Solution:

Curve: $\quad r=a(1-\cos \theta)$

Perimeter of given curve is,

$\mathbf{S}=2 \times \int_{0}^{\pi} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} ) d \theta$

$\frac{d r}{d \theta}=a(\sin \theta) \Rightarrow\left(\frac{d r}{d \theta}\right)^{2}=a^{2} \sin ^{2} \theta$

$r^{2}+\left(\frac{d r}{d \theta}\right)^{2}=a^{2}\left[1-2 \cos \theta+\cos ^{2} \theta\right]+a^{2} \sin ^{2} \theta$

$\begin{aligned} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} &=\sqrt{2} \mathrm{a}(1-\cos \theta)^{1 / 2} \\ &=\sqrt{2} a \sqrt{2} \sin \left(\frac{\theta}{2}\right) \\ \therefore \quad \mathrm{S} &=2 \int_{0}^{\pi} \sqrt{2} a \sqrt{2} \sin \left(\frac{\theta}{2}\right) \mathrm{d} \theta \\ &=4 \mathrm{a} \int_{0}^{\pi} \sin \left(\frac{\theta}{2}\right) d \theta \end{aligned}$

$=4 a\left[-2 \cos \left(\frac{\theta}{2}\right)\right]_{0}^{\pi}$

$s=8 a$