# Surface area by the revolution of cycloid

How to find the surface area of the solid generated by the revolution of the cycloid about $x$-axis?

I know the formula to find out the surface area but I'm getting the point that in the formula why we take the integration limit as 0 to $2\pi$.

The parametric equation of the cycloid is $$x(t)=r(t-\sin t) \, \quad y(t)=r(1-\cos t) \quad \mbox{for t\in[0,2\pi].}$$ Its surface of revolution around the $x$-axis is given by $$S:=2\pi \int_0^{2\pi}y(t) \sqrt{x'(t)^2+y'(t)^2}dt.$$ Then $$x'(t)=r(1-\cos t) \ , \ y'(t)=r\sin t \implies x'(t)^2+y'(t)^2=2r^2(1-\cos t)=4r^2\sin^2(t/2)$$ and we find that \begin{align} S&=2\pi \int_0^{2\pi}r(1-\cos t) \cdot 2r\sin(t/2)dt =8\pi r^2\int_0^{2\pi}\sin^3(t/2)dt\\ &=16\pi r^2\int_0^{\pi}\sin^3(s)ds =16\pi r^2\int_0^{\pi}(1-\cos^2(s))d(-\cos(s))\\ &=16\pi r^2\left[\cos(s)-\frac{1}{3}\cos^3(s)\right]_{\pi}^0=\frac{64\pi r^2}{3}. \end{align}