11.178. A triangle with sides equal to $a, b$ and $c$ rotates alternately around each of its sides. Find the ratio of the volumes of the figures obtained in this way.
## Solution.

Let the volumes of the solids of revolution around sides $a, b, c$ be $V_{a}, V_{b}$, $V_{c}$; then $V_{a}=\frac{1}{3} \pi h_{a}^{2} a, V_{b}=\frac{2}{3} \pi h_{b}^{2} b, V_{c}=\frac{1}{3} \pi h_{c}^{2} c$, where $h_{a}, h_{b}, h_{c}$ are the corresponding heights.

Considering that $a h_{a}=b h_{b}=c h_{c}=2 S$, we have $V_{a}=\frac{2}{3} \pi S h_{a}, V_{b}=\frac{2}{3} \pi S h_{b}$, $V_{c}=\frac{2}{3} \pi S h_{c}$ or $V_{a}=\frac{4}{3} \pi S^{2} \cdot \frac{1}{a}, V_{b}=\frac{4}{3} \pi S^{2} \cdot \frac{1}{b}, V_{c}=\frac{4}{3} \pi S^{2} \cdot \frac{1}{c}$. Finally, $V_{a}: V_{b}: V_{c}=\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$.

Answer: $\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$.