Line Integrals: How to Integrate Functions Over Paths

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: The Fundamental Theorem of Calculus

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
Your next lesson will play in 10 seconds
  • 0:04 Making a Coil
  • 0:52 Line Integration
  • 2:30 Using Parametrizations
  • 4:42 Example
  • 5:38 Lesson Summary
Save Save Save

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Speed Speed

Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Aaron Miller

Aaron teaches physics and holds a doctorate in physics.

Many real-world functions are three dimensional, as we live in a 3D world. In this article, you will learn how to integrate 3D functions over general paths through space. This is a basic skill needed for real science and engineering applications.

Making a Coil

Imagine you are designing a coil. You know roughly how long it needs to be, its radius, and approximately how many turns it makes around its axis. But without an actual model to uncoil into a straight wire and measure, could you predict how much metal you would need for your design?

Whether you are trying to predict the length or the mass of metal needed, answering this question requires some kind of integral that adds up contributions from small segments of the coil. However, because it is a 3D object, integration techniques from single-variable calculus cannot produce an answer.

A prediction would require you to evaluate a line integral, an integral of a function defined on a 3D curve in space. In this article, we will introduce the machinery needed to perform a line integral, and we will show you how to set up and evaluate such an integral in practice.

Line Integration

Evaluation of a line integral, or path integral, begins with defining the path of integration, denoted C, which stands for 'curve.' The path of integration is a continuous curve through a space along which a given function f is integrated. A line integral has the same geometric interpretation as an ordinary integral from single-variable calculus; it's the total area under f. However, the difference is that the integration is no longer constrained to a coordinate axis (i.e., it's not the area between f(x) and the x axis, but rather, the area between f and curve C).

Line integration is most easily visualized for a 2D function f(x, y). Such a function is depicted as a surface with height z = f(x, y) above the xy plane.

Equation 1: Notation for a line integral

You can see a curve C in the xy plane, and the line integral is defined as the area under f directly above C. The conventional notation for the line integral of f over C involves an integration measure ds, rather than the dx you see in single-variable calculus.

Recall that dx in a single-variable integral is defined as an infinitesimal length along the x axis. Analogously, ds is an infinitesimal arc length along the curve C. If f is given as a function of arc length s, then you can evaluate this integral directly using techniques from single-variable calculus. However, in applications, f(s) is rarely known, and f is more likely to be given as a function of coordinates, for example f(x, y).

How do we find an actual value for a line integral? Using a construction called parametrization.

Using Parametrizations

A parametrization of a curve assigns values of a parameter t along the integration path C by specifying functions of t for each coordinate, x(t), y(t), etc. These functions are constructed in such a way that the curve C is traced out by these functions as t varies over a defined interval. In Figure 1, you can see a parametrization of the curve y = x^2 between (0, 0) and (2, 4) as t varies between 0 and 1.

Figure 1: A parametrization of y=x^2.

Here is the mathematical statement of the same parametrization:

Equation 2: A parametrization of y=x^2

You can check that for a given value of t in the interval [0, 1] the ordered pair (x(t), y(t)) is always a point on the curve y = x^2. For example, if t = 0.25, then x(0.25) = 0.5 and y(0.25) = 0.25, so that y(0.25) = (x(0.25))^2. This is how we know the above parametrization is for the curve y = x^2 and not some other curve.

To unlock this lesson you must be a Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use

Become a member and start learning now.
Become a Member  Back
What teachers are saying about
Try it now

Earning College Credit

Did you know… We have over 220 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Used by over 30 million students worldwide
Create an account