Question

Find the surface area of the solid generated by revolving one arc of the cycloid $x=a(t+\sin t),y=a(1+\cos t)$ about its base ($x$-axis)

Answer 1

$y=10\Rightarrow 1+\cos t=0\Rightarrow \cos t=-1\Rightarrow t=-\pi ,\pi$
$x=a(t+\sin t)\Rightarrow \frac{dx}{dt}=a(1+\cos t)$
$y=a(1+\cos t)\Rightarrow \frac{dt}{dx}=-a\sin t$
$\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}=\sqrt{a^2(1+\cos t{)}^2{a}^2{\sin }^{2}t}$
$=\sqrt{a^2[1+{\cos }^{2}t+2\cos t+{\sin }^{2}t]}$
$=a\sqrt{2+2\cos t}=\sqrt{2}a\sqrt{(1+\cos t)}$
$=\sqrt{2}.a\sqrt{2{\cos }^{2}t/2}=2a\cos \frac{t}{2}$
$\therefore$ Surface Area $={\int }_{-\pi }^{\pi }2\pi \cdot \sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$
$={\int }_{-\pi }^{\pi }2\pi a(1+\cos t)\cdot 2a\cos \frac{t}{2}dt$
$={\int }_{-\pi }^{\pi }2\pi a\cdot 2\cdot {\cos }^2\frac{t}{2}\cdot 2a\cos \frac{t}{2}dt$
$=16\pi a^2{\int }_0^{\pi }{\cos }^3\frac{t}{2}dt$
$=16\pi a^2{\int }_0^{\pi }2{\cos }^{3}xdx$ ....$[$Take $\frac{t}{2}=x]$
$=32\pi a^2{I}_3=32\pi a^2\times \frac{2}{3}$
$=\frac{64\pi a^2}{3}$ sq. units