# Line Integral

In Calculus, a line integral is an integral in which the function to be integrated is evaluated along a curve. A line integral is also called the path integral or a curve integral or a curvilinear integral. In this article, we are going to discuss the definition of the line integral, formulas, examples, and the application of line integrals in real life.

## Line Integral Definition

A line integral is integral in which the function to be integrated is determined along a curve in the coordinate system. The function which is to be integrated may be either a scalar field or a vector field. We can integrate a scalar-valued function or vector-valued function along a curve. The value of the line integral can be evaluated by adding all the values of points on the vector field.

## Line Integral Formula

The line integral for the scalar field and vector field formulas are given below:

**Line integral Formula for Scalar Field**

For a scalar field with function f: U ⊆ R^{n} → R, a line integral along with a smooth curve, C ⊂ U is defined as:

∫_{C} f(r) ds = \(\int_{a}^{b}\) f[r(t)] |r’(t)| dt

Here, r: [a, b]→C is an arbitrary bijective parametrization of the curve.

r (a) and r(b) gives the endpoints of C and a < b.

**Line integral Formula for Vector Field**

For a vector field with function, F: U ⊆ R^{n} → R^{n}, a line integral along with a smooth curve C ⊂ U, in the direction “r” is defined as:

∫_{C} F(r). dr = \(\int_{a}^{b}\) F[r(t)] . r’(t)dt.

Here, “.” represents the dot product.

## Application of Line Integral

Line integral has several applications. A line integral is used to calculate the surface area in the three-dimensional planes. Some of the applications of line integrals in the vector calculus are as follows:

- A line integral is used to calculate the mass of wire.
- It helps to calculate the moment of inertia and centre of mass of wire.
- It is used in Ampere’s Law to compute the magnetic field around a conductor.
- In Faraday’s Law of Magnetic Induction, a line integral helps to determine the voltage generated in a loop.
- Line integral helps to calculate the work done by a force on a moving object in a vector field.

### Line Integral Example

Go through the line integral example given below:

**Example:** Evaluate the line integral ∫_{C} F. dr where F(x, y, z) = [P(x, y, z), Q(x, y, z), R(x, y, z)] = (z, x, y), and C is defined by the parametric equations, x = t^{2}, y = t^{3} and z = t^{2} , 0 ≤ t ≤ 1.

**Solution:**

Given that, the function, F(x, y, z) = [P(x, y, z), Q(x, y, z), R(x, y, z)] = (z, x, y)

Parametric equations: x = t^{2}, y = t^{3} and z = t^{2} , 0 ≤ t ≤ 1.

We know that,

∫_{C} F. dr = ∫_{C} P dx + Q dy + R dz

∫_{C} F. dr = \(\int_{0}^{1}\) z(t) x’(t)dt + x(t) y’(t)dt + y(t) z’(t)dt

= \(\int_{0}^{1}\) t^{2} (2t)dt + t^{2} (3t^{2})dt + t^{3} (2t) dt

= \(\int_{0}^{1}\) 2t^{3} dt + 3t^{4} dt + 2t^{4}dt

= \(\int_{0}^{1}\) (5t^{4} + 2t^{3}) dt

= \(\left ( 5\frac{t^{5}}{5}+2\frac{t^{4}}{4} \right )_{0}^{1}\)

Substitute the limits, we get,

∫_{C} F. dr = 3/2

Therefore, the line integral for the given function is 3/2.

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