Line integral of a vector field

We can do more than just take the path integral of a scalar-valued function. We can take a line integral (or path integral) of a vector-valued function $F$ , which, as you will see in a moment, we typically call a vector field.

The line integral of a vector field will play a crucial role in the rest of the class. Out of the four fundamental theorems that form the cornerstones of the second half the course, three of the them involve line integrals of vector fields. (And the fourth involves surface integrals of vector fields, which are closely related to line integrals of vector fields.) I cannot emphasize too strongly the importance of these integrals.

Let’s return to our slinky. Imagine that you put a small magnetized bead on your slinky (the bead has a small hole in it, so it can slide along the slinky). Next, imagine that you put a large magnet to the left of the slink, as shown by the large green square. The magnet will induce a magnetic field $F (x, y, z)$ , shown by the green arrows. For this example, we’re using the constant vector field $F (x, y, z) = (- 1 ∕ 2, 0, 0)$ .

Just so that the picture looks better (i.e., so the scale is the same in each direction), we changed our slinky function to $c (t) = (cos t, sin t, t ∕ (3 π))$ , for $0 \leq t \leq 6 π$ . The function $c (t)$ is a parametrization of the slinky. The function $c (t)$ can be thought of as giving the position of the bead for any given $t$ .

Since the bead is magnetized, the magnetic field exerts a force on the bead. Of course, there’s the issue of the magnetic field causing the bead to move (you know: force equal mass time accelaration, so the force should make the bead accelerate). We’re going to ignore any effects of the magnetic force on the position of the bead; instead the position of the bead at time $t$ is constained to be $c (t)$ .

As you change $t$ (using the slider) to move the bead (the magenta dot), work is done by the magnetic field on the bead. Work equals forces times distance. So, if the magnetic field were aligned with the direction of movement, calculating the work would be simple.

However, the bead does not move in the direction of the magnetic field. It is constrained to move along the slinky. The work is calculated from the component of the force in the direction of movement. For example, when the movement is 90 degrees from the direction of the force, the magnetic field does no work.

First, what is the direction of movement? It is the direction given by the velocity $c^{'} (t)$ . Denote by $T$ the unit vector in the direction of movement: $T (t) = c^{'} (t) ∕ | | c^{'} (t) | |$ . Since $c^{'} (t)$ (and therefore $T$ ) is tangent to the path, we refer to $T$ as the unit tangent vector. It is shown by the blue vector in the below figure.

The component of the force in the direction of movement is simply the dot product of the force $F$ at the point $c (t)$ (shown by the green vector) with the tangent vector $T (t)$ , i.e., it is $F (c (t)) \cdot T (t)$ . For shorthand, we’ll write it as $F \cdot T$ . This number is displayed by the cyan mark on the green slider. Recall that the dot product is zero if the two vectors are orthogonal, is negative if their angle is greater than 90 degrees, and is positive if their angle is less than 90 degrees.

If we multiply the (component of) force $F \cdot T$ by distance, we get work. Consequently, we can think of $F \cdot T$ as a scalar-valued function that gives work per unit length along the slinky. To get the total work, we simply need to take the path integral of this scalar-valued function. In other words, the total work is

\begin{array}{l} Work = \int_{C} F \cdot T d s, \end{array}

where $C$ is the path of the slinky. This integral is basically adding up the product of force ( $F \cdot T$ ) and distance ( $d s$ ) along the slinky, which is work.

Remember the formula for the path integral of a scalar-valued function $f$ in terms of the parameterization $c (t)$ ,

\begin{array}{l} \int_{a}^{b} f (c (t)) | | c^{'} (t) | | d t \end{array}

When we replace $f$ with $F \cdot T$ , we get

\begin{array}{l} \int_{C} F \cdot T d s = \int_{a}^{b} F (c (t)) \cdot T (t) | | c^{'} (t) | | d t, \end{array}

where for our parametrization $c (t)$ of the slinky, $a = 0$ and $b = 6 π$ .

We can simplify this further. Remember that $T (t) = c^{'} (t) ∕ | | c^{'} (t) | |$ . Consquently, the two occurrences of $| | c^{'} (t) | |$ cancel, and we are left with

\begin{array}{l} \int_{C} F \cdot T d s = \int_{a}^{b} F (c (t)) \cdot c^{'} (t) d t . \end{array}

We usually write $T d s$ as $d s$ (as explained in lecture). Hence, the important formula to know is that the line integral of a vector field is

\begin{array}{l} \int_{C} F \cdot d s = \int_{a}^{b} F (c (t)) \cdot c^{'} (t) d t . \end{array}

where the curve $C$ is parametrized by the function $c (t)$ for $a \leq t \leq b$ .

Multivariable calculus and vector analysis

A set of on-line readings

Beta Math ML version

Line integral of a vector field

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

4d. Path integrals and line integrals

4e. Surface integrals

5. The fundamental theorems of vector calculus

6. Review material