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Line Integrals

Formulas & Definitions
Core Definition

An integral where the function is evaluated along a curve CC. For a vector field F\mathbf{F} along a curve parameterized by r(t)\mathbf{r}(t) on t∈[a,b]t \in [a, b]:

∫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

Concept Explanation

Line integrals evaluate quantities (like potential changes, fluid flow, or mass) along a specific spatial trajectory. If a vector field F\mathbf{F} is conservative (gradient of a scalar potential), the line integral is path-independent and depends only on endpoints.

Visual / Geometric Intuition

Interactive Desmos Graph

Engineering Applications

Widely used to calculate mechanical work done by variable force fields along trajectories, analyze circulation in fluid mechanics (lift equations in flight), and solve voltage changes along wire loops in electromagnetics.

Example Problem

Evaluate the line integral ∫CF⋅dr\int_{C} \mathbf{F} \cdot d\mathbf{r} for the field F(x,y)=⟨y,−x⟩\mathbf{F}(x, y) = \langle y, -x \rangle along the straight line path r(t)=⟨t,t⟩\mathbf{r}(t) = \langle t, t \rangle for t∈[0,1]t \in [0, 1].

Solution

Step 1: Compute dr=r′(t) dtd\mathbf{r} = \mathbf{r}'(t) \, dt

r′(t)=⟨1,1⟩  ⟹  dr=⟨1,1⟩ dt\mathbf{r}'(t) = \langle 1, 1 \rangle \implies d\mathbf{r} = \langle 1, 1 \rangle \, dt

Step 2: Substitute r(t)\mathbf{r}(t) into the vector field F\mathbf{F}

F(r(t))=⟨y(t),−x(t)⟩=⟨t,−t⟩\mathbf{F}(\mathbf{r}(t)) = \langle y(t), -x(t) \rangle = \langle t, -t \rangle

Step 3: Evaluate the dot product F(r(t))⋅r′(t)\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)

⟨t,−t⟩⋅⟨1,1⟩=t(1)+(−t)(1)=0\langle t, -t \rangle \cdot \langle 1, 1 \rangle = t(1) + (-t)(1) = 0

Step 4: Integrate

∫CF⋅dr=∫010 dt=0\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{1} 0 \, dt = 0

Connections & References