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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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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.

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.

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.

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

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

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.

Here is the mathematical statement of the same parametrization:

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.

You can construct many parametrizations of the same curve. The next question is, how does a parametrization help us evaluate line integrals? As we said above, a parametrization represents integration path *C* as several coordinate functions of a single parameter *t*. Therefore, the integrand *f(x, y)* of a line integral can be recast as a single-variable function on the curve by substituting a parametrization's coordinate functions:

This is the beauty of constructing a parametrization. We have represented a multidimensional function as a single-variable function on a specific curve *C*. However, our work is not quite finished. In order to use single-variable integration techniques, we need to change the variable *ds* to *dt* using the definition of arc length, which is:

Now we can finally obtain a single-variable integral by substituting into the original definition of the line integral. The full equation for a line integral over a curve in the *xy* plane in terms of a parametrization *x(t)* and *y(t)* is:

If the curve is not planar, but is fully three-dimensional, the equation for the line integral is in terms of three parametrization functions: *x(t)*, *y(t)*, and *z(t)*.

Let's integrate the following:

*C* is the curve *y* = *x*^2 from (0, 0) to (2, 4). In the previous section, we already provided a parametrization of this curve. We now need to plug the parametrization's coordinate functions into Equation 4 above and integrate.

First we transform the integrand by substituting the functions from Equation 2:

Then we take the derivatives of *x(t)* and *y(t)* to plug into Equation 4:

Therefore, the line integral transforms into the single-variable integral of *t* over the interval 0 to 1 (since the parametrization in Equation 2 is defined from 0 to 1). The integral can be evaluated:

What if you didn't have a parametrization? In some cases, you might have to be creative and make up your own. Rest assured, no matter what parametrization you come up with, you will get the same value for the line integral. Evaluating line integrals is independent of the parametrization. Just try to keep it simple.

In this lesson, we walked through the steps of evaluating a line integral to find the area under the curve of a multivariable function along a multidimensional curve through a space. A **line integral** is an integral of a function defined on a 3D curve in space. The key to obtaining an integral that can actually be evaluated using single-variable techniques was constructing a **parametrization**, a mapping from a variable *t* onto a curve *C* using coordinate functions *x(t)*, *y(t)*, etc.

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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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