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Line Integral — Definition, Formula & Examples

A line integral is the integral of a function evaluated along a curve, accumulating values of the function as you travel the path. It generalizes ordinary definite integrals from straight intervals to arbitrary curves in two or three dimensions.

Given a smooth curve CC parameterized by r(t)\mathbf{r}(t) for a≤t≤ba \le t \le b, the line integral of a scalar field ff over CC is ∫Cf ds=∫abf(r(t)) ∥r′(t)∥ dt\int_C f\,ds = \int_a^b f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt. For a vector field F\mathbf{F}, the line integral is ∫CF⋅dr=∫abF(r(t))⋅r′(t) dt\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt.

Key Formula

∫CF⋅dr=∫abF(r(t))⋅r′(t) dt\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt
Where:
  • CC = The curve along which you integrate
  • F\mathbf{F} = A vector field defined along the curve
  • r(t)\mathbf{r}(t) = Parameterization of the curve for t in [a, b]
  • r′(t)\mathbf{r}'(t) = Derivative of the parameterization (tangent vector)

How It Works

To evaluate a line integral, first parameterize the curve CC using a parameter tt. For a scalar line integral, compute ∥r′(t)∥\|\mathbf{r}'(t)\| (the speed), multiply by f(r(t))f(\mathbf{r}(t)), and integrate over tt. For a vector line integral, take the dot product of the vector field with r′(t)\mathbf{r}'(t) and integrate. The scalar version measures quantities like mass along a wire with varying density; the vector version measures work done by a force along a path.

Worked Example

Problem: Evaluate the line integral ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} where F=⟨2x,3y⟩\mathbf{F} = \langle 2x, 3y \rangle and CC is the line segment from (0,0)(0,0) to (1,2)(1,2).
Parameterize the curve: Let r(t)=⟨t, 2t⟩\mathbf{r}(t) = \langle t,\, 2t \rangle for 0≤t≤10 \le t \le 1. Then r′(t)=⟨1,2⟩\mathbf{r}'(t) = \langle 1, 2 \rangle.
r(t)=⟨t, 2t⟩,r′(t)=⟨1, 2⟩\mathbf{r}(t) = \langle t,\, 2t \rangle, \quad \mathbf{r}'(t) = \langle 1,\, 2 \rangle
Substitute into the integrand: Along the curve, F(r(t))=⟨2t, 6t⟩\mathbf{F}(\mathbf{r}(t)) = \langle 2t,\, 6t \rangle. The dot product with r′(t)\mathbf{r}'(t) is 2t(1)+6t(2)=14t2t(1) + 6t(2) = 14t.
F⋅r′(t)=2t+12t=14t\mathbf{F} \cdot \mathbf{r}'(t) = 2t + 12t = 14t
Integrate: Evaluate the integral from 0 to 1.
∫0114t dt=7t2∣01=7\int_0^1 14t\,dt = 7t^2 \Big|_0^1 = 7
Answer: The line integral equals 77.

Why It Matters

Line integrals are essential for computing work done by a force field in physics and for calculating circulation and flux in fluid dynamics. They appear directly in Green's Theorem, Stokes' Theorem, and the Divergence Theorem, which form the backbone of vector calculus used throughout engineering and physics courses.

Common Mistakes

Mistake: Forgetting to include ∥r′(t)∥\|\mathbf{r}'(t)\| in scalar line integrals or confusing dsds with dtdt.
Correction: Remember that ds=∥r′(t)∥ dtds = \|\mathbf{r}'(t)\|\,dt. For scalar integrals you must multiply by the speed factor; for vector integrals you use r′(t) dt\mathbf{r}'(t)\,dt directly (no magnitude).

Related Terms

  • Gradient — Gradient fields have path-independent line integrals
  • Green's Theorem — Converts a line integral to a double integral
  • Divergence — Measures flux density related to line/surface integrals
  • Divergence Theorem — Generalizes line integral results to higher dimensions
  • Curvature — Uses arc length parameterization from line integrals
  • Partial Derivative — Needed to check if a field is conservative