Tutorials › Calculus III › Constant Integrands and Curve Length

Home  /  Tutorials  /  Scalar Line Integrals

Scalar Line Integrals · Tutorial 736 of 999

Constant Integrands and Curve Length

Derive the constant-integrand rule for scalar line integrals and use it to calculate weighted lengths on several common curves.

Intermediate 9 min read

What You'll Learn

  • Define curve length using the speed of a parametrization.
  • Prove the constant-integrand identity for scalar line integrals.
  • Evaluate constant-weight integrals along a line segment.
  • Relate a circular arc’s integral to its radius and angle.
  • Add lengths and integrals across piecewise smooth curves.
  • Identify why omitting speed or substituting chord length gives an incorrect result.

Introduction

A scalar line integral with respect to arc length accumulates values along a curve, with each small piece weighted by its length. When the scalar value is constant along the curve, the calculation has a particularly simple interpretation: the integral is that constant multiplied by the curve’s length.

This rule turns a line integral into a length calculation. It applies to line segments, circular arcs, and curves made of several smooth pieces, provided the scalar integrand has the same constant value along the curve.

Curve Length and Scalar Line Integrals

Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be a piecewise continuously differentiable parametrization of a curve. Its speed at parameter value \(t\) is \(\|\mathbf{r}'(t)\|\). The length of the parametrized curve is the integral of its speed over the parameter interval.

$$ L(C)=\int_a^b \|\mathbf{r}'(t)\|\,dt. $$

For a scalar field \(f\), its scalar line integral with respect to arc length is defined by integrating the field along the parametrization, weighted by speed:

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

These definitions use the same speed factor. The length integral adds up the curve’s infinitesimal lengths; the scalar line integral adds up those lengths after weighting each by the scalar value at that point.

Theorem: If \(C\) has finite length and \(f\) has the constant value \(c\) at every point of \(C\), then $$ \int_C f\,ds=cL(C). $$ In particular, when the integrand is the constant function \(c\), \(\int_C c\,ds=cL(C)\). The result applies to piecewise continuously differentiable curves. The constant \(c\) may be positive, zero, or negative.

The condition is that the integrand be constant on the curve. It need not have that value at points away from the curve. The length is nonnegative, but the integral’s sign follows the sign of \(c\).

Why the Identity Holds

For a smooth portion of a parametrized curve, the definition of the line integral gives \(f(\mathbf{r}(t))=c\). Substituting that value into the defining integral lets us factor out the constant. What remains is exactly the definition of length.

$$ \begin{aligned} \int_C f\,ds &=\int_a^b f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\,dt\\ &=\int_a^b c\|\mathbf{r}'(t)\|\,dt\\ &=c\int_a^b \|\mathbf{r}'(t)\|\,dt\\ &=cL(C). \end{aligned} $$

For a piecewise continuously differentiable curve, split the parameter interval at its finitely many breakpoints. Apply the same argument on each smooth piece, then add the integrals. The lengths of the pieces add to the total length, so the result remains \(cL(C)\). This argument also shows why the identity does not require the curve itself to be a straight line.

Worked Example: A Constant Weight on a Line Segment

Worked Example: A Constant Weight on a Line Segment

Evaluate \(\int_C -2\,ds\), where \(C\) is the line segment from \((1,-2)\) to \((4,2)\).

Parametrize the segment by moving from its starting point to its endpoint at constant rate:

$$ \mathbf{r}(t)=(1+3t,-2+4t),\qquad 0\le t\le 1. $$

At \(t=0\), this gives \((1,-2)\), and at \(t=1\), it gives \((4,2)\). Differentiate and calculate the speed:

$$ \mathbf{r}'(t)=(3,4), \qquad \|\mathbf{r}'(t)\|=\sqrt{3^2+4^2}=5. $$

The length is therefore

$$ L(C)=\int_0^1 5\,dt=5(1-0)=5. $$

Since the integrand is the constant \(-2\), the theorem gives

$$ \int_C -2\,ds=(-2)(5)=-10. $$

Directly from the definition, the same integral is \(\int_0^1(-2)(5)\,dt=\int_0^1-10\,dt=-10\). The endpoint displacement is \((4-1,2-(-2))=(3,4)\), whose length is also \(\sqrt{3^2+4^2}=5\), confirming the curve length and the final result.

Worked Example: A Constant Weight on a Circular Arc

Worked Example: A Constant Weight on a Circular Arc

Evaluate \(\int_C 4\,ds\) along the counterclockwise circular arc of radius \(3\), from angle \(0\) to angle \(\pi/3\), centered at the origin.

A parametrization is

$$ \mathbf{r}(t)=(3\cos t,3\sin t), \qquad 0\le t\le \frac{\pi}{3}. $$

Differentiating gives \(\mathbf{r}'(t)=(-3\sin t,3\cos t)\). Its speed is constant:

$$ \|\mathbf{r}'(t)\| = \sqrt{9\sin^2t+9\cos^2t} = \sqrt{9} = 3. $$

Thus the arc length is

$$ L(C)=\int_0^{\pi/3}3\,dt =3\left(\frac{\pi}{3}-0\right) =\pi. $$

Multiplying by the constant integrand gives

$$ \int_C 4\,ds=4L(C)=4\pi. $$

For a direct check, the definition yields \(\int_0^{\pi/3}4(3)\,dt=12(\pi/3)=4\pi\). This agrees with the length calculation.

Worked Example: A Constant Weight on a Piecewise Curve

Worked Example: A Constant Weight on a Piecewise Curve

Let \(C\) consist of two segments in \(\mathbb{R}^3\): first from \((0,0,0)\) to \((2,0,0)\), then from \((2,0,0)\) to \((2,1,2)\). Evaluate \(\int_C 3\,ds\).

On the first segment, use \(\mathbf{r}_1(t)=(2t,0,0)\) for \(0\le t\le1\). Its derivative is \((2,0,0)\), so its speed is \(2\). Its length is

$$ L_1=\int_0^1 2\,dt=2. $$

For the second segment, use \(\mathbf{r}_2(t)=(2,t,2t)\) for \(0\le t\le1\). This starts at \((2,0,0)\) and ends at \((2,1,2)\). Its derivative is \((0,1,2)\), so its speed is \(\sqrt{0^2+1^2+2^2}=\sqrt{5}\). Therefore,

$$ L_2=\int_0^1\sqrt{5}\,dt=\sqrt{5}. $$

The total curve length is the sum of the lengths of its pieces:

$$ L(C)=L_1+L_2=2+\sqrt{5}. $$

It follows that

$$ \int_C 3\,ds=3(2+\sqrt{5})=6+3\sqrt{5}. $$

To check by direct integration, the first piece contributes \(\int_0^1 3(2)\,dt=6\), and the second contributes \(\int_0^1 3\sqrt{5}\,dt=3\sqrt{5}\). Their sum is \(6+3\sqrt{5}\), as required.

When the Field Is Constant Only Along the Curve

The theorem concerns the values of the field on \(C\), not everywhere in its domain. For example, the field \(f(x,y)=x^2+y^2\) is not constant throughout the plane. However, on the circle \(x^2+y^2=4\), it has the constant value \(4\). Along the full circle, whose length is \(2\pi(2)=4\pi\), the theorem gives

$$ \int_C (x^2+y^2)\,ds=4(4\pi)=16\pi. $$

The restriction of the field to the curve is what determines whether the constant-integrand rule applies. If the field varies along the curve, it cannot be replaced by a single constant in this way.

Common Pitfalls

A frequent error is to omit the speed factor and integrate the constant only with respect to the parameter. On the circular arc above, this incorrect calculation would be

$$ \int_0^{\pi/3}4\,dt=\frac{4\pi}{3}. $$

That is not the scalar line integral with respect to arc length: the definition includes the speed \(3\). The corrected calculation is

$$ \int_0^{\pi/3}4\|\mathbf{r}'(t)\|\,dt = \int_0^{\pi/3}4(3)\,dt = 4\pi. $$

Another error is to use the straight-line distance between the endpoints as the length of a curved arc. The endpoints of this arc are \((3,0)\) and \((3/2,3\sqrt{3}/2)\). Their straight-line distance is \(\sqrt{(3/2)^2+(3\sqrt{3}/2)^2}=3\), whereas the arc length is \(\pi\). The integral uses the length of the specified curve, not the distance between its endpoints.

Key takeaway: First determine the length of the curve that is actually specified. If the integrand has constant value \(c\) along that curve, multiply that length by \(c\). In a parametrization, the speed factor is essential to computing the length.

Check Your Understanding

Use the definitions and the constant-integrand identity to answer these questions.

  1. What is the definition of the length of a parametrized curve in terms of its speed?
  2. A line segment has length \(7\). What is the value of \(\int_C -3\,ds\)?
  3. For a circular arc of radius \(2\) and angle \(\pi/2\), what is its length, and what is \(\int_C 5\,ds\)?
  4. A piecewise curve has pieces of lengths \(3\) and \(4\). If the integrand is constantly \(2\), what is the integral over the full curve?
  5. Why is \(\int_a^b c\,dt\) generally not enough to compute \(\int_C c\,ds\) from a parametrization?

Previous

Next

Back to Tutorials