Chapter 8

Arc Length and Scalar Line Integrals

In Chapter 7, the magnitude $‖ ⃗ 𝑟' (𝑡) ‖$ measured the speed of a moving point. We now use that same quantity to measure distance along the curve. Once we know how to add up little pieces of length, we can also weight those pieces by a density or some other scalar quantity.

8.1Arc Length

One way to estimate the length of a curve is to choose points along it, join successive points with line segments, and add the lengths of those segments. The estimate improves as we use more and shorter segments.

Figure 8.1 Join a handful of points on the curve and measure the polygon instead. With six chords the answer is visibly short—the polygon cuts every corner and misses the loop almost entirely; with fifty it is short by less than you can see. Each side of that polygon is one

Δ ⃗ 𝑟

, and shrinking it is what turns

‖ Δ ⃗ 𝑟 ‖

into

𝑑 𝑠 = ‖ ⃗ 𝑟' (𝑡) ‖ 𝑑 𝑡

. This cubic has no elementary formula for its length, so the number the sum climbs toward is one you could not have written down another way.

Definition 8.1 (Infinitesimal Arc Length). If $⃗ 𝑟 (𝑡)$ is a parametric curve, its infinitesimal arc length is measured by

𝑑 𝑠 = ‖ ⃗ 𝑟' (𝑡) ‖ 𝑑 𝑡

This makes sense: after all the derivative $⃗ 𝑟' (𝑡)$ is the velocity, $‖ ⃗ 𝑟' (𝑡) ‖$ is the speed, and $𝑑 𝑡$ is an infinitesimal length of time. Thus, the product $‖ ⃗ 𝑟' (𝑡) ‖ 𝑑 𝑡$ is an infinitesimal bit of distance - a small length along the curve. To take this infinitesimal information and get something useful out - we need to integrate along the curve.

Definition 8.2 (Arc Length). If $⃗ 𝑟 (𝑡)$ is a parametric curve, its length between $𝑡 = 𝑎$ and $𝑡 = 𝑏$ is given by

𝐿 = \int 𝑏 𝑎 𝑑 𝑠 = \int 𝑏 𝑎 ‖ ⃗ 𝑟' (𝑡) ‖ 𝑑 𝑡

Figure 8.2 The arrow is the velocity; the graph is its length. The two are drawn to the same scale on purpose—stand the arrow upright and it is the height of the graph, which is the whole content of the word "speed." As the point moves it draws the graph out behind it, and the area it has swept out is the distance it has travelled. Send it around twice and the picture on the right never changes while the area on the left keeps climbing, which is why length belongs to the traversal and not only to the image.

Example 8.3 (Arc Length of a Helix). Find the arc length of $⃗ 𝑟 (𝑡) = (c o s (𝑡), s i n (𝑡), 𝑡)$ from $𝑡 = 0$ to $𝑡 = 2 𝜋$ . First, we need to find the velocity $⃗ 𝑟'$ :

⃗ 𝑟' (𝑡) = ⟨ - s i n (𝑡), c o s (𝑡), 1 ⟩

Next, we need to take this velocity and find the speed:

‖ ⃗ 𝑟' (𝑡) ‖ = \sqrt (- s i n 𝑡) 2 + (c o s 𝑡) 2 + 1 = \sqrt 2

Finding arc length is just integrating this over the domain:

\int 2 𝜋 0 ‖ ⃗ 𝑟' (𝑡) ‖ 𝑑 𝑡 = \int 2 𝜋 0 \sqrt 2 𝑑 𝑡 = 2 𝜋 \sqrt 2

Arc-length integrals can be challenging because of the square root, but with some algebra and integration techniques, much can be learned.

From this idea, we can define the arc-length function, which measures the length of a curve $⃗ 𝑟 (𝑡)$ from a starting point $𝑡 = 𝑎$ :

Definition 8.4 (The Arc-Length Function). If $⃗ 𝑟 (𝑡)$ is a parametric curve, for any given starting point $𝑡 = 𝑎$ we may define the arc-length function which measures the length of curve between $𝑎$ and $𝑡$ :

𝑠 (𝑡) = \int 𝑡 𝑎 ‖ ⃗ 𝑟' (𝑢) ‖ 𝑑 𝑢

(Note we have changed the variable of integration so that $𝑡$ is not used in two different contexts)

What is the arc-length function for the helix in our earlier example, starting from $𝑡 = 0$ ? Since $‖ ⃗ 𝑟' (𝑡) ‖ = \sqrt 2$ , we see that

𝑠 (𝑡) = \int 𝑡 0 \sqrt 2 𝑑 𝑢 = \sqrt 2 𝑡

This tells us that after $𝑡$ seconds, we have traced out $\sqrt 2 𝑡$ units of arc length. How could we reparameterize this curve so that its arc-length function is just $𝑠 (𝑡) = 𝑡$ (tracing out $𝑡$ units of arc in $𝑡$ units of time)?

Definition 8.5 (Unit Speed Curve). A curve $⃗ 𝑐 (𝑡)$ is unit speed if $‖ ⃗ 𝑐' (𝑡) ‖ = 1$ for all $𝑡$ . If the motion begins at $𝑡 = 0$ , this means that after $𝑡$ seconds the curve has traversed $𝑡$ units of length. For this reason, we also call unit-speed curves arc-length-parameterized curves.

In our example, to make the helix unit speed we need to slow it down by a factor of $\sqrt 2$ : that is, we need $⃗ 𝑟 (𝑡 / \sqrt 2)$ :

⃗ 𝑟 (𝑡 \sqrt 2) = (c o s 𝑡 \sqrt 2, s i n 𝑡 \sqrt 2, 𝑡 \sqrt 2)

To trace the same portion of the helix, the new parameter runs from $0$ to $2 𝜋 \sqrt 2$ . Differentiating with the chain rule confirms that the new speed is $1$ .

More generally, suppose $⃗ 𝑐 (𝑢) = ⃗ 𝑟 (𝜙 (𝑢))$ is a reparameterization. The chain rule gives

⃗ 𝑐' (𝑢) = ⃗ 𝑟' (𝜙 (𝑢)) 𝜙' (𝑢),

‖ ⃗ 𝑐' (𝑢) ‖ = ‖ ⃗ 𝑟' (𝜙 (𝑢)) ‖ | 𝜙' (𝑢) | .

If $𝜙$ traces the original parameter interval exactly once, a substitution shows that the two parameterizations have the same length. The absolute value makes this true even when the direction is reversed. But if the reparameterization retraces part of the curve or goes around it several times, that distance is counted each time.

8.2Scalar Line Integrals

Suppose a thin wire follows a curve $𝐶$ and has linear density $𝜌$ . A small piece of wire with length $Δ 𝑠$ has mass approximately $𝜌 Δ 𝑠$ . Adding these small masses and taking a limit leads to the scalar line integral

\int 𝐶 𝜌 𝑑 𝑠 .

Figure 8.3 Cut the wire into pieces of equal length. Each piece has its own density, so it has its own mass

𝜌 Δ 𝑠

—and on the right, its own rectangle of width

Δ 𝑠

and height

𝜌

. The bead and the rectangle are the same number, so the mass of the wire is the area under its density graph, and

\int 𝐶 𝜌 𝑑 𝑠

is an ordinary single-variable integral once the wire is straightened out against

𝑠

. Choose

𝜌 = 1

and the graph flattens to height one: its area is

𝐿

, and the line integral has become the arc length. Traverse the wire backwards and the graph flips while its area does not move.

We can denote the domain $𝐶$ as a subscript just like we do for double and triple integrals. When $𝐶$ is a closed curve one may optionally modify the integral sign to denote this, writing $\oint 𝐶 𝜌 𝑑 𝑠$ . Our first goal is to try and figure out how to compute this in terms of integrals we know how to do. First: recall that we can represent a curve $𝐶$ by parameterizing it, writing it as the image of a function $⃗ 𝑐 (𝑡) = (𝑥 (𝑡), 𝑦 (𝑡))$ in the plane, or $⃗ 𝑐 (𝑡) = (𝑥 (𝑡), 𝑦 (𝑡), 𝑧 (𝑡))$ in 3 dimensions. We saw above how to express a small bit of arc length along a parametric curve:

𝑑 𝑠 = ‖ ⃗ 𝑐' (𝑡) ‖ 𝑑 𝑡

To evaluate a density $𝜌 (𝑥, 𝑦, 𝑧)$ along the curve $⃗ 𝑐 (𝑡)$ , we simply plug the curve into the density. This gives a concrete quantity to integrate.

Definition 8.6 (Scalar Line Integral). If $⃗ 𝑐 (𝑡)$ , $𝑎 \leq 𝑡 \leq 𝑏$ , parameterizes a curve $𝐶$ , then

\int 𝐶 𝜌 𝑑 𝑠 = \int 𝑏 𝑎 𝜌 (⃗ 𝑐 (𝑡)) ‖ ⃗ 𝑐' (𝑡) ‖ 𝑑 𝑡 .

Note the simplest line integrals are just when $𝜌 = 1$ . This is the integral just adding up the infinitesimal arc lengths $\int 𝐶 𝑑 𝑠 = \int ‖ ⃗ 𝑐' ‖ 𝑑 𝑡$ . We already met this integral long ago - this gives the arc length of the curve!

Figure 8.4 The same wire, lying flat, with its density standing up over it. The strip above one piece is

Δ 𝑠

wide and

𝜌

tall, so its area is the term the sum was built from—and the area of the whole sheet is

\int 𝐶 𝜌 𝑑 𝑠

. It is the same number as the figure above, arrived at by hanging the graph over the curve instead of straightening the curve out under the graph.

If $𝐿 (𝐶)$ is the length of the wire, its average density is its total mass divided by its total length:

𝜌 a v g = 1 𝐿 (𝐶) \int 𝐶 𝜌 𝑑 𝑠 .

Density is the main interpretation, but the same definition works for any scalar quantity assigned to the points of a plane or space curve.

Example 8.7 (A Straight-Line Integral). Find $\int 𝐶 𝑥 𝑦 𝑑 𝑠$ for $𝐶$ the diagonal of the unit square going from $(0, 0)$ to $(1, 1)$ .

We parameterize the diagonal by

⃗ 𝑐 (𝑡) = (𝑡, 𝑡), 0 \leq 𝑡 \leq 1 .

Along the curve, $𝑥 𝑦 = 𝑡 2$ , while

‖ ⃗ 𝑐' (𝑡) ‖ = ‖ (1, 1) ‖ = \sqrt 2 .

Therefore

\int 𝐶 𝑥 𝑦 𝑑 𝑠 = \int 10 𝑡 2 \sqrt 2 𝑑 𝑡 = \sqrt 2 3 .

Example 8.8 (A Circular Integral). Evaluate $\oint 𝐶 𝑥 2 𝑑 𝑠$ for $𝐶$ the circle of radius $2$ , traversed clockwise.

A clockwise parameterization is

⃗ 𝑐 (𝑡) = (2 c o s 𝑡, - 2 s i n 𝑡), 0 \leq 𝑡 \leq 2 𝜋 .

Here $𝑥 2 = 4 c o s 2 𝑡$ and $‖ ⃗ 𝑐' (𝑡) ‖ = 2$ , so

\oint 𝐶 𝑥 2 𝑑 𝑠 = \int 2 𝜋 0 8 c o s 2 𝑡 𝑑 𝑡 = 8 𝜋 .

Using a counterclockwise parameterization gives exactly the same integral.

This illustrates an important difference between scalar line integrals and the vector line integrals we will meet later: scalar line integrals are orientation-independent. Reversing a parameterization changes the sign of its parameter derivative, but not its magnitude $‖ ⃗ 𝑐' (𝑡) ‖$ .

Sometimes a curve $𝐶$ might be piecewise, and has a corner where the edges join up. In this case, we evaluate the line integral by doing each segment of the curve separately, and adding the results:

\int 𝐶 1 \cup 𝐶 2 𝜌 𝑑 𝑠 = \int 𝐶 1 𝜌 𝑑 𝑠 + \int 𝐶 2 𝜌 𝑑 𝑠

The pieces must cover the desired curve between them, but they do not need to be parameterized in matching directions. Reversing any one piece leaves its scalar line integral unchanged.

Figure 8.5 A path in three pieces, each carrying its own sheet, and the three areas simply add. The middle side is deliberately traced from its far end backwards, and it makes no difference at all: a sheet has an area whichever way you sweep it out. The pieces have to cover the path once between them; they do not have to agree about which way they are going.

Example 8.9 (A Piecewise Curve). Find the line integral of $𝑥$ along the curve $𝐶$ that traces the triangle with vertices $(0, 0)$ , $(1, 0)$ , and $(0, 1)$ , starting at the origin and going counterclockwise.

We break this into three curves:

⃗ 𝑐 1 (𝑡) = (𝑡, 0) ⃗ 𝑐 2 (𝑡) = (1 - 𝑡, 𝑡) ⃗ 𝑐 3 (𝑡) = (0, 1 - 𝑡)

Each segment is traced from $𝑡 = 0$ to $𝑡 = 1$ . We then just compute the line integral along each and add up the results.

On $𝐶 1$ , the integrand is $𝑥 = 𝑡$ and the speed is $1$ . On $𝐶 2$ , the integrand is $𝑥 = 1 - 𝑡$ and the speed is $\sqrt 2$ . On $𝐶 3$ , the integrand is $𝑥 = 0$ . Thus,

\int 𝐶 𝑥 𝑑 𝑠 = \int 10 𝑡 𝑑 𝑡 + \int 10 (1 - 𝑡) \sqrt 2 𝑑 𝑡 + \int 10 0 𝑑 𝑡 = 12 + \sqrt 2 2 .

It is straightforward to generalize this to integrals along space curves instead of curves in the plane:

Example 8.10 (A Space-Curve Integral). Compute the line integral $\int 𝐶 𝑓 𝑑 𝑠$ for $𝑓 (𝑥, 𝑦, 𝑧) = 𝑦 s i n 𝑧$ and $𝐶$ the helix $𝑥 = c o s 𝑡$ , $𝑦 = s i n 𝑡$ and $𝑧 = 𝑡$ for $𝑡 \in [0, 2 𝜋]$ .

Along the helix,

𝑓 (⃗ 𝑐 (𝑡)) = s i n 𝑡 s i n 𝑡 = s i n 2 𝑡,

and from our earlier arc-length calculation,

‖ ⃗ 𝑐' (𝑡) ‖ = \sqrt 2 .

Therefore

\int 𝐶 𝑓 𝑑 𝑠 = \sqrt 2 \int 2 𝜋 0 s i n 2 𝑡 𝑑 𝑡 = 𝜋 \sqrt 2 .

Scalar line integrals measure quantities accumulated against length, so their values do not depend on which direction the curve is traversed. Later, vector line integrals will measure work and circulation; for those integrals, orientation will matter.