1
$\begingroup$

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 $2\pi$.

Please, help me out!

$\endgroup$
3
$\begingroup$

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 $$ \begin{align} S&=2\pi \int_0^{2\pi}r(1-\cos t) \cdot 2r\sin(t/2)dt =8\pi r^2\int_0^{2\pi}\sin^3(t/2)dt\\ &=16\pi r^2\int_0^{\pi}\sin^3(s)ds =16\pi r^2\int_0^{\pi}(1-\cos^2(s))d(-\cos(s))\\ &=16\pi r^2\left[\cos(s)-\frac{1}{3}\cos^3(s)\right]_{\pi}^0=\frac{64\pi r^2}{3}. \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.