Question Number 25744 by rita1608 last updated on 13/Dec/17
$${find}\:{the}\:{surface}\:{area}\:{of}\:{the}\:{solid}\: \\ $$$${formed}\:{by}\:{the}\:{rotation}\:{of}\:{the}\:{arc}\:{of}\: \\ $$$${the}\:{cycloid}\:{x}={a}\left({t}+{sin}\:{t}\right),\: \\ $$$${y}={a}\left(\mathrm{1}+{cost}\right)\:{about}\:{x}\:{axis} \\ $$
Answered by ajfour last updated on 14/Dec/17
![(dx/dt)=y (dy/dx)=((−sin t)/((1+cos t))) (√(1+((dy/dx))^2 )) =(√(1+((sin^2 t)/((1+cos t)^2 )))) =((√2)/(1+cos t)) S=2∫_0 ^( π) (2πy)(√(1+((dy/dx))^2 ))dx =4π∫_0 ^( π) [a(1+cos t)]^2 [((√2)/(1+cos t))]dt =4(√2)πa^2 ∫_0 ^( π) (1+cos t)dt =4(√2)π^2 a^2 .](Q25776.png)
$$\frac{{dx}}{{dt}}={y}\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{−\mathrm{sin}\:{t}}{\left(\mathrm{1}+\mathrm{cos}\:{t}\right)} \\ $$$$\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{1}+\frac{\mathrm{sin}\:^{\mathrm{2}} {t}}{\left(\mathrm{1}+\mathrm{cos}\:{t}\right)^{\mathrm{2}} }}\: \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+\mathrm{cos}\:{t}} \\ $$$${S}=\mathrm{2}\int_{\mathrm{0}} ^{\:\:\pi} \left(\mathrm{2}\pi{y}\right)\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=\mathrm{4}\pi\int_{\mathrm{0}} ^{\:\:\pi} \left[{a}\left(\mathrm{1}+\mathrm{cos}\:{t}\right)\right]^{\mathrm{2}} \left[\frac{\sqrt{\mathrm{2}}}{\mathrm{1}+\mathrm{cos}\:{t}}\right]{dt} \\ $$$$\:\:=\mathrm{4}\sqrt{\mathrm{2}}\pi{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\:\:\pi} \left(\mathrm{1}+\mathrm{cos}\:{t}\right){dt} \\ $$$$\:\:=\mathrm{4}\sqrt{\mathrm{2}}\pi^{\mathrm{2}} {a}^{\mathrm{2}} \:. \\ $$
Commented by rita1608 last updated on 14/Dec/17
$${why}\:\mathrm{2}\:{is}\:{multiplied}\:,\:{in}\:{some}\: \\ $$$${questions}\:\mathrm{2}\:{is}\:{not}\:{multiplied}\:{plss}\:{tell} \\ $$$${me}\:{when}\:\mathrm{2}\:{is}\:{multiplied} \\ $$
Commented by ajfour last updated on 15/Dec/17
$${if}\:{there}\:{exists}\:{a}\:{line}/{plane}\:{of} \\ $$$${symmetrtry}\:{we}\:{can}\:{find}\:{only} \\ $$$${half}\:{the}\:{length}/{surface}\:{area}/ \\ $$$${volume}\:{and}\:{double}\:{it}. \\ $$