Scalar Line Integral

Let γ:[a,b]↦X\gamma: [a,b]\mapsto X be a path of class C1C^1 where X⊂RnX\subset \mathbb{R}^n and f:X↦Rf:X\mapsto \mathbb{R} be a continuous function. The scalar line integral of ff along γ\gamma is defined by

∫γf ds=∫abf(γ(t))∥γ′(t)∥ dt\int_\gamma f\, ds = \int_a^b f(\gamma(t))\Vert \gamma'(t) \Vert \, dt

Vector Line Integral

∫γF⋅ ds=∫abF(γ(t))⋅γ′(t) dt\int_\gamma \mathbf{F} \cdot \, d\mathbf{s} = \int_a^b \mathbf{F}(\gamma(t)) \cdot \gamma'(t) \, dt

Interpretation: If F\mathbf{F} is a vector field of force, then ∫γF⋅ ds\int_\gamma \mathbf{F}\cdot\, d\mathbf{s} gives the work done on an object when it travels through the vector field along γ\gamma.

Equivalently, if F=(F1,F2,…,Fn)\mathbf{F}=(F_1, F_2, \dots, F_n), we can write

∫γF⋅ds=∫ab(F1 dx1+F2 dx2+…,+Fn dxn)\int_\gamma \mathbf{F}\cdot d\mathbf{s} = \int_a^b (F_1\, dx_1 + F_2\, dx_2 + \dots, + F_n\, dx_n)

Parameterization. Let CC be a curve in Rn\mathbb{R}^n. A parameterization of CC is a path γ:[a,b]↦Rn\gamma:[a,b]\mapsto \mathbb{R}^n of class C1C^1 such that the image of γ\gamma is CC and γ\gamma is injective except at finitely many points.

  • closed path: γ:[a,b]↦R\gamma:[a,b]\mapsto\mathbb{R} such that γ(a)=γ(b)\gamma(a)=\gamma(b)
  • simple path: no point intersection, i.e., γ\gamma is injective
  • closed/simple curve: the parameterization is closed or simple.

Re-parameterization. Let γ1:[a,b]↦Rn\gamma_1:[a,b]\mapsto \mathbb{R}^n and γ2:[c,d]↦Rn\gamma_2:[c,d]\mapsto\mathbb{R}^n both be paths of class C1C^1. We say that γ2\gamma_2 is a reparameterization of γ1\gamma_1 if there exists a bijective function ϕ:[c,d]↦[a,b]\phi:[c,d]\mapsto [a,b] of class C1C^1 such that γ2=γ1∘ϕ\gamma_2=\gamma_1 \circ \phi and the inverse ϕ−1:[a,b]↦[c,d]\phi^{-1}:[a,b]\mapsto[c,d] is of class C1C^1

  • we say γ2\gamma_2 is orientation-preserving if ϕ(c)=a,ϕ(d)=b\phi(c)=a, \phi(d)=b
  • … orientation-reversing if ϕ(c)=b,ϕ(d)=a\phi(c)=b, \phi(d)=a

Proposition about Integral over Parameterization

If γ2\gamma_2 is a re-parameterization of γ1\gamma_1, for function f:X↦Rf:X\mapsto\mathbb{R}, we have

∫γ1f ds=∫γ2f ds\int_{\gamma_1} f \, ds = \int_{\gamma_2} f\, ds

For vector line integral, the conclusion is similar.

  • If it’s orientation-preserving, then

∫γ1F⋅ds=∫γ2F⋅ds\int_{\gamma_1} \mathbf{F}\cdot d\mathbf{s}=\int_{\gamma_2} \mathbf{F}\cdot d\mathbf{s}

  • If it’s orientation-reversing, then

∫γ1F⋅ds=−∫γ2F⋅ds\int_{\gamma_1} \mathbf{F}\cdot d\mathbf{s}=-\int_{\gamma_2} \mathbf{F}\cdot d\mathbf{s}

Integral over Curve

Let CC be a curve and one of its parameterization γ1\gamma_1 and define ∫Cf ds=∫γ1f ds\int_C f\, ds=\int_{\gamma_1} f\, ds. If CC has an orientation, we let γ1\gamma_1 to preserve this orientation, and define ∫CF⋅ds=∫γ1F⋅ds\int_C \mathbf{F}\cdot d\mathbf{s} = \int_{\gamma_1}\mathbf{F}\cdot d\mathbf{s}

If CC is closed or it’s the union of finite number of closed curves, we use

∮Cf ds:=∫Cf ds∮CF⋅ds:=∫CF⋅ds\oint_C f\,ds:=\int_C f\, ds \\ \oint_C \mathbf{F}\cdot d\mathbf{s}:=\int_C \mathbf{F}\cdot d\mathbf{s}

Orientation of Curve

Let C=∂DC=\partial D be the boundary of a closed and simple region DD in R2\mathbb{R}^2. We say that CC is positive oriented if DD always lies on the left when traversing CC.

Green’s Theorem

∮CF⋅ds=∬D(∂F2∂x−∂F1∂y) dxdy\oint_C \mathbf{F}\cdot d\mathbf{s}=\iint_D \Big( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \Big)\, dxdy