A line integral adds up quantities along a curve instead of over an interval or a flat region. It lets you measure accumulated mass, charge, temperature, or work along a path that may bend through two or three dimensions. This matters in physics because real motion often follows curved paths, and forces or fields can change from point to point.
Line integrals connect geometry, calculus, and physical meaning in one calculation.
To compute a line integral, you usually parameterize the curve as r(t), then rewrite the integrand and distance element in terms of t. For a scalar field f, the integral ∫C f ds accumulates field values weighted by small arc lengths. For a vector field F, the integral ∫C F · dr measures how much the field pushes in the direction of travel, which is work when F is a force.
The sign and value can depend on direction, path shape, and whether the vector field is conservative.
Key Facts
- Scalar line integral: ∫C f ds = ∫a^b f(r(t)) |r'(t)| dt
- Vector line integral: ∫C F · dr = ∫a^b F(r(t)) · r'(t) dt
- Work by a force field: W = ∫C F · dr
- Arc length element: ds = |r'(t)| dt
- Reversing direction leaves ∫C f ds unchanged but changes ∫C F · dr to its negative.
- If F = ∇φ is conservative, then ∫C F · dr = φ(B) - φ(A), so the integral depends only on endpoints.
Vocabulary
- Line integral
- An integral that accumulates a scalar or vector quantity along a curve.
- Parameterization
- A description of a curve using a vector function r(t) that gives position as a parameter changes.
- Scalar field
- A function that assigns a single number, such as temperature or density, to each point in space.
- Vector field
- A function that assigns a vector, such as force or velocity, to each point in space.
- Conservative field
- A vector field whose line integral between two points is independent of the path and can be written as the gradient of a potential function.
Common Mistakes to Avoid
- Using dt instead of ds in a scalar line integral, which is wrong because scalar accumulation along a curve must include the speed factor |r'(t)|.
- Forgetting the dot product in ∫C F · dr, which is wrong because only the component of the vector field tangent to the path contributes to work.
- Ignoring the direction of traversal for a vector line integral, which is wrong because reversing the path changes the sign of ∫C F · dr.
- Using endpoints only for every vector field, which is wrong because endpoint shortcuts apply only to conservative fields.
Practice Questions
- 1 Compute ∫C f ds for f(x, y) = x + y along the line segment from (0, 0) to (3, 4).
- 2 Let F(x, y) = <2x, y> and C be r(t) = <t, t^2> for 0 ≤ t ≤ 2. Compute ∫C F · dr.
- 3 A force field is perpendicular to the tangent direction at every point of a moving particle's path. Explain what the line integral ∫C F · dr equals and why.