A line integral (also called path integral) is an integral where we integrate a function of two or three variables over a curve C. It is a generalization of the usual one variable integration.

A line integral of two variables is of the form \int_{C}Pdx +Qdy where, C is the curve over which we want to integrate and P=P(x,y) \text{ and } Q=Q(x,y) are functions of the two variables x and y.

We now explain the procedure to evaluate a line integral over a given curve C and then give some examples.

### How to evaluate/calculate line integrals?

- Parametrize the curve C in terms of a single variable, say ‘t’. For example, if the curve is the unit circle on the plane, x^2 + y^2 = 1, we can parametrise it as x=sin(t), y=cos(t) \text{ for } 0\leq t\leq 2\pi.
- Substitute the values of “x” and “y” in the integral.
- Express “dx” and “dy” in terms of “dt”.
- Evaluate the integral by treating it as a usual one variable integral in terms of “t”.

To summarise, we should convert the line integral to an ordinary integral with respect to the parameter and evaluate it. Let us now look at some examples of evaluating line integrals.

#### Example 1 of evaluating line integrals:

Evaluate the line integral \int_{C}xydx+(x/y)dy, where C is the given curve, C: x = t^2, y = 2t, 0 ≤ t ≤ 1.

** Solution**: We have that, x(t) = t^2 \implies dx=2t .

Also, y(t) = 2t \implies dy=2 .

Substituting the values of x, y, dx and dy in terms of t we get,

Required line integral = \int_{0}^{1} [2t^3*2t + (t/2)*2] dt \\ = \int_{0}^{1} [4t^4 + t] dt \\ = [4(t^5/5) + (t^2/2)]_0^1 = 0.8+0.5 =1.3 .

#### Example 2 of evaluating line integrals:

Evaluate the line integral \int_{C}y^3dx+xydy, where C is the given curve, C: x = t^3, y = t, 0 ≤ t ≤ 3.

** Solution**: We have that, x(t) = t^3 \implies dx=3t^2 .

Also, y(t) = t \implies dy=1 .

Substituting the values of x, y, dx and dy in terms of t we get,

Required line integral = \int_{0}^{3} [t^3*3t^2 + t^4*1] dt \\ = \int_{0}^{3} [3t^5 + t^4] dt \\ = [3(t^6/6) + (t^5/5)]_0^3 = 364.5+48.6 = 413.1.

### Evaluating line integrals of the form \int_{C} F(x,y)ds

We can evaluate integrals of the form \int_{C} F(x,y)ds by putting ds= \sqrt{x'(t)^2 + y'(t)^2}dt .

Here ds represents the arc length element,

**Example**: Evaluate the line integral where C is the given curve: \int_{C}(x/y)ds C: x = t^3, y = t^4, 0 ≤ t ≤ 4.

** Solution**: We have that, x(t) = t^3 \implies dx=3t^2 .

Also, y(t) = t^4 \implies dy=4t^3 .

Substituting the values of x, y, and ds in terms of t we get,

Required line integral = \int_{0}^{4} (1/t)\sqrt{3t^2+4t^3} dt \\ = \int_{0}^{4} \sqrt{\frac{3t^2+4t^3}{t^2}} dt \\ = \int_{0}^{4} (3+4t)^{1/2} dt = [\frac{(3+4t)^{3/2}}{3/2}]_0^4 = 12.937 .