Line Integrals: Examples and Practice

A line integral adds up a quantity along a curve C instead of along an interval. There are two kinds. The scalar line integral ∫_C f ds adds f times arc length, giving things like the mass of a wire. The vector line integral ∫_C F · dr adds the part of a vector field along the curve, giving the work done by a force.

Before this lesson: Vector Fields, Vector Functions and Space Curves and Arc Length of a Space Curve

What a line integral measures

Cut the curve C into many tiny pieces. For a scalar line integral, multiply the value of f on each piece by the piece’s length ds, then add. If f is the density of a wire in grams per centimeter, the total is the mass of the wire.

For a vector line integral, F is a force and dr is a tiny step along the curve. The dot product F · dr keeps only the part of the force that points along the step. Adding these gives the work done by F on an object that moves along C. A force pushing with the motion does positive work; a force pushing against it does negative work.

Along y = x² (Example 3), the field F = ⟨y, x²⟩ points partly along the path everywhere, so the work is positive: 5/6.

Line integral formulas

Describe C with a vector function r(t) = ⟨x(t), y(t)⟩ or ⟨x(t), y(t), z(t)⟩, for a ≤ t ≤ b. Then ds = |r′(t)| dt and dr = r′(t) dt.

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

\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}\big(\mathbf{r}(t)\big) \cdot \mathbf{r}'(t)\,dt = \int_C P\,dx + Q\,dy

Parametrize the curve: write r(t) and the interval a ≤ t ≤ b, going in the given direction.
Find r′(t). For f ds, also find |r′(t)|.
Substitute x(t), y(t) (and z(t)) into f or F.
Multiply by |r′(t)| or dot with r′(t), then integrate from a to b.

Common parametrizations: the segment from point A to point B is r(t) = A + t(B − A), 0 ≤ t ≤ 1. The circle of radius a, counterclockwise, is r(t) = ⟨a cos t, a sin t⟩, 0 ≤ t ≤ 2π. The graph y = g(x) is r(t) = ⟨t, g(t)⟩.

Line integrals examples

Example 1: A scalar line integral along a segment

Evaluate $\int_C (x + y)\,ds$ , where C is the segment from (0, 0) to (3, 4).

\mathbf{r}(t) = \langle 3t, 4t \rangle, \quad 0 \le t \le 1

Parametrize the segment: A + t(B − A).

\mathbf{r}'(t) = \langle 3, 4 \rangle

Differentiate.

|\mathbf{r}'(t)| = \sqrt{9 + 16} = 5

Find the speed.

x + y = 3t + 4t = 7t

Substitute x(t) and y(t).

\int_C (x + y)\,ds = \int_0^1 7t \cdot 5\,dt

Set up: f times |r′(t)|.

= \Big[\dfrac{35t^2}{2}\Big]_0^1 = \dfrac{35}{2}

Integrate and substitute.

Answer: 35/2 = 17.5. Check: the segment has length 5 and the average value of x + y on it is 3.5, and 5 × 3.5 = 17.5.

Example 2: A scalar line integral on a semicircle

Evaluate $\int_C x^2\,ds$ , where C is the upper half of the unit circle.

\mathbf{r}(t) = \langle \cos t, \sin t \rangle, \quad 0 \le t \le \pi

Parametrize the upper half.

|\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1

Find the speed: r′(t) = ⟨−sin t, cos t⟩.

\int_C x^2\,ds = \int_0^{\pi} \cos^2 t\,dt

Substitute x = cos t and ds = 1 dt.

= \int_0^{\pi} \dfrac{1 + \cos 2t}{2}\,dt

Use the power-reducing identity (see half-angle formulas).

= \Big[\dfrac{t}{2} + \dfrac{\sin 2t}{4}\Big]_0^{\pi}

Integrate.

= \dfrac{\pi}{2}

Substitute: sin 2π = sin 0 = 0.

Answer: π/2 ≈ 1.571.

Example 3: Work along a parabola

Evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ , where F(x, y) = ⟨y, x²⟩ and C is the parabola y = x² from (0, 0) to (1, 1).

\mathbf{r}(t) = \langle t, t^2 \rangle, \quad 0 \le t \le 1

Parametrize the graph with x = t.

\mathbf{r}'(t) = \langle 1, 2t \rangle

Differentiate.

\mathbf{F}\big(\mathbf{r}(t)\big) = \langle t^2, t^2 \rangle

Substitute: y = t² and x² = t².

\mathbf{F} \cdot \mathbf{r}'(t) = t^2(1) + t^2(2t)

Take the dot product.

= t^2 + 2t^3

Multiply.

\int_0^1 (t^2 + 2t^3)\,dt = \Big[\dfrac{t^3}{3} + \dfrac{t^4}{2}\Big]_0^1

Integrate.

= \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{5}{6}

Substitute and add.

Answer: 5/6.

Example 4: Circulation around a circle

Find the work done by F(x, y) = ⟨−y, x⟩ on a particle that moves once counterclockwise around the circle x² + y² = 4.

\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle, \quad 0 \le t \le 2\pi

Parametrize the circle counterclockwise.

\mathbf{r}'(t) = \langle -2\sin t, 2\cos t \rangle

Differentiate.

\mathbf{F}\big(\mathbf{r}(t)\big) = \langle -2\sin t, 2\cos t \rangle

Substitute: −y = −2 sin t and x = 2 cos t.

\mathbf{F} \cdot \mathbf{r}'(t) = 4\sin^2 t + 4\cos^2 t

Take the dot product.

= 4

Use sin² t + cos² t = 1.

W = \int_0^{2\pi} 4\,dt = 8\pi

Integrate.

Answer: W = 8π ≈ 25.13. The field swirls counterclockwise, the same way the particle moves, so it helps the motion all the way around. Going clockwise would give −8π.

Example 5: Work on a bead along a twisted wire

A bead slides along a wire shaped like r(t) = ⟨t, t², t³⟩ for 0 ≤ t ≤ 1, with distances in meters. A force F(x, y, z) = ⟨y, z, x⟩ newtons acts on it. Find the work done by the force.

\mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle

Differentiate.

\mathbf{F}\big(\mathbf{r}(t)\big) = \langle t^2, t^3, t \rangle

Substitute: y = t², z = t³, x = t.

\mathbf{F} \cdot \mathbf{r}'(t) = t^2 + 2t^4 + 3t^3

Take the dot product: t²(1) + t³(2t) + t(3t²).

W = \int_0^1 (t^2 + 3t^3 + 2t^4)\,dt

Set up the work integral.

= \dfrac{1}{3} + \dfrac{3}{4} + \dfrac{2}{5}

Integrate each term and substitute t = 1.

= \dfrac{20 + 45 + 24}{60} = \dfrac{89}{60}

Add over the common denominator 60.

Answer: W = 89/60 ≈ 1.48 joules.

Common mistakes

Mistake	Example of the error	Fix
Leaving out \|r′(t)\| in f ds	Writing ∫ 7t dt in Example 1 and getting 7/2	ds = \|r′(t)\| dt. Here \|r′(t)\| = 5.
Using \|r′(t)\| in F · dr	Multiplying F by the speed instead of taking a dot product	dr = r′(t) dt, so dot F with the vector r′(t).
Ignoring direction	Going clockwise in Example 4 and still reporting 8π	Reversing C changes the sign of ∫ F · dr (but not of ∫ f ds).
Differentiating a component wrong	Using r′(t) = ⟨1, t²⟩ for r(t) = ⟨t, t²⟩	Differentiate each component: r′(t) = ⟨1, 2t⟩.
Wrong t-interval	Using 0 ≤ t ≤ 2π for the upper semicircle	The upper half of the circle is 0 ≤ t ≤ π.

Related methods

The differential form P dx + Q dy

Line integrals are often written ∫_C P dx + Q dy. Substitute dx = x′(t) dt and dy = y′(t) dt. This is the same integral as ∫_C F · dr with F = ⟨P, Q⟩.

Path independence

For most fields, the value of ∫ F · dr depends on the path, not just on its endpoints. For gradient fields it doesn’t: then ∫_C ∇f · dr = f(end) − f(start). That shortcut is the fundamental theorem for line integrals, in the next lesson.

Piecewise curves

If C is made of several smooth pieces, such as the sides of a triangle, parametrize each piece separately and add the results.

Practice problems

Evaluate $\int_C y\,ds$ , where C is the segment from (0, 0) to (6, 8).
Evaluate $\int_C xy\,ds$ , where C is the quarter of the circle x² + y² = 4 in the first quadrant.
Evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ , where F = ⟨x, y⟩ and C is r(t) = ⟨t, t²⟩, 0 ≤ t ≤ 2.
Evaluate $\int_C y\,dx + x\,dy$ , where C is the segment from (0, 0) to (2, 3).
Find the work done by F = ⟨y, −x⟩ once counterclockwise around the unit circle.
Evaluate $\int_C (x^2 + y^2 + z^2)\,ds$ , where C is the helix r(t) = ⟨cos t, sin t, t⟩, 0 ≤ t ≤ 2π.
A thin wire lies along the upper half of the circle x² + y² = 9, in centimeters. Its density is δ(x, y) = y grams per centimeter. Find its mass.
If ∫_C F · dr = 5 and ∫_C f ds = 7, find both integrals along −C, the same curve traced backward.
Evaluate ∫_C F · dr for F = ⟨y², xy⟩ from (0, 0) to (1, 1) along (a) the line y = x and (b) the parabola y = x².
A drone with weight force F = ⟨0, 0, −19.6⟩ newtons flies along r(t) = ⟨4t, 3t, 10t⟩, 0 ≤ t ≤ 1, in meters. Find the work done by gravity.

Show answers

\mathbf{r}(t) = \langle 6t, 8t \rangle, \quad |\mathbf{r}'(t)| = \sqrt{36 + 64} = 10

Parametrize the segment and find the speed.

\int_0^1 8t \cdot 10\,dt = \Big[40t^2\Big]_0^1 = 40

Substitute y = 8t and integrate.

Answer: 40

\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle, \quad 0 \le t \le \dfrac{\pi}{2}, \quad |\mathbf{r}'(t)| = 2

Parametrize the quarter circle.

\int_0^{\pi/2} 4\cos t\sin t \cdot 2\,dt = 8\int_0^{\pi/2} \sin t\cos t\,dt

Substitute xy = 4 cos t sin t.

= 8 \cdot \Big[\dfrac{\sin^2 t}{2}\Big]_0^{\pi/2} = 4

Integrate with u-substitution, u = sin t.

Answer: 4

\mathbf{F} \cdot \mathbf{r}'(t) = \langle t, t^2 \rangle \cdot \langle 1, 2t \rangle = t + 2t^3

Substitute and take the dot product.

\int_0^2 (t + 2t^3)\,dt = \Big[\dfrac{t^2}{2} + \dfrac{t^4}{2}\Big]_0^2 = 2 + 8 = 10

Integrate and substitute t = 2.

Answer: 10

x = 2t, \; y = 3t, \; dx = 2\,dt, \; dy = 3\,dt

Parametrize the segment, 0 ≤ t ≤ 1.

y\,dx + x\,dy = 3t(2\,dt) + 2t(3\,dt) = 12t\,dt

Substitute and combine like terms.

\int_0^1 12t\,dt = 6

Integrate.

Answer: 6

\mathbf{F} \cdot \mathbf{r}'(t) = \langle \sin t, -\cos t \rangle \cdot \langle -\sin t, \cos t \rangle

Use r(t) = ⟨cos t, sin t⟩, 0 ≤ t ≤ 2π.

= -\sin^2 t - \cos^2 t = -1

Take the dot product and use sin² t + cos² t = 1.

W = \int_0^{2\pi} (-1)\,dt = -2\pi

Integrate.

Answer: −2π ≈ −6.28. This field swirls clockwise, against the motion.

|\mathbf{r}'(t)| = |\langle -\sin t, \cos t, 1 \rangle| = \sqrt{2}

Find the speed.

x^2 + y^2 + z^2 = \cos^2 t + \sin^2 t + t^2 = 1 + t^2

Substitute.

\int_0^{2\pi} (1 + t^2)\sqrt{2}\,dt = \sqrt{2}\Big[t + \dfrac{t^3}{3}\Big]_0^{2\pi}

Set up and integrate.

= \sqrt{2}\Big(2\pi + \dfrac{8\pi^3}{3}\Big) \approx 125.82

Substitute t = 2π: (2π)³ = 8π³.

Answer: √2 (2π + 8π³/3) ≈ 125.82

\mathbf{r}(t) = \langle 3\cos t, 3\sin t \rangle, \quad 0 \le t \le \pi, \quad |\mathbf{r}'(t)| = 3

Parametrize the upper half.

m = \int_0^{\pi} 3\sin t \cdot 3\,dt = 9\Big[-\cos t\Big]_0^{\pi}

Mass = ∫ δ ds.

= 9(1 + 1) = 18

Substitute.

Answer: 18 g

\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r} = -5

Reversing the direction flips the sign of the work.

\int_{-C} f\,ds = \int_C f\,ds = 7

Arc length ds is always positive, so f ds doesn’t change.

Answer: −5 and 7

(a)\; \mathbf{r} = \langle t, t \rangle: \; \int_0^1 (t^2 \cdot 1 + t^2 \cdot 1)\,dt = \dfrac{2}{3}

On y = x, F = ⟨t², t²⟩ and r′ = ⟨1, 1⟩.

(b)\; \mathbf{r} = \langle t, t^2 \rangle: \; \int_0^1 (t^4 \cdot 1 + t^3 \cdot 2t)\,dt = \int_0^1 3t^4\,dt = \dfrac{3}{5}

On y = x², F = ⟨t⁴, t³⟩ and r′ = ⟨1, 2t⟩.

Answer: (a) 2/3; (b) 3/5. Different paths with the same endpoints give different values here.

10.

\mathbf{F} \cdot \mathbf{r}'(t) = \langle 0, 0, -19.6 \rangle \cdot \langle 4, 3, 10 \rangle = -196

Take the dot product: only the z-parts multiply to a nonzero number.

W = \int_0^1 (-196)\,dt = -196

Integrate.

Answer: −196 J. Gravity does negative work because the drone climbs 10 m: −19.6 × 10 = −196.

Line integrals FAQ

What is the difference between ds and dr in a line integral?

ds = |r′(t)| dt is a tiny length, a number. dr = r′(t) dt is a tiny step, a vector. Use ds with a scalar function f, and use dr with a vector field F and a dot product.

Does the direction of the curve matter?

For ∫ F · dr, yes: reversing the curve changes the sign. For ∫ f ds, no: lengths are positive in either direction.

How do you find work with a line integral?

Work is W = ∫_C F · dr. Parametrize the path, substitute into F, dot with r′(t) and integrate over the t-interval.

What does a line integral of a scalar function represent?

With f = 1 it is the length of C. With a density f, it is the mass of a wire shaped like C. Dividing by the length gives the average value of f along C.

Summary

Scalar line integral: ∫_C f ds = ∫ f(r(t)) |r′(t)| dt.
Vector line integral: ∫_C F · dr = ∫ F(r(t)) · r′(t) dt = ∫ P dx + Q dy.
Parametrize C first, in the direction of travel.
Work = ∫ F · dr. Reversing C changes its sign; ∫ f ds does not change.
In general, ∫ F · dr depends on the path.

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