Line Integrals

Quick answer

A line integral adds up a quantity along a curve instead of along an interval. The scalar line integral ∫f ds weighs each small piece of the curve by its length, giving masses of wires and averages along paths. The vector line integral ∫F · dr adds up the component of a field along the direction of motion, giving the work a force does on a moving object. Both are computed by parametrizing the curve and turning the line integral into an ordinary integral in t.

What you'll learn

Compute a scalar line integral with ds = |r′(t)| dt
Compute the work integral of a vector field along a curve
Explain how orientation affects each kind of line integral
Interpret line integrals as mass, length and work

Integrating along a curve

A single integral runs along a straight interval of the $x$ -axis. A line integral runs along a curve $C$ , which may bend through the plane or through space. Cut $C$ into short pieces, multiply a quantity on each piece by the piece’s length $\Delta s$ , and add. The limit is

\int_C f(x, y)\,ds

the line integral of $f$ with respect to arc length. If $f$ is the density of a wire bent along $C$ , in mass per unit length, the integral is the mass of the wire. With $f = 1$ it is the length of $C$ .

To compute it, parametrize $C$ as $\mathbf{r}(t)$ for $a \le t \le b$ . A small change $dt$ moves the point a distance $|\mathbf{r}'(t)|\,dt$ , the same arc length element as before:

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

The answer does not depend on which parametrization is used, as long as it traces $C$ once.

Work along a path

A constant force $\mathbf{F}$ moving an object along a straight displacement $\mathbf{d}$ does work $\mathbf{F} \cdot \mathbf{d}$ : only the component of the force along the motion counts. For a varying field and a curved path, cut the path into short pieces. On each piece the displacement is about $\Delta\mathbf{r}$ and the work about $\mathbf{F} \cdot \Delta\mathbf{r}$ . Adding and taking the limit gives the line integral of $\mathbf{F}$ along $C$ :

W = \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}\big(\mathbf{r}(t)\big) \cdot \mathbf{r}'(t)\,dt

In components, with $\mathbf{F} = (P, Q)$ and $d\mathbf{r} = (dx, dy)$ , the same integral is written $\int_C P\,dx + Q\,dy$ .

Why direction matters for work but not for mass

Arc length is always positive, so a wire has the same mass whichever end you start from: $\int_C f\,ds$ ignores direction. Work does not. Walking the path backward reverses every displacement $\Delta\mathbf{r}$ , so every term $\mathbf{F} \cdot \Delta\mathbf{r}$ changes sign. If $-C$ is the curve traced the other way,

\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}

A curve with a chosen direction is oriented, and every work integral needs one. For a closed curve, the usual orientation is counterclockwise, and the integral is written $\oint_C \mathbf{F} \cdot d\mathbf{r}$ . When $\mathbf{F}$ is a fluid’s velocity, this is the circulation: how strongly the flow runs around $C$ .

The field (x², −xy) along a quarter circle

Worked examples

Example 2 — Work against the field

Find the work done by $\mathbf{F}(x, y) = (x^2,\ -xy)$ on a particle moving counterclockwise along the quarter circle from $(1, 0)$ to $(0, 1)$ .

With $\mathbf{r}(t) = (\cos t,\ \sin t)$ , $0 \le t \le \tfrac{\pi}{2}$ :

\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = (\cos^2 t)(-\sin t) + (-\cos t\sin t)(\cos t) = -2\cos^2 t\,\sin t

W = \int_0^{\pi/2} -2\cos^2 t\,\sin t\,dt = \Big[\tfrac{2}{3}\cos^3 t\Big]_0^{\pi/2} = -\frac{2}{3}

The work is negative: the field pushes against the motion, as the figure shows. Traveling from $(0, 1)$ back to $(1, 0)$ , the work is $+\tfrac{2}{3}$ .

One turn of the helix

Common mistakes

Practice problems

Evaluate $\int_C x\,ds$ along the segment from $(0, 0)$ to $(3, 4)$ .

Answer

$\tfrac{15}{2}$

Full solution

With $\mathbf{r}(t) = (3t, 4t)$ , $|\mathbf{r}'(t)| = 5$ , so $\int_0^1 3t \cdot 5\,dt = \tfrac{15}{2}$ .
Find the length of one turn of the helix $(\cos t,\ \sin t,\ t)$ , $0 \le t \le 2\pi$ .

Answer

$2\pi\sqrt{2}$

Full solution

$|\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}$ , so the length is $\int_0^{2\pi} \sqrt{2}\,dt$ .
A wire lies along the upper half of the unit circle, with density $y$ at the point $(x, y)$ . Find its mass.

Answer

$2$

Full solution

$\int_C y\,ds = \int_0^\pi \sin t\,dt = 2$ .
Evaluate $\int_C (x + y)\,ds$ along the segment from $(0, 0)$ to $(1, 1)$ .

Answer

$\sqrt{2}$

Full solution

With $\mathbf{r}(t) = (t, t)$ , $ds = \sqrt{2}\,dt$ , so the integral is $\int_0^1 2t\sqrt{2}\,dt = \sqrt{2}$ .
Find the work done by $\mathbf{F} = (x,\ y)$ along the segment from $(1, 0)$ to $(0, 1)$ .

Answer

$0$

Full solution

With $\mathbf{r}(t) = (1 - t,\ t)$ , $\mathbf{F} \cdot \mathbf{r}' = (1 - t)(-1) + t = 2t - 1$ , and $\int_0^1 (2t - 1)\,dt = 0$ . Both endpoints are at distance $1$ from the origin, and $\mathbf{F} = \nabla\tfrac{x^2 + y^2}{2}$ .
Find the circulation of $\mathbf{F} = (y,\ -x)$ around the unit circle, counterclockwise.

Answer

$-2\pi$

Full solution

$\mathbf{F} \cdot \mathbf{r}' = \sin t(-\sin t) + (-\cos t)\cos t = -1$ , so the integral is $-2\pi$ . This field circles clockwise, against the direction of travel.
Evaluate $\int_C x^2\,dx + y^2\,dy$ along $y = x^2$ from $(0, 0)$ to $(1, 1)$ .

Answer

$\tfrac{2}{3}$

Full solution

With $\mathbf{r}(t) = (t, t^2)$ : $\int_0^1 \big(t^2 + t^4 \cdot 2t\big)\,dt = \tfrac{1}{3} + \tfrac{1}{3} = \tfrac{2}{3}$ .
Without integrating again, find the work done by $\mathbf{F} = (x^2,\ -xy)$ along the quarter circle from $(0, 1)$ to $(1, 0)$ .

Answer

$\tfrac{2}{3}$

Full solution

This is the path of Example 2 traced backward, so the work changes sign: $-\left(-\tfrac{2}{3}\right)$ .
A student computes the work of Example 4 as $\int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,|\mathbf{r}'(t)|\,dt$ . On the unit circle they still get $2\pi$ . Why is the method wrong anyway?

Answer

The factor $|\mathbf{r}'(t)|$ does not belong in a work integral; here it happens to equal $1$ .

Full solution

The vector $\mathbf{r}'(t)$ already carries the speed. On a circle of radius $2$ , with $\mathbf{r}(t) = (2\cos t, 2\sin t)$ , the correct circulation is $8\pi$ , while the extra factor $2$ would give $16\pi$ .

Frequently asked questions

What is a line integral?

An integral taken along a curve. The scalar version ∫f ds adds up f times arc length; the vector version ∫F · dr adds up the tangential component of a field times arc length.

How do you compute ∫f ds?

Parametrize the curve as r(t) for a ≤ t ≤ b. Then ds = |r′(t)| dt, and ∫f ds = ∫ f(r(t)) |r′(t)| dt from a to b.

How do you compute ∫F · dr?

With the same parametrization, dr = r′(t) dt, so ∫F · dr = ∫ F(r(t)) · r′(t) dt from a to b.

Does the direction of the curve matter?

For ∫F · dr, yes: reversing the direction changes the sign. For ∫f ds, no: arc length is positive either way.

What does ∫F · dr mean physically?

When F is a force, it is the work done by the force on an object moving along the curve. When F is a velocity field and the curve is closed, it is the circulation of the flow around the curve.

What to learn next

The Fundamental Theorem for Line IntegralsFor a gradient field, a line integral is the change in the potential between the endpoints. Test whether a field is conservative and find its potential.

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