Line Integral (original) (raw)

Last Updated : 10 Oct, 2025

A **Line Integral is used to evaluate a function along a curve or path. It helps calculate quantities like work or flux over a specific route, often applied in engineering. Line integrals are the reverse of differentiation, also known as anti-differentiation.

In generic words, a 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.

Line-Integral

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:

\oint f(\mathbf{r})\, ds = \int_{a}^{b} f[\mathbf{r}(t)] \, \lvert \mathbf{r}'(t) \rvert \, dt

Where,

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:

\oint \mathbf{F}(\mathbf{r}) \cdot d\mathbf{s} = \int_{a}^{b} \mathbf{F}[\mathbf{r}(t)] \cdot \mathbf{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,

\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C \langle P, Q, R \rangle \cdot \langle dx, dy, dz \rangle**∫ C

**OR

\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (P \, dx + Q \, dy + R \, dz)

r'(t) can also be written as,
\frac{d\mathbf{r}}{dt} \, dt = \left( \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} + \frac{dz}{dt} \hat{k} \right)

\mathbf{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}

For any function \mathbf{F} = A \hat{i} + B \hat{j} + C \hat{k},

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:

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:

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

**Also Check

Solved Examples on Line Integral

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

**Solution:

Given the integral:

∫C(x+y) ds

along the curve x2 + y2 = 9 from (0, 3) to (3,0):

x = 3cos⁡ t, y = 3 sin⁡ t

with t going from π/2 (at (0, 3)) to 0 (at (3,0)).

\frac{dx}{dt} = -3 \sin t, \quad \frac{dy}{dt} = 3 \cos t

ds = \sqrt{(-3 \sin t)^2 + (3 \cos t)^2} \, dt = 3 \, dt

0∫π/2 (3cos⁡ t + 3sin ⁡t)⋅3 dt = 9 0∫π/2 (cos ⁡t + sin ⁡t) dt

= −9 π/2∫0 (cos⁡ t + sin ⁡t)

π/2∫0 cos⁡ t dt = 1, π/2∫0 sin⁡ t dt = 1

⇒−9(1 + 1) = −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:

Calculate the work done by the force field f(x, y) = (y, 0) on the particle moving along (x = sin(t)), (y = cos(t)) from (t = 0) to (t = 1).

Step 1: Parametrize the force field along the path: F(t) = (cos(t), 0)

Step 2: Calculate the derivatives of the path: \frac{dx}{dt} = \cos(t), \quad \frac{dy}{dt} = -\sin(t)

Step 3: The work done is the line integral:

W = \int_0^1 \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} dt = \int_0^1 \cos(t) \cdot \cos(t) + 0 \cdot (-\sin(t)) dt = \int_0^1 \cos^2(t) dt

Step 4: Use the identity:

\cos^2(t) = \frac{1 + \cos(2t)}{2}

Step 5: The integral becomes:

W = \int_0^1 \frac{1 + \cos(2t)}{2} dt = \frac{1}{2} \int_0^1 (1 + \cos(2t)) dt = \frac{1}{2} \left[ t + \frac{\sin(2t)}{2} \right]_0^1 = \frac{1}{2} \left(1 + \frac{\sin(2)}{2}\right)

Numerical value:

W \approx 0.5 + \frac{0.909297}{4} = 0.7273

**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.

**Solution:

Given:

**Step 1: Express F along the curve r(t)

We have F=⟨z,y,x⟩

Along the curve:

So: F(r(t))=⟨t3, t, t2⟩

**Step 2: Find r′(t)

r(t) = ⟨t2, t, t3⟩

\mathbf{r}'(t) = \left\langle \frac{d}{dt}(t^2), \frac{d}{dt}(t), \frac{d}{dt}(t^3) \right\rangle = \langle 2t, 1, 3t^2\rangle

**Step 3: Write the line integral

\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \ dt

Calculate the dot product:

F(r(t)) ⋅ r′(t) = ⟨t3, t, t2⟩⋅⟨2t,1,3t2⟩ = t3⋅2t + t⋅1 + t2⋅3t2

= 2t4 + t + 3t4 = (2t4 + 3t4) + t = 5t4 + t

**Step 4: Integrate from 0 to 1

0∫1​ (5t4 + t)dt = 0∫1​ 5t4dt + 0∫1​ tdt = 5 0∫1​ t4dt + 0∫1​ tdt

= 5 \left[ \frac{t^5}{5} \right]_0^1 + \left[ \frac{t^2}{2} \right]_0^1 = 5 \cdot \frac{1}{5} + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}a