58. A convex broken line of length $d$ rotates around a straight line connecting its ends. What can be the area of the surface of revolution thus obtained?
58. Let $A_{1} A_{2} \ldots A_{n}$ be a given broken line, $A_{1} A_{n}$ - a straight line passing through its ends, $B$ - a point that divides the perimeter of the broken line in half (Fig. 137). The segment to which point $B$ belongs is $A_{i} A_{i+1}$.

The surface obtained by rotating the broken line around the line $A_{1} A_{n}$ will consist of cones, truncated cones, and possibly cylinders (cones and truncated cones can degenerate into circles and rings in special cases; see Fig. 138). If $S_{j}$ is the lateral surface area of the body formed by rotating the segment $A_{j} A_{j+1}$, then $S_{j}=\pi l_{j}\left(r_{j}+r_{j+1}\right)$, where $l_{j}$ is the length of the generating line $A_{j} A_{j+1}$, and $r_{j}$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-169.jpg?height=235&width=751&top_left_y=240&top_left_x=332)

Fig. 137.

$r_{j+1}$ are the distances of points $A_{j}$ and $A_{j+1}$ from the axis of rotation. Let $\rho_{j}$ denote the length of the broken line $A_{1} A_{2} \ldots A_{j}$; then, obviously, $\rho_{j}>r_{j}$. Since

$$
l_{j}=\rho_{j+1}-\rho_{j}
$$

Then

$$
S_{j}=\pi l_{j}\left(r_{j}+r_{j+1}\right)<\pi\left(\rho_{j+1}-\rho_{j}\right)\left(\rho_{j}+\rho_{j+1}\right)=\pi\left(\rho_{j+1}^{2}-\rho_{j}^{2}\right)
$$

Consider the surface area obtained by rotating the broken line $A_{1} B$ around $A_{1} A_{n}$. It is equal to the sum of the areas $S_{1}, S_{2}, \ldots, S_{i-1}$

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-169.jpg?height=348&width=751&top_left_y=872&top_left_x=332)

Fig. 138.

and the area obtained by rotating the segment $A_{i} B$. Adding the inequalities (*) corresponding to $j=1,2, \ldots, i-1$ term by term. The expressions on the right-hand sides of these inequalities are the areas of circular rings with inner and outer radii $\rho_{j}$ and $\rho_{j+1}$, respectively. Therefore, the area of the considered part of the surface does not exceed the area of a circle with radius $d / 2$ (Fig. 139), which is $\pi d^{2} / 4$. Similarly, we find that the area of the surface obtained by rotating the broken line $B A_{i+1} A_{i+2} \ldots A_{n-1} A_{n}$ also does not exceed $\pi d^{2} / 4$; therefore, the area of the entire surface does not exceed $\pi d^{2} / 2$.

The area of the entire surface is equal to $\pi d^{2} / 2$ only in the case when the entire broken line lies on a straight line perpendicular to the axis of rotation (then all inequalities become equalities); but in such a case, the broken line itself degenerates into a doubly covered segment.

However, by considering broken lines of the form $A_{1} A_{2} A_{3}$, where $A_{1} A_{2}=A_{8} A_{3}$ and $A_{1} A_{2}$ is sufficiently small (Fig. 140), we can obtain a surface of the considered type, the area of which can be made arbitrarily close to $\pi d^{2} / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-170.jpg?height=391&width=388&top_left_y=343&top_left_x=226)

Fig. 139.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-170.jpg?height=484&width=307&top_left_y=253&top_left_x=715)

Fig. 140.