Functional integral on curve and surface

4.4 Functional integral on curve and surface

It is non-oriented and done for functions.

Curve integral of the first kind

Assuming a curve $γ \subseteq R^{m}$ whose parametric representative is $x = x (t), t \in [a, b]$ where $x$ is piecewisely of $C^{1}$ with respect to $t$ . Assuming a function $ρ : γ \to R$ , if when

max ∥Pk − Pk−1∥ → 0 1≤k≤n

the limit of

∑ ρ (ξk)∥Pk − Pk−1∥ k

exists, then the functional integral on curve $γ$ is

∫ ∑ ρ(x)dl = lim ρ(ξk)∥Pk − Pk −1∥ γ k

where $d l$ is the infinitesimal arc length. It is not strict! See Figure 4.3 for counter example, where 2 points on the dashed black line could be arbitrarily close.

Definition 4.4.1 Arc length. Given a partition of the curve $Π : a = t_{0} < t_{1} < \dots < t_{n} = b$ , $P_{k} = γ (t_{k})$ , the total length of the broken line is $\sum_{k = 1}^{n} ∥ γ (t_{k}) - γ (t_{k - 1}) ∥$ . If

n ∑ L = sup ∥γ(tk) − γ(tk−1)∥ < + ∞ Π k=1

then $γ$ is called a rectifiable curve, $L$ is called the (arc) length of $γ$ .

Proposition 4.4.2 Mark $λ (Π) = max_{1 \leq k \leq n} (t_{k} - t_{k - 1})$ , if

lim L (Π) = L λ(Π)→0

then $γ$ is rectifiable and $L = L (γ)$ , i.e. $\forall ϵ > 0$ , $\exists δ (ϵ) > 0$ , such that for any partition $Π$ of $[a, b]$ , $λ (Π) < δ (ϵ) \Rightarrow | L (Π) - L | < ϵ$ .

Theorem 4.4.3 If curve $γ$ has a regular parametric representative of $C^{1}$ , i.e. $x (t), t \in [a, b]$ where $x^{'} (t) \neq 0, \forall t$ . Here $γ = {x (t) | a \leq t \leq b}$ , and $x$ is a one-to-one correspondence of $(a, b)$ with $γ$ . Then $γ$ is a rectifiable curve, and its length is $\int_{a}^{b} ∥ x^{'} (t) ∥ d t$ .

Definition 4.4.4 Define $l (t) = \int_{a}^{t} ∥ x^{'} (s) ∥ d s$ where $∥ l^{'} (t) ∥ = ∥ x^{'} (t) ∥ > 0$ , its inverse function is $t = t (l)$ , then $x (t) = x (t (l)) = \tilde{x} (l)$ is called the parametric representative of $γ$ under the arc-length parameter $l$ .

Definition 4.4.5 Assuming $f \in C (γ)$ , define the infinitesimal arc-length $d l = ∥ x^{'} (t) ∥ d t$ , then

∫ ∫ b f dl ≜ f(x (t))∥x ′(t)∥dt γ a ◟ --◝◜--◞ dl

Theorem 4.4.6 The integral defined above has nothing to do with the selection of parametric representative of $γ$ .

Proof Assuming $t = t (s)$ and $t^{'} (s) > 0, \forall s$ and where $s \in [α, β] \mapsto t \in [a, b]$ , then $\begin{aligned} \int_{[α, β]} f (x (t (s))) ‖ \frac{d}{d s} x (t (s)) ‖ d s & = \int_{[a, b]} f (x (t)) ∥ x^{'} (t) ∥ | t^{'} (s) | | det \frac{\partial s}{\partial t} | d t \\ = \int_{[a, b]} f (x (t)) ∥ x^{'} (t) ∥ d t \end{aligned}$

They are equivalent. $◻$

Example 4.4.1 Given $γ : r = 2 (1 + \cos θ), - π \leq θ \leq π$ in polar coordinate, seek its length and center of figure (centroid, the center of mass when evenly distributed).

Length.

∫ π ∘ -------------- L = x ′(𝜃)2 + y′(𝜃)2d𝜃 − π

where

\text{[math]}

Therefore,

∫ π 𝜃 L = 4 ∖left|cos -∖right|d𝜃 = 16 − π 2

It is trivial that $y_{C} = 0$ ,

1 ∫ π ∘ -------------- 8 xC = --- x(𝜃) x′(𝜃)2 + y ′(𝜃)2d𝜃 =-- 16 −π 5

$d l$ in different coordinate systems.

∘ ------------ dl = ∥x′(t)∥dt = ⟨x ′(t),x′(t)⟩dt

Assuming in another coordinate system, the coordinate is $u$ , and

x = ϕ(u ), u(t) = ϕ−1(x (t)) → ˜γ = ϕ−1(γ)

So the infinitesimal length is

┌│ -------m-----------m----------------- ┌│ -m-------------------------------- │∘ ∑ ∂x--dui-∑ ∂x--duj- │∘ ∑ dui-duj- ∂x---∂x- dl = ∖left⟨ ∂u dt , ∂u dt ∖right ⟩dt = dt dt ∖lef t⟨∂u ,∂u ∖right⟩dt i=1 i j=1 j i,j=1 i j

Express it in quadratic form, we have $\begin{aligned} d l & = \sqrt{(\begin{array}{c} u_{1}^{'} (t) & \dots & u_{m}^{'} (t) \end{array}) {(⟨ \frac{\partial x}{\partial u_{i}}, \frac{\partial x}{\partial u_{j}} ⟩)}_{m \times m} (\begin{array}{c} u_{1}^{'} (t) \\ ⋮ \\ u_{m}^{'} (t) \end{array})} d t \\ = \sqrt{(\begin{array}{c} d u_{1} & \dots & d u_{m} \end{array}) \underset{Metric (Gram) matrix}{\underset{⏟}{{(⟨ \frac{\partial x}{\partial u_{i}}, \frac{\partial x}{\partial u_{j}} ⟩)}_{m \times m}}} (\begin{array}{c} d u_{1} \\ ⋮ \\ d u_{m} \end{array})} \end{aligned}$

Example 4.4.2 Polar coordinate in 2-dimension.

\text{[math]}

We have

( ) ( ) ∂x-= cos𝜃 , ∂x- = − rsin𝜃 ∂r sin 𝜃 ∂𝜃 rcos 𝜃

Therefore, the Gram matrix is

\text{[math]}

and

√ ∘ -------------- dl = = (dr)2 + (rd𝜃)2

Surface integral of the first kind

Definition 4.4.7 Assuming a surface $Σ \subseteq R^{m}$ , its regular parametric representative of $C^{1}$ is $x : D \to R^{m}$ where $D \in R^{2}$ , $(u, v) \in D \mapsto x (u, v) \in Σ$ , and $\frac{\partial x}{\partial u} (u, v), \frac{\partial x}{\partial v} (u, v)$ are linear independent $\forall (u, v) \in D$ . Then $Σ = {x (u, v) | (u, v) \in D}$ .

How to calculate the area of the surface? One method is to divide the surface by triangles (Figure 4.4(a)), and seek the supremum of the total area of the triangles. Yet Schwartz constructed a counter example in the

19^{th}

century that even for a cylindrical surface, an arbitrary partition could make the supremum of the total area become

+ \infty

. Kunihiko proved in his works that if the smallest angle of all triangles is greater than a given positive, then this method is valid.

Another method is to divide the surface by grids (Figure 4.4(b)). The shape in red shadow is a parallelogram composed by $\frac{\partial x}{\partial u} d u, \frac{\partial x}{\partial v} d v$ . Then we have $\begin{aligned} d σ & = ‖ \frac{\partial x}{\partial u} d u ‖ ‖ \frac{\partial x}{\partial v} d v ‖ \sin θ = \sqrt{{‖ \frac{\partial x}{\partial u} ‖}^{2} {‖ \frac{\partial x}{\partial v} ‖}^{2} - {‖ \frac{\partial x}{\partial u} ‖}^{2} {‖ \frac{\partial x}{\partial v} ‖}^{2} \cos^{2} θ} d u d v \\ = \sqrt{\underset{E}{\underset{⏟}{⟨ \frac{\partial x}{\partial u}, \frac{\partial x}{\partial u} ⟩}} \underset{G}{\underset{⏟}{⟨ \frac{\partial x}{\partial v}, \frac{\partial x}{\partial v} ⟩}} - \underset{F^{2}}{\underset{⏟}{{⟨ \frac{\partial x}{\partial u}, \frac{\partial x}{\partial v} ⟩}^{2}}}} d u d v \end{aligned}$

Therefore,

√ ---------- dσ = EG − F 2dudv

Theorem 4.4.8 The definition above has nothing to do with the selection of parametric representative of $Σ$ .

Proof Assuming $\tilde{x} (t_{1}, t_{2}) = x (u_{1} (t_{1}, t_{2}), u_{2} (t_{1}, t_{2}))$ where $(t_{1}, t_{2}) \mapsto (u_{1}, u_{2})$ is a diffeomorphism of $C^{1}$ , then $\begin{aligned} d σ & = \sqrt{det {(⟨ \frac{\partial \tilde{x}}{\partial t_{i}}, \frac{\partial \tilde{x}}{\partial t_{j}} ⟩)}_{2 \times 2}} d t_{1} d t_{2} \\ = \sqrt{det {(⟨ \sum_{k = 1}^{2} \frac{\partial x}{\partial u_{k}} \frac{\partial u_{k}}{\partial t_{i}}, \sum_{l = 1}^{2} \frac{\partial x}{\partial u_{l}} \frac{\partial u_{l}}{\partial t_{j}} ⟩)}_{2 \times 2}} d t_{1} d t_{2} \\ = \sqrt{det \frac{\partial (u_{1}, u_{2})}{\partial (t_{1}, t_{2})} det {(⟨ \frac{\partial x}{\partial u_{k}}, \frac{\partial x}{\partial u_{l}} ⟩)}_{2 \times 2} det \frac{\partial (u_{1}, u_{2})}{\partial (t_{1}, t_{2})}} d t_{1} d t_{2} \\ = \sqrt{det {(⟨ \frac{\partial x}{\partial u_{k}}, \frac{\partial x}{\partial u_{l}} ⟩)}_{2 \times 2}} d u_{1} d u_{2} \end{aligned}$

$◻$

Figure 4.4: Calculate the area of the surface

Example 4.4.3 Given $Σ : z = f (x, y)$ where $f \in C^{1}$ , $(x, y) \in D \subseteq R^{2}$ , seek the area of $Σ$ .

Assuming $x = (x, y, f (x, y))$ , then

∂x ∂x ---=, ---= ∂x ∂y

we have

∂f 2 ∂f 2 ∂f ∂f E = 1 + ∖left(---∖right) , G = 1 + ∖left(---∖right) , F = ------ ∂x ∂y ∂x ∂y

then

∘ ----------------------------------------- √ ---------- ∂f ∂f ∘ ----------- dσ = EG − F2dxdy = 1 + ∖lef t(---∖right)2 + ∖lef t(--∖right )2dxdy = 1 + ∥ ∇f ∥2dxdy ∂x ∂y

So the area of $Σ$ is

∫ ∫ ∘ ----------2 A = 1 + ∥∇f ∥ dxdy D

Since the gradient vector points at the fastest direction of change of $f$ , and its magnitude equals to the rate of change, the formula above also has a geometric meaning.

Example 4.4.4 Assuming surface $Σ$ is derived by a curve $y = f (x) > 0$ in $(x, y)$ plain rotating around $x$ axis by $2 π$ , seek the area of $Σ$ .

Assuming $x = (x, f (x) \cos θ, f (x) \sin θ)$ , we have

′ 2 2 E = 1 + (f (x)) , G = f (x), F = 0

then

∘ ------------------ ∘ ------------ dσ = [1 + (f′(x))2]f 2(x )dxd𝜃 = 1 + (f′(x ))2dxd 𝜃

So the area of $Σ$ is

∫ b∫ 2π ∘ ------------ ∫ ∫ A = f(x) 1 + (f′(x))2dxd 𝜃 = 2π f (x )dl = 2πydl a 0 γ γ

Example 4.4.5 Given $Σ : x^{2} + y^{2} + z^{2} = R^{2}$ , longitude $ϕ \in [ϕ_{1}, ϕ_{2}] \subseteq [0, 2 π]$ , latitude $θ \in [θ_{1}, θ_{2}] \subseteq [0, π]$ , seek the area of $Σ$ .

Assuming $x = (R \sin θ \cos ϕ, R \sin θ \sin ϕ, R \cos θ)$ , we have

2 2 2 E = R , G = R sin 𝜃, F = 0

So the area of $Σ$ is

∫ ∫ 𝜃2 ϕ2 2 2 A = R sin 𝜃dϕd 𝜃 = R (ϕ2 − ϕ1 )(cos 𝜃1 − cos 𝜃2) 𝜃1 ϕ1

Particularly, take $ϕ_{1} = 0, ϕ_{2} = 2 π, θ_{1} = 0, θ_{2} = π$ , we get the area of a spherical surface $A = 4 π R^{2}$ . Generally, we have

Am (R ) = dVm-(R ) dR

Surface in higher dimension. Assuming

Σ \subseteq R^{m}

is a

k

-dimension surface, its regular representative of

C^{1}

x =

where all tangent vectors $\frac{\partial x}{\partial u_{i}}$ are linear independent. Define

┌│ ------------------------------------------ dσ = ││ det ∖lef t(∖lef t⟨ ∂x-,-∂x-∖right⟩∖right)k×kdu1 ⋅⋅⋅duk ∘ ∂ui ∂uj ◟----------------◝ ◜----------------◞ Gram matrix

Assuming $f : Σ \to R$ is continuous, define

∫ ∫ ∘ ------------------------------------------ ∂x---∂x- f dσ = f (x (u1,...,uk)) det ∖lef t(∖lef t⟨ ∂u ,∂u ∖right⟩∖right)k×kdu1 ⋅⋅⋅duk Σ D i j

Note The definition above has nothing to do with the selection of regular representative.

$k = 1$ , the surface degenerates to a curve $x (t)$ , at this time
$∘ --′----′---- ′ dσ = ⟨x(t),x (t)⟩ = ∥x (t)∥dt = dl$
$k = 2$ , theorem is proved.
$k = m$ , $\begin{aligned} d σ & = \sqrt{det [{(\frac{\partial (x_{1}, . . ., x_{m})}{\partial (u_{1}, . . ., u_{m})})}^{T} (\frac{\partial (x_{1}, . . ., x_{m})}{\partial (u_{1}, . . ., u_{m})})]} \\ = | det \frac{\partial (x_{1}, . . ., x_{m})}{\partial (u_{1}, . . ., u_{m})} | d u_{1} \dots d u_{m} = d x_{1} \dots d x_{m} \end{aligned}$

Example 4.4.6 Assuming $Σ \subseteq R^{m + 1} : x_{m + 1} = f (x_{1}, . . ., x_{m})$ where $f \in C^{1}$ and $(x_{1}, . . ., x_{m}) \in D \subseteq R^{m}$ , here $Σ$ is called a hypersurface in $R^{m + 1}$ . Seek the ( $m$ -dimension) area of $Σ$ .

Assuming $x = (x_{1}, . . ., x_{m}, f (x_{1}, . . ., x_{m}))$ , we have

-∂x- == ei +-∂f-em+1 ∂xi ∂xi

then

∖lef t⟨ ∂x-,-∂x-∖right⟩ = ∖left⟨e + ∂f-e ,e + -∂f-e ∖right⟩ = δ + ∂f--∂f-- ∂xi ∂xj i ∂xi m+1 j ∂xj m+1 ij ∂xi ∂xj

Hence,

∖left(∖left⟨-∂x-, ∂x-∖right⟩∖right) = ∖left(δ + -∂f-∂f--∖right) = I + ∇f (∇f )T ∂xi ∂xj m×m ij ∂xi ∂xj m×m m

For any vector $v$ perpendicular to $\nabla f$ , we have

[Im + ∇f (∇f )T]v = v

and

[Im + ∇f (∇f )T ]∇f = (1 + ∥∇f ∥2)∇f

Therefore,

T ∏m 2 2 det(Im + ∇f (∇f ) ) = λi = 1◟-⋅ ⋅◝⋅◜⋅ ⋅ 1◞ ⋅(1 + ∥∇f ∥ ) = 1 + ∥∇f ∥ i=1 m −1

Terminally,

∘ ----------2 d σ = 1 + ∥∇f ∥ dx1⋅⋅⋅dxm

An example. $Σ : x_{1}^{2} + \dots + x_{m + 1}^{2} = R^{2}$ .

∘ ------------------- ∘ ---------- xm+1 = ± R2 − x2− ⋅ ⋅⋅ − x2 = ± R2 − ∥xˆ∥2 1 m

then we have $\begin{aligned} d σ & = \sqrt{1 + {‖ \nabla (\pm \sqrt{R^{2} - ∥ \hat{x} ∥^{2}}) ‖}^{2}} d x_{1} \dots d x_{m} \\ = \sqrt{1 + {‖ \frac{1}{2} \frac{\mp 2 \hat{x}}{\sqrt{R^{2} - ∥ \hat{x} ∥^{2}}} ‖}^{2}} d x_{1} \dots d x_{m} \\ = \frac{R}{\sqrt{R^{2} - ∥ \hat{x} ∥^{2}}} d x_{1} \dots d x_{m} \end{aligned}$

Terminally, $\begin{aligned} A_{m} (R) & = \int_{B_{m} (R)} \frac{R}{\sqrt{R^{2} - ∥ \hat{x} ∥^{2}}} d x_{1} \dots d x_{m} \\ = \int_{- R}^{R} d x_{m} \int_{x_{1}^{2} + \dots + x_{m - 1}^{2} \leq R^{2} - x_{m}^{2}} \frac{R}{\sqrt{R^{2} - ∥ \hat{x} ∥^{2}}} d x_{1} \dots d x_{m - 1} \\ = \int_{- R}^{R} \frac{R d x_{m}}{\sqrt{R^{2} - x_{m}^{2}}} \int_{B_{m - 1} (\sqrt{R^{2} - x_{m}^{2}})} \frac{\sqrt{R^{2} - x_{m}^{2}} d x_{1} \dots d x_{m - 1}}{\sqrt{R^{2} - x_{m}^{2} - x_{1}^{2} - \dots - x_{m - 1}^{2}}} \\ = 2 \int_{0}^{R} \frac{R}{\sqrt{R^{2} - x_{m}^{2}}} A_{m - 1} (\sqrt{R^{2} - x_{m}^{2}}) d x_{m} \end{aligned}$

Recall

∫ R ∘ --2----2- Vm (R ) = 2 0 Vm −1∖left( R − xm∖right )dxm

Notice that

dV ∫ R R dV ∘ --------- ---m-(R) = 2 ∘------------m-−1∖left( R2 − x2m ∖right)dxm dR 0 R2 − x2m dR

Since $A_{1} (R) = 2 π R = V_{2}^{'} (R)$ , so we have

Am (R ) = V′ (R ) m− 1