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Question Number 226351 by Lara2440 last updated on 26/Nov/25
Prove Mo^ bious String is Not a Orientated Surface. σ(u,θ)= { (((1−u∙sin((1/2)θ))cos(θ))),(((1−u∙sin((1/2)θ))sin(θ))),((u∙cos((1/2)θ))) :} , −(1/2)≤u≤(1/2) , 0≤θ≤2π
$$\mathrm{Prove}\:\mathrm{M}\ddot {\mathrm{o}bious}\:\mathrm{String}\:\mathrm{is}\:\mathrm{Not}\:\mathrm{a}\:\mathrm{Orientated}\:\mathrm{Surface}. \\ $$$$\sigma\left({u},\theta\right)=\begin{cases}{\left(\mathrm{1}−{u}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right)\mathrm{cos}\left(\theta\right)}\\{\left(\mathrm{1}−{u}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right)\mathrm{sin}\left(\theta\right)}\\{{u}\centerdot\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)}\end{cases}\:\:,\:−\frac{\mathrm{1}}{\mathrm{2}}\leq{u}\leq\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{0}\leq\theta\leq\mathrm{2}\pi \\ $$
Answered by Lara2440 last updated on 26/Nov/25
To show that this Surface is Oriented, we have to compare the direction of the normal vector. if the normal vector has same direction it is a orientated surface. we fix t=0 for a convenience the calculate. normal vector (n^� /(∣∣n^� ∣∣))=^(def) (((∂σ/∂u)×(∂σ/∂θ))/(∣∣(∂σ/∂u)×(∂σ/∂θ)∣∣)) (n^� /(∣∣n^� ∣∣))=−cos(θ)cos((1/2)θ)e_1 ^� −sin((1/2)θ)sin(θ)e_2 ^� +cos((1/2)θ)e_3 ^� start point is N^→ (0,0) and End point N^→ (0,2π) (the reason for θ=2π is that the cycle of returning to it′s place after one turn is 2π) compare each Start End direction N^→ (0,0)=(−1,0,0) N^→ (0,2π)=(1,0,0) ∴ σ is not orientated
$$\: \\ $$$$\mathrm{To}\:\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{Surface}\:\mathrm{is}\:\mathrm{Oriented}, \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{compare}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{vector}. \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{vector}\:\mathrm{has}\:\mathrm{same}\:\mathrm{direction} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{orientated}\:\mathrm{surface}. \\ $$$$\mathrm{we}\:\mathrm{fix}\:{t}=\mathrm{0}\:\mathrm{for}\:\mathrm{a}\:\mathrm{convenience}\:\mathrm{the}\:\mathrm{calculate}. \\ $$$$\mathrm{normal}\:\mathrm{vector}\:\frac{\hat {\boldsymbol{\mathrm{n}}}}{\mid\mid\hat {\boldsymbol{\mathrm{n}}}\mid\mid}\overset{\mathrm{def}} {=}\frac{\frac{\partial\sigma}{\partial{u}}×\frac{\partial\sigma}{\partial\theta}}{\mid\mid\frac{\partial\sigma}{\partial{u}}×\frac{\partial\sigma}{\partial\theta}\mid\mid} \\ $$$$\:\: \\ $$$$\frac{\hat {\boldsymbol{\mathrm{n}}}}{\mid\mid\hat {\boldsymbol{\mathrm{n}}}\mid\mid}=−\mathrm{cos}\left(\theta\right)\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\hat {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\mathrm{sin}\left(\theta\right)\hat {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} +\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\hat {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\mathrm{start}\:\mathrm{point}\:\mathrm{is}\:\overset{\rightarrow} {\boldsymbol{\mathrm{N}}}\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{and}\:\mathrm{End}\:\mathrm{point}\:\overset{\rightarrow} {\boldsymbol{\mathrm{N}}}\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$$$\left(\mathrm{the}\:\mathrm{reason}\:\mathrm{for}\:\theta=\mathrm{2}\pi\:\mathrm{is}\:\mathrm{that}\:\mathrm{the}\:\mathrm{cycle}\:\mathrm{of}\:\mathrm{returning}\:\mathrm{to}\right. \\ $$$$\left.\mathrm{it}'\mathrm{s}\:\mathrm{place}\:\mathrm{after}\:\mathrm{one}\:\mathrm{turn}\:\mathrm{is}\:\mathrm{2}\pi\right) \\ $$$$\mathrm{compare}\:\mathrm{each}\:\mathrm{Start}\:\mathrm{End}\:\mathrm{direction} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{N}}}\left(\mathrm{0},\mathrm{0}\right)=\left(−\mathrm{1},\mathrm{0},\mathrm{0}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{N}}}\left(\mathrm{0},\mathrm{2}\pi\right)=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\: \\ $$$$\therefore\:\sigma\:\mathrm{is}\:\mathrm{not}\:\mathrm{orientated}\: \\ $$

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