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Counting Repeated Traversals in Scalar Integrals

See why integrating a scalar field with respect to arc length counts every traversal, even when the path covers the same geometric curve more than once.

Intermediate 10 min read

What You'll Learn

  • Distinguish a curve’s geometric trace from a path’s traversal history.
  • State when repeating a closed traversal multiplies a scalar line integral.
  • Explain why arc-length integrals do not change sign when a path is reversed.
  • Compute integrals for repeated circular traversals and retraced segments.
  • Account for different traversal speeds without changing the contribution of a complete pass.
  • Identify the error in counting a repeatedly traced curve only once.

One Geometric Curve, More Than One Traversal

A path can travel over the same geometric curve more than once. For example, a particle can circle a wire twice before stopping at its starting point. The wire as a set of points has not changed, but the path has visited each point twice.

This distinction matters for a scalar line integral. The integral \(\int_C f\,ds\) accumulates the value of \(f\) along the path’s motion, weighted by arc length. If the path makes a second complete pass over the same curve, that pass contributes again. The answer is not determined just by which points lie on the curve; it also depends on how many times the path traverses them.

Here \(f\) is a scalar field, \(ds\) is an element of arc length, and a parametrized curve \(\mathbf r(t)\), \(a\le t\le b\), gives the calculation

$$ \int_C f\,ds = \int_a^b f(\mathbf r(t))\,\|\mathbf r'(t)\|\,dt. $$

This formula applies when the path is piecewise smooth and \(f\) is continuous along it. The speed \(\|\mathbf r'(t)\|\) is nonnegative, so the integral adds contributions along the path rather than assigning them a sign based on direction.

Definition: The geometric trace of a path is the set of points it visits. A traversal is one pass along a specified curve. A scalar line integral with respect to arc length counts the contributions from every traversal, including repeated visits to the same points.

The Repeated-Traversal Rule

Suppose a piecewise smooth path makes \(k\) complete passes over the same closed curve, and each pass follows that curve once. Let \(I\) be the scalar line integral for one pass. Then the integral along the full path is \(kI\). This is true whether the passes are made at the same speed or at different speeds, provided each pass covers the same curve once.

Theorem: If a path consists of \(k\) complete traversals of a piecewise smooth closed curve \(C\), and \(f\) is continuous along \(C\), then $$ \int_{\text{full path}} f\,ds = k\int_C f\,ds. $$ The integral on the right is the value for one traversal. The formula counts repeated traversals; it does not treat the path as a set of points visited only once.

To see why the rule holds, first consider a single parametrization \(\mathbf r(t)\), \(a\le t\le b\), of one pass around \(C\), with \(\mathbf r(a)=\mathbf r(b)\). Let \(L=b-a\). Concatenate \(k\) copies of this parametrization by using, on the \(j\)-th interval, the curve \(\mathbf R(t)=\mathbf r(t-jL)\), for \(a+jL\le t\le a+(j+1)L\), where \(j=0,\ldots,k-1\). Each interval contributes the same integral.

$$ \begin{aligned} \int_{\text{full path}} f\,ds &= \sum_{j=0}^{k-1} \int_{a+jL}^{a+(j+1)L} f(\mathbf r(t-jL))\, \|\mathbf r'(t-jL)\|\,dt\\ &= \sum_{j=0}^{k-1} \int_a^b f(\mathbf r(u))\,\|\mathbf r'(u)\|\,du\\ &= k\int_a^b f(\mathbf r(u))\,\|\mathbf r'(u)\|\,du. \end{aligned} $$

In the second line, the substitution \(u=t-jL\) shifts each interval back to the parameter interval for one pass. The conclusion follows because the integrand on every pass is the same as on the first pass.

The timing of a pass does not affect its contribution either. If the curve is followed using a smooth, strictly increasing parameter change \(t=\phi(u)\), then the new velocity is \(\mathbf r'(\phi(u))\phi'(u)\). Since \(\phi'(u)>0\), its speed is \(\|\mathbf r'(\phi(u))\|\phi'(u)\). Substitution in the integral therefore gives the original one-pass integral. This explains why a pass completed quickly and a pass completed slowly count equally when both cover the same curve once.

Worked Example: Two Passes Around a Circle

Worked Example: Two Passes Around a Circle

Let \(C\) be the circle of radius \(2\), and let \(f(x,y)=x^2+y^2+x\). Find the integral when the circle is traversed twice, using \(\mathbf r(t)=(2\cos t,2\sin t)\) for \(0\le t\le4\pi\).

First, evaluate the field along the parametrized circle:

$$ f(\mathbf r(t)) =(2\cos t)^2+(2\sin t)^2+2\cos t =4+2\cos t. $$

The velocity is \(\mathbf r'(t)=(-2\sin t,2\cos t)\), so the speed is

$$ \|\mathbf r'(t)\| = \sqrt{4\sin^2t+4\cos^2t} =2. $$

Thus the integral over the two passes is

$$ \begin{aligned} \int_0^{4\pi}(4+2\cos t)\,2\,dt &=\int_0^{4\pi}(8+4\cos t)\,dt\\ &=\left[8t+4\sin t\right]_0^{4\pi}\\ &=32\pi+0\\ &=32\pi. \end{aligned} $$

For one pass, the same calculation over \(0\le t\le2\pi\) gives \(16\pi\). Two passes give \(2(16\pi)=32\pi\), as calculated directly. The cosine term contributes zero on each complete turn, while the constant term contributes again on the second turn.

Direction and Repeated Visits

Because the arc-length element uses speed, reversing direction does not introduce a negative sign. On a line segment, for example, moving from \(x=0\) to \(x=2\) and then back from \(x=2\) to \(x=0\) gives two positive contributions. The second pass is not subtracted from the first.

This is different from a signed accumulation in which direction matters. For a scalar line integral with respect to \(ds\), the quantity being accumulated is the field value times a nonnegative length element. If \(f\) itself is negative somewhere, that can make the contribution negative there; reversing the direction of travel does not change the sign of that contribution.

Worked Example: A Segment Traversed Forward and Back

Let the path go from \((0,0)\) to \((2,0)\), then back to \((0,0)\), and let \(f(x,y)=1+x\). Compute the integral along the full path and compare it with one pass over the segment.

On the outward pass, use \(\mathbf r_1(t)=(t,0)\), \(0\le t\le2\). Its speed is \(1\), and \(f(\mathbf r_1(t))=1+t\). Therefore,

$$ \int_{\text{outward}} f\,ds = \int_0^2(1+t)\,dt = \left[t+\frac{t^2}{2}\right]_0^2 =2+2 =4. $$

On the return pass, use \(\mathbf r_2(t)=(4-t,0)\), \(2\le t\le4\). Its velocity is \((-1,0)\), so its speed is still \(1\). The field value is \(1+(4-t)=5-t\), giving

$$ \int_{\text{return}} f\,ds = \int_2^4(5-t)\,dt = \left[5t-\frac{t^2}{2}\right]_2^4 = (20-8)-(10-2) =4. $$

The full path integral is \(4+4=8\). One pass along the geometric segment contributes \(4\), so the forward-and-back path contributes twice that. As a check, both pass integrals are positive and equal even though the second pass moves in the opposite direction.

Different Speeds, the Same Number of Passes

A parameter interval does not directly tell you how many times a curve has been traversed. What matters is the change in the curve’s position and, for a scalar arc-length integral, the distance accumulated along the way. A parametrization may spend more parameter time on one part of the curve than another, but that does not create extra length if the path still makes just one pass.

Worked Example: Two Circular Passes at Nonconstant Speed

Consider the unit circle parametrized by

$$ \mathbf r(t)=(\cos\theta(t),\sin\theta(t)), \qquad \theta(t)=2\pi(t+t^2), \qquad 0\le t\le1. $$

Since \(\theta(0)=0\) and \(\theta(1)=4\pi\), this path makes two complete turns. Its angular speed is \(\theta'(t)=2\pi(1+2t)>0\), which is not constant. Find the integral of \(f(x,y)=1+x\) along the path.

The field along the path is \(1+\cos\theta(t)\). Differentiating the position gives a speed of \(\theta'(t)\), because the circle has radius \(1\) and \(\theta'(t)>0\). Thus

$$ \begin{aligned} \int_{\text{path}} f\,ds &=\int_0^1 \bigl(1+\cos\theta(t)\bigr)\theta'(t)\,dt\\ &=\int_0^{4\pi}(1+\cos u)\,du\\ &=\left[u+\sin u\right]_0^{4\pi}\\ &=4\pi. \end{aligned} $$

The substitution \(u=\theta(t)\) is valid because \(\theta'(t)>0\), and the limits are \(u=0\) and \(u=4\pi\). On one turn, the integral is \(\int_0^{2\pi}(1+\cos u)\,du=2\pi\). Two turns give \(4\pi\), confirming the repeated-traversal rule despite the varying speed.

Counting Traversals in a Calculation

When a parametrization covers a curve several times, its full parameter interval includes every pass. It is often helpful to split the interval into one-pass pieces and compute the contribution on each piece. For a path that retraces only part of its trace, split it at the points where the motion changes from one piece to another; count each pass over a piece separately.

1
Identify one pass.
Find the parameter interval corresponding to a single traversal of the curve or segment.
2
Compute its contribution.
Use \(f(\mathbf r(t))\|\mathbf r'(t)\|\) and integrate over that interval.
3
Count the passes.
If the same traversal occurs \(k\) times, multiply its one-pass integral by \(k\). If different pieces are traversed different numbers of times, add their contributions with the appropriate counts.
4
Check the path, not just its endpoints.
A closed path has matching start and end points, but that fact alone does not say how many times it went around.

A Common Counting Error

A frequent mistake is to replace the path by its geometric trace and calculate only one contribution for every point on that trace. This loses information when the path repeats. For instance, in the circle example, one pass contributed \(16\pi\). Treating the twice-traversed circle as though it were traversed once would give

$$ \int_0^{2\pi}(4+2\cos t)\,2\,dt=16\pi, $$

which is not the integral along the specified path. The path has parameter interval \(0\le t\le4\pi\), so its integral is \(32\pi\). The correction is to account for both complete turns, either by integrating over the full interval or by doubling the one-turn result.

Key takeaway: For a scalar line integral with respect to arc length, repeated traversal repeats the contribution. The geometric trace may be unchanged, but the path integral is counted with traversal multiplicity.

Why the Distinction Matters

The distinction between a curve and a path is useful whenever a model describes an actual route or motion. A wire may be traced more than once by a moving sensor; a particle may loop around a track; or a path may retrace a segment before continuing. In each case, \(\int f\,ds\) accumulates the field along the motion, so a repeated visit contributes again.

Before evaluating a scalar line integral, therefore, check not only which geometric points appear, but also how the path moves through them. A simple closed curve traversed once and the same curve traversed three times have the same geometric trace, but their integrals differ by a factor of three.

Check Your Understanding

Use the role of arc length and traversal count to answer these questions.

  1. If a closed curve is traversed four times, how is its scalar line integral related to the one-pass integral?
  2. Why does reversing direction on a segment not negate the scalar line integral with respect to \(ds\)?
  3. A unit circle is traversed twice with \(f(x,y)=1+x\). What is the integral, given that its one-pass value is \(2\pi\)?
  4. What information is lost if a repeatedly traversed path is replaced by its geometric trace?
  5. How can you check whether a parametrization of a closed curve makes one turn or several?

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