Line Integrals of Vector Fields

In this section, we will explore the concept of line integrals of vector fields.

Table of Contents

• Table of Contents
• Introduction - Moving in a Force Field
• Example Problem: Work in Circular Motion
• More on Path Directions
  • Scalar Line Integrals on Reversed Curves
  • Vector Line Integrals on Reversed Curves
• Summary and Next Steps
• Appendix: Forces and Work

Introduction - Moving in a Force Field

Imagine a carboard box in someone's driveway. It's sitting still when, suddenly, the wind picks up - it's a hurricane! The carboard box flies off the driveway and into the air, where gravity takes over and pulls it back down. When it goes back down, it doesn't just fall straight down, the wind still pushes it left and right.

The curve in the visual

I used quite a complex curve in the visual above:

It's actually quite difficult to create a curve like this, because it needs to satisfy a few conditions:

• It must start at a given point and end at a given point.
• It must look chaotic, but not too chaotic.
• It must not go off the screen.

The initial point is and the final point is . I made range from to to automatically satisfy the first condition in the -direction. I added to make the final point correct.

Then, I just kept adding and subtracting terms until I got something that looked good. Each term gets to make sure that at the boundaries they are , hence the curve still starts and ends at the correct points.

The force we are considering is gravity (ignore forces from the wind, just note that it affects the box's motion). Recall that the work done due to a force is the dot product of the force and the distance moved:

(For a quick refresher on forces and work, see the appendix at the end of this page.)

However, we aren't just moving in a straight line - we are moving along a curve, due to the wind pushing the box around. Instead, we can break the curve into small segments and sum the work done along each segment.

Options
Segments

Let's call each segment , where is the segment number (starting from and going to ). For each segment, the work done is:

So the total work done is just the sum of these segmented works:

In calculus, we take these discrete concepts and take limits to make them continuous. As the segments get smaller and smaller, the work done becomes an infinitesimal work:

An infinitesimal change in the curve, can be written as:

Substituting this back into the equation for the infinitesimal work, we get:

Finally, the total work done along the curve is the integral of the infinitesimal work:

Example Problem: Work in Circular Motion

Suppose the cardboard box gets caught in a whirlwind and starts moving in a circle of radius . The force acting on the box is given by:

If the circle is centered at (check the figure below):

1. Construct a parametric equation for motion along the circle in both counterclockwise and clockwise directions.
2. Find the work done as the box moves around the circle once in the counterclockwise direction.
3. Find the work done as the box moves around the circle once in the clockwise direction.
4. Observe the difference in the work done in the two cases.

(Source)

1. First, let's construct the parametric equation for motion along the circle.

   For the counterclockwise direction, let's first start with a circle at the origin. Let represent the angle from the positive -axis. The height is given by and the width is given by . Then, we can translate this circle to the center at . The parametric equation is then:

   For the clockwise direction, we can just negate the term in the -coordinate:

2. To find the work done in the counterclockwise direction, we need to evaluate the integral:

   Based on the definitions for the force and the curve, we can evaluate the force as:

   And the derivative of the curve as:

   Thus, the integral becomes:

   We can evaluate the dot product and integrate to find the work done:

3. We can follow the same steps for the clockwise direction, but there's another way to do this. Given that the clockwise direction is essentially just the negative of the counterclockwise direction, then the dot product would be negative.

   To see this, take the infinitesimal vector and compare it to :

   Therefore:

   Meaning, the total work done in the clockwise direction is just the negative of the work done in the counterclockwise direction:

4. The difference in the work done in the two cases is . The takeaway is that if you reverse the direction of the curve, the work done is the negative of the work done in the original direction.

More on Path Directions

We saw in the example problem that reversing the direction of the curve results in the negative of the work done in the original direction. Let's explore this concept further and see if it holds for any curve.

First, let's consider a curve defined as , where .

Next, consider the reverse of this curve, . In order to parameterize this reversed curve, we can think about reversing the direction of the input parameter. That is, instead of , we can consider . A quick check shows that this is indeed the reverse of the original curve:

1. When , the parameter for the reversed curve is .
2. When , the parameter for the reversed curve is .

Therefore, the reverse of the curve is:

Now, let's consider how the line integral changes when we reverse the direction of the curve.

Scalar Line Integrals on Reversed Curves

The line integral of a scalar field along a curve , as previously defined, is:

For the reversed curve, for now we just write instead of :

Recall what this visually means. If we construct a 3D plot of the scalar field where and are the input parameters and is the output, then the line integral is the area under the curve in this 3D plot:

Options
Segments

Now think about it: even if we reverse the direction of the curve, the area under the curve in the 3D plot doesn't change. Therefore, the line integral of a scalar field along a reversed curve is the same as the line integral along the original curve:

More rigorous, please!

To show this more rigorously, we can substitute the parameterization of the reversed curve into the integral:

(Notice that the boundaries of the integral do not swap, since we already accounted for the reversal in the parameterization.)

Make the substitution :

The boundaries of the integral change as well:

(This should start to look like a reverse in the integral limits.)

Substitute these into the integral:

This is the exact same integral as the un-reversed curve, just with a different name for the integration variable;

Vector Line Integrals on Reversed Curves

We have already seen that the line integral of a vector field reverses when the curve is reversed.

Just like before, start by writing expressions for the line integral of a vector field along a curve.:

For the reversed curve, we can write the line integral as:

In the line integral, every infinitesimal step of the way, we are taking the dot product of the vector field and the derivative of the curve. If we reverse the direction of the curve, the derivative of the curve also reverses, meaning the dot product is negated. Thus, reversing the curve negates the line integral of a vector field along the curve:

Thus:

More rigorous, please!

To show this more rigorously, we can substitute the parameterization of the reversed curve into the integral.

Recall that the parameterization of the reversed curve is . The infinitesimal vector is then :

Since our vector field now acts on the reversed curve, we must evaluate it at the reversed curve:

Finally, let's put this all together in the integral:

Now, we can make the substitution . As seen previously, this swaps the limits of the integral and the sign of the differential:

Summary and Next Steps

In this section, we explored the concept of line integrals of vector fields.

Here are the key points to remember:

• The line integral of a vector field along a curve is the integral of the dot product of the vector field and the derivative of the curve.

• The work done along a curve is the line integral of the force along the curve.

• If the curve is reversed, the line integral of a scalar field remains the same, while the line integral of a vector field is negated.

In the next section, we shall extend the Fundamental Theorem of Calculus to line integrals.

Appendix: Forces and Work

When we talk about forces, we often talk about the work done by a force. Imagine a force pushing on a ball, and the ball moves a distance (magnitude ).

Let's say the force is gravity - , and it is pushing on a ball of mass . The work done is defined by the product of the force and the distance moved:

However! This is not the complete picture. Gravity acts in the downward direction, but the ball is moving in a different direction:

In that case, the work done considers only the distance moved in the direction of the force, in this case, units:

This is otherwise known as the dot product of the force and the distance moved. The complete formula for the work done in a straight line for a constant force is then: