Line Integrals of Scalar and Vector Fields: Work, Circulation and Path Independence

A line integral sums a quantity along a curve: the scalar line integral ∫_C f ds integrates a function with respect to arc length, and the vector line integral ∫_C F · dr integrates the tangential component of a vector field, giving the work done by a force along the path. In brief: for F = (y, x) and the path r(t) = (t, t²), 0 ≤ t ≤ 1, the work is ∫₀¹ (t² · 1 + t · 2t) dt = ∫₀¹ 3t² dt = 1. Because this F is the gradient of xy, the answer depends only on the endpoints: xy at (1, 1) minus xy at (0, 0) = 1.

Line integrals generalize the single integral from an interval to a curve. This guide covers the parametrization of curves, scalar integrals (mass of a wire), vector integrals (work and circulation), conservative fields with potentials, and the test for path independence, with worked examples and a summary table.

What is a scalar line integral?

Let a smooth curve C be parametrized by r(t) = (x(t), y(t)), a ≤ t ≤ b. The arc length element is ds = |r′(t)| dt = √(x′² + y′²) dt. Then ∫_C f ds = ∫_a^b f(r(t)) |r′(t)| dt. With f = 1 it gives the length of C. With f = linear density, it gives the mass of a wire. The value does not depend on the direction of travel.

Example 1: the length of a helix

r(t) = (cos t, sin t, t), 0 ≤ t ≤ 2π. |r′| = √(sin² t + cos² t + 1) = √2. Length = 2π√2 ≈ 8.886.

Example 2: the mass of a wire

A semicircular wire of radius 2, r(t) = (2 cos t, 2 sin t), 0 ≤ t ≤ π, with density f = y. |r′| = 2. Mass = ∫₀^π 2 sin t · 2 dt = 4 · 2 = 8.

Example 3: a straight segment

∫_C (x + y) ds over the segment from (0, 0) to (3, 4). Parametrize r(t) = (3t, 4t), 0 ≤ t ≤ 1, |r′| = 5. Then ∫₀¹ 7t · 5 dt = 17.5.

What is a line integral of a vector field?

For F = (P, Q) and a curve C with r(t), ∫_C F · dr = ∫_a^b F(r(t)) · r′(t) dt = ∫ P dx + Q dy. It is the work done by the force F moving along C. When C is closed, the integral ∮ F · dr is called the circulation. Reversing the direction of C changes the sign.

Example 4: work along a parabola

F = (x², xy), C: r(t) = (t, t²), 0 ≤ t ≤ 1. F(r(t)) · r′(t) = (t², t³) · (1, 2t) = t² + 2t⁴. Integral = 1/3 + 2/5 = 11/15.

Example 5: the same endpoints, a different path

Same F, but along the straight segment r(t) = (t, t), 0 ≤ t ≤ 1: F · r′ = (t², t²) · (1, 1) = 2t². Integral = 2/3 ≈ 0.667, which differs from 11/15 ≈ 0.733. For this field, the work depends on the path.

Example 6: circulation around a circle

F = (−y, x), C: r(t) = (cos t, sin t), 0 ≤ t ≤ 2π. F · r′ = (−sin t, cos t) · (−sin t, cos t) = 1. Circulation = 2π. The field rotates around the origin and always pushes along the circle.

What is a conservative field?

A field F is conservative if F = ∇φ for a scalar function φ, called a potential. Then the fundamental theorem for line integrals states that ∫_C F · dr = φ(B) − φ(A), where A and B are the endpoints. Consequently the integral over every closed curve is zero.

Example 7: using a potential

F = (2xy, x²) = ∇(x²y). The work from (0, 0) to (2, 3) along any path is 4 · 3 − 0 = 12.

Example 8: gravity

Near the Earth’s surface F = (0, 0, −mg) = ∇(−mgz). Lifting a mass from height z₁ to z₂ requires the work mg(z₂ − z₁), regardless of the route.

How do you test for path independence?

On a simply connected region (without holes), a field F = (P, Q) with continuous partial derivatives is conservative if and only if ∂Q/∂x = ∂P/∂y. In three dimensions, the condition is curl F = 0. Examples of what happens with a hole are given by the field F = (−y, x)/(x² + y²), which has ∂Q/∂x = ∂P/∂y everywhere except the origin, yet its circulation around the unit circle is 2π. That is why the simply connected assumption matters.

Example 9: testing two fields

F = (2xy, x²): ∂Q/∂x = 2x and ∂P/∂y = 2x. Equal, so F is conservative. G = (y, −x): ∂Q/∂x = −1, ∂P/∂y = 1. Not equal, so G is not conservative; for example, its circulation along the unit circle is ∫ (sin t · (−sin t) + (−cos t)(cos t)) dt = −2π, not zero.

Example 10: a 3D check

F = (yz, xz, xy) has curl F = (x − x, y − y, z − z) = 0, with the potential φ = xyz. The integral from (0, 0, 0) to (1, 2, 3) is 6.

How do you find a potential function?

Integrate P with respect to x to get φ = ∫ P dx + g(y). Differentiate with respect to y, compare with Q to find g′(y), then integrate.

Example 11: finding a potential

F = (2x + y, x + 3y²). ∂Q/∂x = 1 = ∂P/∂y. φ = ∫ (2x + y) dx = x² + xy + g(y). φ_y = x + g′(y) = x + 3y², so g = y³. φ = x² + xy + y³. The integral from (0, 0) to (1, 1) is 1 + 1 + 1 = 3.

Example 12: a closed path

For the same F, the integral around any closed curve is zero. This is faster than parametrizing the path.

Table of methods

Where do line integrals come from?

The concept of work as a force along a path was developed in the early nineteenth century. The notion of a potential was introduced in the work of Lagrange and Laplace on gravitation, and the name “potential” is associated with George Green (1828) and Carl Friedrich Gauss. The relation between circulation and area, studied, was published by Green in an essay in 1828.

Where are line integrals used?

In physics they give the work of electric, magnetic and gravitational forces, the voltage along a wire in a changing magnetic field (Faraday’s law) and the circulation of a fluid. In engineering they compute masses and centers of mass of wires. In thermodynamics, the fact that energy is a state function reflects that its differential is exact.

Which mistakes are common?

Forgetting |r′(t)| in the scalar integral

The arc length element is not dt. Multiply by the speed of the parametrization.

Ignoring orientation

A vector line integral changes sign when the direction is reversed. A scalar one does not.

Applying the potential method to a non-conservative field

Test ∂Q/∂x = ∂P/∂y first. Example 5 shows that two paths with the same endpoints give different answers for a non-conservative field.

Forgetting holes in the domain

Equal partial derivatives guarantee a potential only on a simply connected region.