29.38. Calculate the length of the cycloid branch

$$
x=a(t-\sin t), \quad y=a(1-\cos t), \quad \text { where } \quad 0 \leqslant t \leqslant 2 \pi
$$

## 29.7. Surface Area

Let $f(x)$ be a continuous positive function on the interval $[a, b], x_{0}=a<x_{1}<\ldots<x_{n-1}<b=x_{n}$ - some points on the interval $[a, b]$. Suppose we rotate the curve $y=f(x)$ around the $O x$ axis. The broken line with vertices $A_{k}=\left(x_{k}, f\left(x_{k}\right)\right)$ sweeps out a certain surface, consisting of truncated cones. If the sum of the lateral surface areas of these cones has a limit as the length of the largest segment $A_{k} A_{k+1}$ approaches zero, then this limit is called the surface area of revolution swept out by the curve $y=f(x)$.
29.38. Answer: $8a$. It is clear that

$$
\sqrt{x'^2 + y'^2} = a \sqrt{(1 - \cos t)^2 + \sin^2 t} = 2a \sin \frac{t}{2}
$$

Therefore, the length of the branch of the cycloid is

$$
2a \int_{0}^{2\pi} \sin \frac{t}{2} dt = -\left.4a \cos \frac{t}{2}\right|_{0}^{2\pi} = 8a
$$