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Work and Vector Line Integrals · Tutorial 771 of 999

Work Integrals Versus Scalar Line Integrals

Learn to set up, interpret, and compare vector work integrals and scalar line integrals along a parametrized curve.

Intermediate 9 min read

What You'll Learn

  • Define scalar line integrals using arc length and vector work integrals using displacement.
  • Set up both integrals from a parametrized curve.
  • Explain why speed appears in scalar line integrals but a tangent vector appears in work integrals.
  • Determine how reversing the curve changes each integral.
  • Avoid confusing accumulation per unit length with work by a force.

Two Integrals, Two Different Accumulations

A line integral can describe different kinds of accumulation along a curve. A scalar line integral adds up a scalar quantity per unit of distance travelled. A vector work integral adds up the component of a force in the direction of each small displacement.

The distinction matters even when both integrals are evaluated along exactly the same path. A scalar value has no direction, so its accumulation uses arc length. A force has a direction, so its work depends on how that force aligns with the motion.

Central distinction: A scalar line integral has the form \(\int_C f\,ds\). A vector work integral has the form \(\int_C \mathbf{F}\cdot d\mathbf{r}\). The first accumulates scalar values over distance; the second accumulates force components along displacement.

Definitions from a Parametrization

Let \(C\) be a piecewise smooth curve parametrized by \(\mathbf{r}(t)\), for \(a\leq t\leq b\). On each smooth piece, the velocity vector is \(\mathbf{r}'(t)\), and the speed is \(\lVert\mathbf{r}'(t)\rVert\). The arc-length element is \(ds=\lVert\mathbf{r}'(t)\rVert\,dt\).

Definition: If \(f\) is a continuous scalar field along \(C\), its scalar line integral is
$$ \int_C f\,ds = \int_a^b f(\mathbf{r}(t))\lVert\mathbf{r}'(t)\rVert\,dt. $$
The integral is added over the smooth pieces if the curve is piecewise smooth.

This integral uses the value of \(f\) at each point, multiplied by the small amount of distance travelled there. If \(f\) is a mass per unit length, for example, \(\int_C f\,ds\) gives the total mass distributed along the curve. If \(f\) is nonnegative, the accumulated total is nonnegative. More generally, a scalar line integral can be signed if \(f\) takes negative values.

Definition: If \(\mathbf{F}\) is a continuous vector field along \(C\), its work integral is
$$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$
This is the work done by the force field \(\mathbf{F}\) as a particle moves along the oriented curve \(C\).

Here, \(\mathbf{r}'(t)\,dt\) represents an infinitesimal displacement, not an infinitesimal distance. The dot product selects the component of the force along that displacement. A force perpendicular to the motion contributes zero work at that point; a force in the direction of motion contributes positive work, and a force against the motion contributes negative work.

Why the Formulas Have Different Factors

The formulas follow from what each integral is meant to add. Divide a smooth parameter interval into short subintervals. Over one such subinterval, the distance travelled is approximately \(\lVert\mathbf{r}'(t)\rVert\Delta t\), while the displacement vector is approximately \(\mathbf{r}'(t)\Delta t\).

For a scalar field, the contribution from that small portion is approximately the scalar value times the distance. For a force field, the contribution to work is approximately the force dotted with the displacement. Taking limits of these sums produces the two integral formulas.

$$ \begin{aligned} \text{scalar contribution} &\approx f(\mathbf{r}(t))\lVert\mathbf{r}'(t)\rVert\Delta t,\\ \text{work contribution} &\approx \mathbf{F}(\mathbf{r}(t))\cdot \bigl(\mathbf{r}'(t)\Delta t\bigr). \end{aligned} $$

Equivalently, where \(\mathbf{r}'(t)\ne\mathbf{0}\), define the unit tangent vector \(\mathbf{T}(t)=\mathbf{r}'(t)/\lVert\mathbf{r}'(t)\rVert\). Then the work formula can be written as \(\mathbf{F}\cdot\mathbf{T}\,ds\): it accumulates the tangential component of force per unit distance. This makes the difference especially clear. The scalar integral uses \(f\) itself, whereas work uses the tangential component of \(\mathbf{F}\).

Why the formulas work: In a short interval, arc length is approximately \(\lVert\mathbf{r}'(t)\rVert\Delta t\), while displacement is approximately \(\mathbf{r}'(t)\Delta t\). Multiplying a scalar by the former gives scalar accumulation; dotting a force with the latter gives work. Summing these contributions and taking the limit yields the respective line integrals.

Worked Example: Compare Both Integrals on One Segment

Worked Example: Compare Both Integrals on One Segment

Let \(\mathbf{r}(t)=(t,2t)\), \(0\leq t\leq1\). Along this segment, compare the scalar line integral of \(f(x,y)=x+y\) with the work integral for \(\mathbf{F}(x,y)=(y,x)\).

First find the velocity and speed:

$$ \mathbf{r}'(t)=(1,2), \qquad \lVert\mathbf{r}'(t)\rVert=\sqrt{1^2+2^2}=\sqrt{5}. $$

At the point \(\mathbf{r}(t)=(t,2t)\), the scalar field is \(f(\mathbf{r}(t))=t+2t=3t\). Therefore,

$$ \int_C f\,ds = \int_0^1 3t\sqrt{5}\,dt = 3\sqrt{5}\left[\frac{t^2}{2}\right]_0^1 = \frac{3\sqrt{5}}{2}. $$

For work, evaluate the force along the curve: \(\mathbf{F}(\mathbf{r}(t))=(2t,t)\). Its dot product with the velocity is

$$ \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = (2t,t)\cdot(1,2) = 2t+2t = 4t. $$

Thus,

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_0^1 4t\,dt = \left[2t^2\right]_0^1 = 2. $$

Check: The scalar integral weights \(3t\) by the constant speed \(\sqrt{5}\), giving \(3\sqrt{5}/2\). The work integrand is \(4t\), obtained from the force-displacement dot product, giving \(2\). These are different quantities, so there is no reason for their values to agree.

Worked Example: Scalar Accumulation and Work on a Quarter-Circle

Worked Example: Scalar Accumulation and Work on a Quarter-Circle

Consider the counterclockwise quarter-circle \(\mathbf{r}(t)=(2\cos t,2\sin t)\), \(0\leq t\leq\pi/2\). Compute the scalar line integral for \(f(x,y)=x^2+y^2\) and the work integral for \(\mathbf{F}(x,y)=(-y,x)\).

The velocity and speed are

$$ \mathbf{r}'(t)=(-2\sin t,2\cos t), \qquad \lVert\mathbf{r}'(t)\rVert = \sqrt{4\sin^2t+4\cos^2t} = 2. $$

On this circle, \(x^2+y^2=4\), so \(f(\mathbf{r}(t))=4\). The scalar integral is

$$ \int_C f\,ds = \int_0^{\pi/2}4(2)\,dt = 8\left[\ t\ \right]_0^{\pi/2} = 4\pi. $$

For the work integral, the force along the path is \(\mathbf{F}(\mathbf{r}(t))=(-2\sin t,2\cos t)\), which equals \(\mathbf{r}'(t)\). Therefore,

$$ \begin{aligned} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) &=(-2\sin t)^2+(2\cos t)^2\\ &=4\sin^2t+4\cos^2t\\ &=4. \end{aligned} $$

It follows that

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{\pi/2}4\,dt = 4\left[\ t\ \right]_0^{\pi/2} = 2\pi. $$

Check: The quarter-circle has length \(2(\pi/2)=\pi\), and \(f=4\) everywhere on it, so its scalar integral is \(4\pi\). The force is tangent to the path with magnitude \(2\); its work is \(2\pi\), as the computed integral gives.

Orientation: Distance Does Not Have a Direction

Reversing the direction of traversal leaves arc length unchanged. Consequently, a scalar line integral \(\int_C f\,ds\) has the same value when the curve is traversed in reverse. A work integral changes sign under reversal: each displacement changes direction, so the force-displacement dot product changes sign. This distinction is consistent with the earlier result on orientation and work.

Worked Example: A Nonconstant Speed and a Perpendicular Force

Worked Example: A Nonconstant Speed and a Perpendicular Force

Take \(\mathbf{r}(t)=(t^2,0)\), \(0\leq t\leq1\). This traces the horizontal segment from \((0,0)\) to \((1,0)\), but its speed varies. Compute \(\int_C x\,ds\) and the work of the constant force \(\mathbf{F}(x,y)=(0,1)\).

The velocity and speed are

$$ \mathbf{r}'(t)=(2t,0), \qquad \lVert\mathbf{r}'(t)\rVert=2t \quad (0\leq t\leq1). $$

Along the curve, \(x=t^2\). The scalar line integral is therefore

$$ \int_C x\,ds = \int_0^1 t^2(2t)\,dt = \int_0^1 2t^3\,dt = \left[\frac{t^4}{2}\right]_0^1 = \frac12. $$

For work, the force is perpendicular to the velocity at every point:

$$ \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = (0,1)\cdot(2t,0) = 0. $$

Hence,

$$ \int_C\mathbf{F}\cdot d\mathbf{r} = \int_0^1 0\,dt = 0. $$

Check: The scalar result is positive because \(x\) is nonnegative along the segment and is accumulated over distance. The work is zero because the force has no component in the direction of the horizontal displacement. Both results match their interpretations.

A Common Mistake: Using \(dt\) Instead of \(ds\)

A scalar line integral is not generally \(\int_a^b f(\mathbf{r}(t))\,dt\). That expression accumulates with respect to the parameter, not with respect to distance, and can change if the same path is traversed at a different speed. For a scalar line integral, the speed factor must be included.

For instance, on \(\mathbf{r}(t)=(t,2t)\), \(0\leq t\leq1\), take \(f(x,y)=x\). Omitting the speed would give the incorrect calculation

$$ \int_0^1 f(\mathbf{r}(t))\,dt = \int_0^1 t\,dt = \frac12. $$

But the speed is \(\sqrt{5}\), so the correct scalar line integral is

$$ \int_C f\,ds = \int_0^1 t\sqrt{5}\,dt = \frac{\sqrt{5}}{2}. $$

The correction is to use \(ds=\lVert\mathbf{r}'(t)\rVert dt\). A different, equally important error is to replace a work integral by \(\int_C\lVert\mathbf{F}\rVert\,ds\). That expression accumulates the force magnitude, not its tangential component. Work requires the dot product \(\mathbf{F}\cdot d\mathbf{r}\).

How to Choose the Integral

Start by asking what is being accumulated. If the quantity is a scalar amount per unit length, use \(\int_C f\,ds\). If a force acts while an object moves and the desired quantity is work, use \(\int_C\mathbf{F}\cdot d\mathbf{r}\). Then parametrize the curve, evaluate the field along it, and include the appropriate factor: speed for a scalar line integral, or the velocity vector in the force dot product.

The distinction also clarifies units. If \(f\) has units of mass per unit length, then \(\int_C f\,ds\) has units of mass. If \(\mathbf{F}\) has units of force, then \(\int_C\mathbf{F}\cdot d\mathbf{r}\) has units of force times distance, or work. Dimensional consistency is a useful check on the setup.

Check Your Understanding

Use the definitions and interpretations to answer these questions.

  1. For a scalar field \(f\) and parametrization \(\mathbf{r}(t)\), what factor converts \(dt\) into arc length?
  2. Why does a force perpendicular to the motion contribute zero work?
  3. What happens to \(\int_C f\,ds\) and \(\int_C\mathbf{F}\cdot d\mathbf{r}\) when the orientation of \(C\) is reversed?
  4. On \(\mathbf{r}(t)=(t,3t)\), \(0\leq t\leq1\), set up—not evaluate—the scalar line integral for \(f(x,y)=x\).
  5. Explain why \(\int_C\lVert\mathbf{F}\rVert\,ds\) is not, in general, the work done by \(\mathbf{F}\).

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