12.326. A plane broken line consists of $n$ equal segments, connected in a zigzag pattern at an angle $\alpha$ to each other. The length of each segment of the broken line is $a$. This line rotates around a straight line passing through one of its ends and parallel to the bisector of the angle $\alpha$. Find the surface area of the body of revolution.
## Solution.

The first segment, having a common point with the axis of rotation, describes the lateral surface of a complete cone $S_{1}$, and all

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1082.jpg?height=609&width=633&top_left_y=420&top_left_x=612)

Fig. 12.187 others - the lateral surfaces of truncated cones $S_{2}, S_{3} \ldots S_{\mathrm{n}}$ (Fig. 12.187).

Let $r$ be the radius of the base of the first cone. Then the first truncated cone has base radii $r$ and $2 r$, the second $2 r$ and $3 r$, the third $3 r$ and $4 r$, and so on.

Thus,

$$
\begin{aligned}
& S_{1}=\pi r a \\
& S_{2}=\pi(r+2 r) a=3 \pi r a \\
& S_{3}=\pi(2 r+3 r) a=5 \pi r a \\
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& S_{n}=(2 n-1) \pi r a
\end{aligned}
$$

The surface area of the solid of revolution $S=S_{1}+S_{2}+S_{3}+\ldots+S_{n}=\pi r a(1+3+5+\ldots+2 n-1)=\pi r a \cdot \frac{1+2 n-1}{2} \cdot n=\pi r a n^{2}$.

Since $r=a \sin \frac{\alpha}{2}$, then $S=\pi a^{2} n^{2} \sin \frac{\alpha}{2}$.

Answer: $\pi a^{2} n^{2} \sin \frac{\alpha}{2}$.