Question Number 12577 by tawa last updated on 26/Apr/17
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{generated}\:\mathrm{when}\:\mathrm{the}\:\mathrm{curve}\:\:\mathrm{x}\:=\:\mathrm{a}\left(\theta\:−\:\mathrm{sin}\theta\right),\:\left(\mathrm{1}\:−\:\mathrm{cos}\theta\right) \\ $$$$\theta\:=\:\mathrm{0},\:\theta\:=\:\pi\:\:\mathrm{rotates}\:\mathrm{about}\:\mathrm{x}−\mathrm{axis}\:\mathrm{through}\:\mathrm{2}\pi\:\mathrm{radian}. \\ $$$$\mathrm{Note}:\:\mathrm{1}\:−\:\mathrm{cos}\theta\:=\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right) \\ $$
Answered by mrW1 last updated on 26/Apr/17
![x=a(θ−sin θ) y=1−cos θ A=2π∫_x_1 ^x_2 ydx=2πa∫(1−cos θ)(1−cos θ)dθ =2πa∫4sin^4 (θ/2) dθ =16πa∫_0 ^π sin^4 (θ/2) d(θ/2) =16πa∫_0 ^(π/2) sin^4 t dt =16πa[((sin 4t−8sin 2t+12t)/(32))]_0 ^(π/2) =16πa((12)/(32))×(π/2)=3aπ^2](https://www.tinkutara.com/question/Q12579.png)
$${x}={a}\left(\theta−\mathrm{sin}\:\theta\right) \\ $$$${y}=\mathrm{1}−\mathrm{cos}\:\theta \\ $$$${A}=\mathrm{2}\pi\int_{{x}_{\mathrm{1}} } ^{{x}_{\mathrm{2}} } {ydx}=\mathrm{2}\pi{a}\int\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\left(\mathrm{1}−\mathrm{cos}\:\theta\right){d}\theta \\ $$$$=\mathrm{2}\pi{a}\int\mathrm{4sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}\:{d}\theta \\ $$$$=\mathrm{16}\pi{a}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}\:{d}\frac{\theta}{\mathrm{2}} \\ $$$$=\mathrm{16}\pi{a}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{4}} \:{t}\:{dt} \\ $$$$=\mathrm{16}\pi{a}\left[\frac{\mathrm{sin}\:\mathrm{4}{t}−\mathrm{8sin}\:\mathrm{2}{t}+\mathrm{12}{t}}{\mathrm{32}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{16}\pi{a}\frac{\mathrm{12}}{\mathrm{32}}×\frac{\pi}{\mathrm{2}}=\mathrm{3}{a}\pi^{\mathrm{2}} \\ $$
Commented by tawa last updated on 26/Apr/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 26/Apr/17
![Area generated = 2π∫_0 ^( x_1 ) ydx A=2π∫_0 ^( π) a(1−cos θ)^2 dθ ...(i) =2π∫^( π) _0 a(1+cos θ)^2 dθ ...(ii) (i)+(ii) gives: 2A=2πa∫_0 ^( π) (2+2cos^2 θ)dθ A=πa∫_0 ^( π) (3+cos 2θ)dθ =πa[3θ+((sin 2θ)/2) ]_0 ^π A=3𝛑^2 a .](https://www.tinkutara.com/question/Q12582.png)
$${Area}\:{generated}\:=\:\mathrm{2}\pi\int_{\mathrm{0}} ^{\:{x}_{\mathrm{1}} } {ydx} \\ $$$${A}=\mathrm{2}\pi\int_{\mathrm{0}} ^{\:\pi} {a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} {d}\theta\:\:\:\:…\left({i}\right) \\ $$$$\:\:\:=\mathrm{2}\pi\underset{\mathrm{0}} {\int}^{\:\pi} {a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} {d}\theta\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\:{gives}: \\ $$$$\mathrm{2}{A}=\mathrm{2}\pi{a}\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{2}+\mathrm{2cos}\:^{\mathrm{2}} \theta\right){d}\theta \\ $$$${A}=\pi{a}\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{3}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\ $$$$\:\:\:\:=\pi{a}\left[\mathrm{3}\theta+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}\:\right]_{\mathrm{0}} ^{\pi} \\ $$$${A}=\mathrm{3}\boldsymbol{\pi}^{\mathrm{2}} \boldsymbol{{a}}\:. \\ $$
Commented by tawa last updated on 26/Apr/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$