalar-value function along a curve, obtaining such as the mass of wire from its density. We can also incorporate certain types of vector-valued functions along a curve. These vector-valued functions are the ones whose input and output size are similar and we usually define them as vector fields.

The line integral of the vector field is also interpreted as the amount of work that a force field does on a particle as it moves along a curve.

Application of Line Integral

Line integrals can be used to find the three-dimensional surface areas. It is an extension of simple integrals and is mostly applicable for curvy surfaces.
In the field of classical mechanics, line integrals are used to calculate the work done by an object of mass m, moving in a gravitational field.
Also, if one wants to figure out how many calories a swimmer might burn in swimming along a certain route, provided the currents in all areas can be accurately predicted. The total work that he needs to do would vary upon the strength and direction of the current. So, a line integral over his route will help to determine the total work done or calories that a swimmer will burn in swimming along the desired path.
In electrical engineering, line integrals are used to determine the exact length of power cable needed to connect two substations that could be miles away from each other.
Space flight engineers regularly use line integrals for long missions. While launching exploratory satellites, they consider the path of the different orbiting velocities of Earth and the planet the probe is targeted for.
Line integrals are also used to find the velocity and trajectory of an object, predict the position of the planets, and understand electromagnetism in depth.
They are also used in statistics to evaluate survey data and help draw out useful strategies. In chemistry, line integrals are used to determine the rate of reaction and know some necessary information regarding radioactive decay reactions.
A line integral can also be used to calculate the mass of a wire, its moment of inertia, and the center of mass of the wire. It is also used to calculate the magnetic field around a conductor when using Ampere’s law.
Ampere’s law states that the line integral of a magnetic field B around a closed path ‘C’ is equal to the total current flowing through the area bounded by boundary ‘C’.

Step-by-Step Guide to solving Line Integrals

Identify the function f(x,y,z) in the given function and the curve ‘C’ over which the integration will take place. If it is a problem involving the work done on an object, then f(x,y,z) represents the force on the object.
Determine its parametric equations which are represented as x(t), y(t), z(t). The given equation would be a function of x, y, and z. So, it first needs to be translated into its parametric form x(t), y(t), and z(t).

For example, the equation of a circle is given as,
x² + y² = r².
Its parametric form would be,
x(t)= rcos(t) and y(t)= rsin(t), where t: 0→2π.

However, after determining the parametric equation, you move along the curve in the opposite direction as ‘t’ increases, the value of the line integral is multiplied by -1.

Difference between Line Integrals and Definite Integrals

Definite integral involves infinite summation of infinitesimal elements between two given boundaries called limits. It is an open integral which means the starting and ending boundaries need not be the same. It can be an integration of over a line, surface, volume, etc.

Line integral on the other hand is a closed integral which has a particular direction of travel in the direction of the given function. Most line integrals are definite integrals but the reverse is not necessarily true. Examples of line integrals are stated below. This will help you understand the concept more clearly.

Line Integral Examples with Solutions

The line integral example given below helps you to understand the concept clearly.

1. Find the line integral of

[_{c}int (1+x^2y)ds]

Where C is considered as an ellipse

r(t) = (2cos t) + (3sin t)

for 0 ≤ t ≤ 2π

Solution:

We calculate,

ds =[sqrt{(-2sint)^2+(3cost)^2dt}]

= [sqrt{4sin^2t + 9cost^2t}]

We have the integral

[_{0}int^{2pi} (1+(2cost)^2(3sint))] [sqrt{4sin^2t + 9cost^2t}]

Hence, we get line integral = 15.87.

2. Evaluate [{_c}int 4x^3 ds]

where ‘C’ is the line segment from (1,2) to (-2,-1).

Solution:

Here is the parameterization of the curve

r(t) = (1-t) (1, 2) + t (-2,-1)

( 1-3t, 2 – 3t)

For, 0≤ t ≤ 1.

Note: we are changing the direction of the curve and this will also change the parameterization of the curve, so we can ensure that we start/end at the proper point.

Here, you can see the line integral

[{_c}int4x^3ds]=[{_0}int^{1} 4(1-3t)^3] [sqrt{9+9dt}] (substituted the parametric form)

= [12sqrt{2(-1/12)(1-3t)^4 {_0}int{^1}}] (integrated the term (1-3t)³)

= [12sqrt{2(-1/12)[1-3(1)]^4 -[1-3(0)]^4}] (applying the limits)

= [12sqrt{2(-1/12)[16-1]}]

= [12sqrt{2(-5/4)}]

= [-15sqrt{2}]

= -21.213

Quiz Time

1. The line integral is used to calculate

Force
Length
Area
Volume

2. The integral form of potential and field relation is given by the line integral.

True
False

3. The value of c∫dI along a circle of radius 2 units is

A. Zero

B. 2π

C. 4π

D. 8π

Tips to understand Line Integrals:

Understanding line integrals takes time. It isn’t like you read this post and have fully understood line integrals. Read and reread the above article and make your own revision notes.

Also, go through solved examples and understand the concept applied at each step. Once students understand the fundamentals, then it becomes easier to solve problems of line integrals.

Practice often. Every problem of line integral is different and requires consistent practice to understand the chapter well. The more you practice different kinds of problems, the easier it would be to write during the exams.