$x = a (\theta + sin \theta ), \ y = a (1 – cos \theta)$

Solution:

Given curve: Cycloid $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$

The length of given curve is:

$\mathrm{S}=\int_{\theta_{1}}^{\theta_{2}} \sqrt{\left(\frac{d x}{d \theta}\right)^{2}+\left(\frac{d y}{d \theta}\right)^{2}} \mathrm{d} \theta$

$\begin{aligned} \frac{d x}{d \theta}=a(1+\cos \theta) & \\ \frac{d y}{d \theta}=a \sin \theta \\ \therefore\left(\frac{d x}{d \theta}\right)^{2}+\left(\frac{d y}{d \theta}\right)^{2} &=a^{2}\left[1+2 \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta\right] \\ &=2 a^{2}[1+\cos \theta] \\ &=4 a^{2}\left[\cos ^{2} \theta / 2\right] \end{aligned}$

$\therefore \sqrt{\left(\frac{d x}{d \theta}\right)^{2}+\left(\frac{d y}{d \theta}\right)^{2}}=2 a \cos \theta / 2$

$\therefore S=\int_{-\pi}^{\pi} 2 a \cos \theta / 2 \mathrm{d} \theta$

$\begin{aligned} &=2 \times \int_{0}^{\pi} 2 a \cos \theta / 2 d \theta \\ &=4 \mathrm{a}[2 \sin \theta / 2]_{0}^{\pi} \\ \therefore \mathrm{s} &=8 \mathrm{a} \end{aligned}$