2.59. The cone is formed by rotating a right-angled triangle with an area of $S$ around one of its legs. Find the volume of the cone if the length of the circumference described by the point of intersection of the triangle's medians during its rotation is $L$.
2.59. The desired volume $V=\frac{1}{3} \pi r^{2} h$. Let the cone be formed by rotating $\triangle A B C$ around the leg $B C$ (Fig. 2.48); then $A C=\boldsymbol{r}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-487.jpg?height=480&width=312&top_left_y=137&top_left_x=46)

Fig. 2.48

$B C=h$. By the condition, $\frac{1}{2} r h=S$; hence $V=\frac{2}{3} \pi r \cdot \frac{1}{2} r h=\frac{2}{3} \pi r S$. Moreover, by the condition, $2 \pi \cdot D N==L$, where $D$ is the point of intersection of the medians, $D N \perp B C$. But $D N: A C=D M: A M=1: 3$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-487.jpg?height=94&width=755&top_left_y=461&top_left_x=384)
Thus, $V=\frac{2}{3} \pi S \cdot \frac{3 L}{2 \pi}=S L$.

Answer: SL.