Question

Find the surface area of the solid generated by revolving the cycloid $x=a(t+\sin t),y=a(1+\cos t)$.

Answer 1

$y=0\Rightarrow 1+\cos t=0$
$\cos t=-1\Rightarrow t=-\pi ,\pi$
$x=a(t+\sin t);y=a(1+\cos t)$
$\frac{dx}{dt}=a(1+\cos t)\frac{dy}{dt}=-a\sin t$
$\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}=\sqrt{a^2(1+\cos t{)}^2+a^2{\sin }^{2}t}$
Surface area $={\int }_{-\pi }^{\pi }2\pi a(1+\cos t)2a\cos \frac{t}{2}dt$
$={\int }_{-\pi }^{\pi }2\pi a.2{\cos }^2\frac{t}{2}.2a\cos \frac{t}{2}dt$
$=16\pi a^2{\int }_0^{\pi }co{s}^3\frac{t}{2}dt$
$=16\pi a^2{\int }_0^{\pi /2}2{\cos }^{3}xdx$
$[$Take $\frac{t}{2}=x]$
$=32\pi a^2{I}_3=32\pi a^2\times \frac{2}{3}$
$=\frac{64}{3}\pi a^2$ sq. units.