Q.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).

Use $S=\int 2\pi y\,ds$ with the cycloid's parametric derivatives, simplify using the half-angle identity $1+\cos t=2\cos^2(t/2)$, and integrate.

1. Parametric derivatives.

   $x=a(t+\sin t)\Rightarrow \dfrac{dx}{dt}=a(1+\cos t)$

   $y=a(1+\cos t)\Rightarrow \dfrac{dy}{dt}=-a\sin t$

2. Arc-length element.

   $\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2=a^2(1+\cos t)^2+a^2\sin^2 t=a^2\big[1+2\cos t+\cos^2t+\sin^2t\big]=a^2[2+2\cos t]=2a^2(1+\cos t)$

   Using $1+\cos t=2\cos^2\dfrac{t}{2}$: this equals $4a^2\cos^2\dfrac{t}{2}$, so

   $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt=2a\left|\cos\dfrac{t}{2}\right|dt$

3. Surface-area formula (revolution about the $x$-axis).

   $S=\displaystyle\int_0^{2\pi}2\pi y\,ds$

   With $y=a(1+\cos t)=2a\cos^2\dfrac{t}{2}$:

   $S=2\pi\displaystyle\int_0^{2\pi}2a\cos^2\dfrac{t}{2}\cdot 2a\left|\cos\dfrac{t}{2}\right|dt=8\pi a^2\displaystyle\int_0^{2\pi}\left|\cos\dfrac{t}{2}\right|^3dt$

4. Substitute $u=t/2$, $dt=2\,du$ (limits $u:0\to\pi$):

   $S=8\pi a^2\cdot 2\displaystyle\int_0^\pi|\cos u|^3\,du=16\pi a^2\displaystyle\int_0^\pi|\cos u|^3\,du$

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