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Jamie Little
Part IA Michaelmas, Lent, Easter Term

Line Integrals

Curves

A curve in space is parameterised by:

r(t)=(x(t),y(t),z(t)),t∈[t1,t2]\mathbf{r}(t) = (x(t), y(t), z(t)), \quad t \in [t_1, t_2]

Orientation: Parameter tt gives direction along the curve.

Closed curve: r(t1)=r(t2)\mathbf{r}(t_1) = \mathbf{r}(t_2)

Definition of Line Integral

Given vector field F(r)\mathbf{F}(\mathbf{r}) and curve Γ\Gamma:

∫ΓF⋅dr=∫t1t2F(r(t))⋅drdt dt\int_\Gamma \mathbf{F} \cdot d\mathbf{r} = \int_{t_1}^{t_2} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}\,dt

Curve Γ through a vector field F, showing small displacement dr and the dot product F·dr along the path

Interpretation: The integral adds up the component of F\mathbf{F} along the curve at each point.

Properties

  1. Parameterisation independent: Different parameterisations of the same curve give the same integral.

  2. Orientation matters: Reversing the curve multiplies by −1-1: ∫−ΓF⋅dr=−∫ΓF⋅dr\int_{-\Gamma} \mathbf{F} \cdot d\mathbf{r} = -\int_\Gamma \mathbf{F} \cdot d\mathbf{r}

  3. Additivity: For Γ=Γ1+Γ2\Gamma = \Gamma_1 + \Gamma_2: ∫Γ=∫Γ1+∫Γ2\int_\Gamma = \int_{\Gamma_1} + \int_{\Gamma_2}

Work Done

Physical interpretation: If F\mathbf{F} is force, the line integral gives work done:

W=∫ΓF⋅drW = \int_\Gamma \mathbf{F} \cdot d\mathbf{r}

Example: Work done by gravity moving an object along a path.

Example

Evaluate ∫ΓF⋅dr\int_\Gamma \mathbf{F} \cdot d\mathbf{r} for F=(2xy,x2)\mathbf{F} = (2xy, x^2) along:

  1. Straight line from (0,0)(0,0) to (1,1)(1,1)
  2. Parabola y=x2y = x^2 from (0,0)(0,0) to (1,1)(1,1)

Line: Parameterise r(t)=(t,t)\mathbf{r}(t) = (t, t), t∈[0,1]t \in [0, 1].

drdt=(1,1)\frac{d\mathbf{r}}{dt} = (1, 1)

F=(2t2,t2)\mathbf{F} = (2t^2, t^2)

F⋅drdt=2t2+t2=3t2\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = 2t^2 + t^2 = 3t^2

∫013t2 dt=[t3]01=1\int_0^1 3t^2\,dt = [t^3]_0^1 = 1

Parabola: Parameterise r(t)=(t,t2)\mathbf{r}(t) = (t, t^2), t∈[0,1]t \in [0, 1].

drdt=(1,2t)\frac{d\mathbf{r}}{dt} = (1, 2t)

F=(2t3,t2)\mathbf{F} = (2t^3, t^2)

F⋅drdt=2t3+2t3=4t3\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = 2t^3 + 2t^3 = 4t^3

∫014t3 dt=[t4]01=1\int_0^1 4t^3\,dt = [t^4]_0^1 = 1

Both paths give the same result — F\mathbf{F} may be conservative.

Conservative Fields

A vector field is conservative if the line integral depends only on endpoints, not the path.

Equivalent conditions:

  1. F=∇ϕ\mathbf{F} = \nabla \phi for some scalar field ϕ\phi (potential)
  2. F⋅dr\mathbf{F} \cdot d\mathbf{r} is an exact differential
  3. ∮CF⋅dr=0\oint_C \mathbf{F} \cdot d\mathbf{r} = 0 for all closed curves CC
  4. ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} (curl-free)

Path Independence

If F=∇ϕ\mathbf{F} = \nabla \phi:

∫ΓF⋅dr=∫Γ∇ϕ⋅dr=ϕ(rend)−ϕ(rstart)\int_\Gamma \mathbf{F} \cdot d\mathbf{r} = \int_\Gamma \nabla \phi \cdot d\mathbf{r} = \phi(\mathbf{r}_\text{end}) - \phi(\mathbf{r}_\text{start})

The line integral equals the potential difference.

Proof:

∇ϕ⋅dr=∂ϕ∂xdx+∂ϕ∂ydy+∂ϕ∂zdz=dϕ\nabla \phi \cdot d\mathbf{r} = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy + \frac{\partial \phi}{\partial z}dz = d\phi

Finding the Potential

If we know F=∇ϕ\mathbf{F} = \nabla \phi and want ϕ\phi:

ϕ(r)=∫r0rF⋅dr+C\phi(\mathbf{r}) = \int_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{F} \cdot d\mathbf{r} + C

Choose any convenient path (e.g., along coordinate axes).

Why Line Integrals Matter

  • Work done by forces: Fundamental in mechanics
  • Circulation: ∮v⋅dr\oint \mathbf{v} \cdot d\mathbf{r} measures rotation in fluids
  • Electromagnetic induction: ∮E⋅dr=−dΦdt\oint \mathbf{E} \cdot d\mathbf{r} = -\frac{d\Phi}{dt} (Faraday’s law)
  • Conservative systems: Potential energy reconstructs from force field
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