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

Interpretation. This gives the weighted length of γ\gamma where ff is the weight.

Vector 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. Let F:X↦Rn\mathbf{F}:X\mapsto\mathbb{R}^n be a continuous function. The vector line integral of F\mathbf{F} along γ\gamma is

∫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(F1x1′+F2x2′+⋯+Fnxn′) dt=∫ab(F1 dx1+F2 dx2+⋯+Fn dxn)\begin{aligned} &\int_\gamma \mathbf{F}\cdot d\mathbf{s}\\= &\int_a^b (F_1x_1' + F_2x_2' + \dots + F_nx_n')\, dt\\= &\int_a^b (F_1\, dx_1 + F_2\, dx_2 + \dots + F_n\, dx_n)\\\end{aligned}

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

Let C=∂DC=\partial D be the boundry of a closed and bounded region DD in R2R^2. Suppose CC is the union of finitely many simple closed piecewise C1C^1 curves and CC is positively oriented. Let F:X↦R2\mathbf{F}:X\mapsto\mathbb{R}^2 be a vector field of class C1C^1 where X⊂R2X\subset\mathbb{R}^2 contains DD. Then

∮∂DF⋅ds=∮∂DF1 dx+F2 dy=∬D(∂F2∂x−∂F1∂y) dxdy\oint_{\partial D} \mathbf{F}\cdot d\mathbf{s}=\oint_{\partial D} F_1\, dx + F_2\, dy=\iint_D \Big( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \Big)\, dxdy

This conclusion can be expressed in Stokes’ Theorem’s form. If we let F=(F1,F2,0)\mathbf{F}=(F_1, F_2, 0), then

∇×F=(0,0,F2x−F1y)\nabla \times \mathbf{F}=(0,0,{F_2}_x-{F_1}_y)

so the form is equivalent to

∮CF⋅ds=∬D(∇×F)⋅k dA\oint_C \mathbf{F}\cdot d\mathbf{s}=\iint_D (\nabla\times\mathbf{F})\cdot\mathbf{k}\, dA

Proof

Proof. We discuss the case when DD is a type 3 elementary region D={(x,y):x∈[a,b],y∈[g(x),h(x)]}D=\{ (x,y):x\in[a,b],y\in[g(x), h(x)] \}.

Parameterize orientation-preserving C1:γ1(x)=(x,g(x))C_1:\gamma_1(x)=(x,g(x)) and orientation-reversing C2:γ2(x)=(x,h(x))C_2:\gamma_2(x)=(x,h(x)).

This shows that

∮CF1 dx=∫C1F1 dx−∫C2F1 dx=∫abF1(x,g(x)) dx−∫abF1(x,h(x)) dx\begin{aligned} \oint_C F_1\, dx &= \int_{C_1} F_1\, dx - \int_{C_2} F_1 \, dx \\ &= \int_a^b F_1(x, g(x))\,dx - \int_a^b F_1(x,h(x))\,dx \end{aligned}

while

∬D−∂F1∂y dA=∫ab∫g(x)h(x)−∂F1∂y dydx=−∫ab(F1(x,h(x))−F1(x,g(x))) dx\begin{aligned} \iint_D -\frac{\partial F_1}{\partial y}\,dA &= \int_a^b \int_{g(x)}^{h(x)} -\frac{\partial F_1}{\partial y} \,dydx\\ &=-\int_a^b \Big( F_1(x,h(x)) - F_1(x,g(x)) \Big)\,dx \end{aligned}

Similarly, we can prove that ∮CF2 dy=∬D∂F2∂x dA\oint_C F_2\,dy=\iint_D\frac{\partial F_2}{\partial x}\,dA. ■\blacksquare

Divergence Theorem in R2\mathbb{R}^2

Let C=∂DC=\partial D be the boundry of a closed and bounded region DD in R2R^2. Suppose CC is the union of finitely many simple closed piecewise C1C^1 curves and CC is positively oriented. Let F:X↦R2\mathbf{F}:X\mapsto\mathbb{R}^2 be a vector field of class C1C^1 where X⊂R2X\subset\mathbb{R}^2 contains DD. If n\mathbf{n} is the outward unit normal vector to DD, then

∮CF⋅n ds=∬D∇⋅F dA\oint_C\mathbf{F}\cdot\mathbf{n}\, ds=\iint_D \nabla\cdot\mathbf{F}\,dA

Proof

Proof. Parameterize CC as γ(t)=(x(t),y(t))\gamma(t)=(x(t), y(t)), positively oriented, then its tangent vector is γ′=(x′,y′)\gamma'=(x',y'). Thus outward unit normal vector n=(y′,x′)∥γ′∥\mathbf{n}=\frac{(y',x')}{\|\gamma'\|}.

So, by definition of ∮\oint, we know

∮CF⋅n ds=∫ab(F⋅n)(γ(t))∥γ′(t)∥ dt=∫ab(F1,F2)⋅(y′,x′)∥γ′∥∥γ′∥ dt=∫abF1y′dt−F2x′dt=∮CF1dy−F2dx=∬D(∂F1∂x−∂(−F2)∂y) dA=∬D(∂F1∂x+∂F2∂y) dA=∬D∇⋅F dA\begin{aligned} \oint_C \mathbf{F}\cdot\mathbf{n}\, ds &=\int_a^b (\mathbf{F}\cdot\mathbf{n})(\gamma(t))\|\gamma'(t)\|\, dt\\ &=\int_a^b (F_1,F_2) \cdot \frac{(y',x')}{\|\gamma'\|} \|\gamma'\|\, dt\\ &=\int_a^b F_1 y'dt-F_2 x'dt\\ &=\oint_C F_1dy-F_2dx\\ &=\iint_D \Big( \frac{\partial F_1}{\partial x}-\frac{\partial (-F_2)}{\partial y} \Big)\, dA\\ &=\iint_D \Big( \frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}) \, dA\\ &=\iint_D \nabla\cdot \mathbf{F}\, dA \end{aligned}