Line Integral

Line Integral is the integral that is evaluated along a line or a curve. Generally, an integral is calculated when we need to determine a quantity’s value such as area, volume, temperature, etc. for a larger part of the body when we have an expression for a smaller part. It is the opposite of differentiation and is also called an anti-derivative of the function. The Line Integral is used in engineering in various fields when we need to determine a function’s value along a line or along a curve. For example, if we need to calculate work done on an electron by a force field along a curve, we can determine it using line integral.

In this article, we will learn about the definition of line integral, its formula, applications, some solved examples based on the calculation of line integral, and some frequently asked questions related to line integral.

Definition of Line Integral

Line Integral is the integral of a function calculated along a line or a curve. It is also known as path integral, curve integral, or curvilinear integral. In generic words, line integral is the sum of the function’s value at the points in the interval on the curve along which the function is integrated.

Formula of Line Integral

Line Integral is calculated for both types of functions/fields, i.e. scalar fields and vector fields. The formula for two cases is written separately as follows:

For scalar Fields

Let f defines a scaler field, then the line integral along a smooth curve C is defined as follows:

∮ f(r)ds = ∫_a^bf[r(t)]|r'(t)|dt

Where,

• f is the Scalar Field
• r(t) is the Parametrical Representation of Curve along which Integral is to be Evaluated
• |r'(t)| represents the Magnitude of Derivative of the Parametric Representation

The LHS of above equation represents the line integral of the function f along curve C. a and b represent the lower limit and upper limit of integration respectively. r is an arbitrary parametrical representation of the curve C.

For Vector Fields

Let F defines a vector field and the curve along which the line integral is to be calculated is C. Then, the expression to calculate the line integral is as follows:

∮ F(r) . ds = ∫_a^b F[r(t)] . [r'(t)]dt

Here, a and b are limits of integration, r is the parametrical representation of the curve C, and ‘ . ‘ represents the dot product between the vectors.

Line Integral in Differential Form

In differential form, a line integral can be expressed as a mathematical expression involving differential terms and functions. For example, a function is represented as F = <P, Q, R>, where P, Q and R are function’s expressions in X, Y and Z directions respectively. Let ‘dr’ represents the differential displacement along the curve C along which integral is to be found, then line integral would be,

∫_C F . dr = ∫_C <P, Q, R> . <dx, dy,d z>

OR

∫_C F . dr = ∫_C (Pdx + Qdy + Rdz)

r'(t) can also be written as,

dr/dt(dt) = (dx/dt.i + dy/dt.j + dz/dt.k).dt

dr = dx.i + dy.j + dz.k

For any function F = Ai + Bj + Ck

F.r'(t) = M.dx + N.dy + P.dz

Evaluating Line Integral

Line Integral can be evaluated using the formulae defined above for a particular problem depending upon whether it is for scalar field or vector field. The basic steps involved in evaluating line integral are listed as follows:

• As a first step, we need to parameterize the curve along which integral is to be found out in specified range, if not given in that form already.
• Find the differentials for the parametric curve obtained, as dx, dy and dz.
• Select the appropriate formula based upon scalar or vector fields.
• Substitute the parametric function and it’s derivative into the formula.
• Evaluate the dot product for vector fields or simply multiply for scalar ones.
• Integrate the obtained expression within specified limits to get the value of line integral.

Fundamental Theorem for Line Integrals

According to Fundamental Theorem for Line Integrals

∫_a^b F'(x) . dx = F(b) – F(a)

Applications of Line Integral

As already mentioned, a Line Integral is used to find a function’s integral along a line or a curve. Here, some practical applications of line integrals are listed as follows:

• It may be used in finding work done by gravitational force to move a particle along a specific path defined by a curve.
• It can be used in finding work done by a force on a moving object.
• It is used in determining rate of reaction as per Law of Mass Action.
• It is also used to calculate the value of magnetic field around a conductor as per Ampere’s Law.
• It is also used to find the voltage generated in a loop due to electromagnetic induction as per Faraday’s Law.

There are many more applications of Line Integrals which one may find in various engineering fields. Below are some solved examples which are involve calculation of line integral. You may refer them to understand the concept of line integral in a better way.

Line Integrals of Vector Valued Functions

Let, r be a vect function defined in t such that,

r(t) = x(t)i + y(t)j a ≤ t ≤ b

It r(t) is differentiable on a smooth curve C then,

∫_C f(x, y) . dr = ∫_a^b f{x(t), y(t)}.√{(x'(t)² + y'(t)²} dt

if, r(t) = x(t)i + y(t)j + x(t)k a ≤ t ≤ b

∫_C f(x, y) . dr = ∫_a^b f{x(t), y(t), z(t)}.√{(x'(t)² + y'(t)²+ z'(t)²} dt

Read More,

• Integration by Parts
• Integration by Substitution Formula
• Integration by Partial Fractions

Examples on Line Integral

Example 1: Evaluate the integral ∮(x+y).ds along the curve x² + y² = 9, from (0,3) to (3,0).

Solution:

Let us parametrically define the curve x² + y² = 9 as x = 3sin(t) and y = 3cos(t). The lower limit and upper limit of the integral in terms of t, then becomes, t = 0 to t = pi/2.

Now, r(t) = ( 3sin(t), 3cos(t) ) , and r'(t) = ( 3cos(t), -3sin(t) )

⇒ |r'(t)| = √((3sin(t))² + (3cos(t))²) = 3

As per the formula, ∮ f(r)ds = ∫_a^bf[r(t)]|r'(t)|dt

∮(x+y)ds = ∫₀^pi/2 (3sin(t) + 3cos(t))(3)dt = 9 × [-cos(t) + sin(t)]₀ ^pi/2 = 9 × 2 = 18

Thus, the required value of line integral comes out to be 18.

Example 2: A force field is represented as following function, F(x,y) = (y,0), find the value of work done by the force on a particle moving in the direction described as x = sin(t) and y = cos(t), when t varies from 0 to 1.

Solution:

Work done by a force F on a moving particle is calculated as,

∮ F(r) . ds = ∫_a^b F[r(t)] . [r'(t)]dt

Here,

r(t) = ( sin(t) ,cos(t) ) and r'(t) = ( cos(t), -sin(t) )

On substituting these values in the expression above, we get,

∮ F(r) . ds = ∫₀¹ ( cos(t), 0) . ( sin(t), cos(t)) dt = ₀ ∫¹ (sin(t) cos(t)) dt = (1/2) ₀ ∫¹ sin(2t)dt

⇒ ∮ F(r) . ds = (1/4) × [cos(0) – cos(2)] = (1/4) × (1.42) = 0.355

Thus, the value of work done comes out to be 0.355 units.

Example 3: For a vector field represented as F = <z,y,x>, evaluate the line integral ∫_C F.dr along the curve C parameterized as < t², t, t³ > for 0 < t < 1.

We know that,

∫_C F . dr = ∫_C <P,Q,R> . <dx,dy,dz>

Here, <P,Q,R> = <z,y,x> and dr = <dx,dy,dz> or dr = <2tdt, dt, 3t²dt>

On substituting these values in above expression, we get,

∫_C F . dr = ∫₀ ¹ (2t⁴ + t + 3t⁴) dt = 7/5

Thus, the required value of line integral comes out to be 7/5.

Line Integral – FAQs

1. What is Line Integral in Calculus with Example?

Line Integral is the integral of a function evaluated along a line or a curve. It is also known as curve integral, path integral or curvilinear integral. For example, if we need to find work done by a force on a moving object, then it can be found be evaluating the line integral of the force field along direction of motion of the object.

2. What Type of Fields, Line Integral can be Calculated?

Line Integral is calculated for both types of functions or fields, i.e. scalar field and vector field. There are separate expressions to determine the value of integral in both cases.

3. What is Line Integral for a Scalar Field?

The expression for determining value of line integral of a scalar function is defined as follows,

∮ f(r)ds = _a ∫^bf[r(t)]|r'(t)|dt

4. What is Line Integral for a Vector Field?

The expression to determine the value of line integral of a scalar function is defined as follows,

∮ F(r) . ds = _a ∫^b F[r(t)] . [r'(t)]dt

5. What is the Difference between Line Integral and Surface Integral?

• Line integral is used to determine a function’s integral along a line or a curve.
• Surface integral is used to determine a function’s integral over a surface.

6. Can a Line Integral be Zero?

Yes, the value of a line integral can be zero. Its value depends upon the function being integrated and the path along which the function is integrated. Therefore, if the function is such that its components in some particular x,y and z directions cancel out each other, then line integral along that path comes out to be zero.

7. When Can a Line Integral be Negative?

The sign of a line integral depends on the orientation of the function and the curve along which function is to be integrated. So, when the curve is traversed in opposite direction with respect to the function to be integrated, the value of line integral comes out to be negative.

8. What is ‘dr’ in Line Integral?

The ‘dr’ signifies a differential displacement vector in the direction of the curve along which value of integral is to be evaluated. As a vector, dr can be expressed as <dx,dy,dz>, which represent infinitesimal displacement along coordinate axes in direction of the curve of line integral.