11.190. The radius of the base of the cone is $R$. Two mutually perpendicular generators divide the lateral surface area of the cone into parts in the ratio $1: 2$. Find the volume of the cone.
Solution.

VO is the height of the given cone, $BA$ and $BC$ are the generators mentioned in the problem (Fig. 11.77). Let $BA = l$. Since $BA = BC$ and $\angle ABC = 90^{\circ}$, then $AC = l \sqrt{2}$. Since the generators $BA$ and $BC$ divide the lateral surface area of the cone in the ratio $1:2$, the length of the arc subtended by $AC$ is $1/3$ of the circumference of the base, and $\angle AOC = 120^{\circ}$. Let $K$ be the midpoint of $AC$. Then $AK = \frac{l \sqrt{2}}{2}$, $\angle AOK = 60^{\circ}$, and $AK = OA \cdot \sin \angle AOK; \frac{l \sqrt{2}}{2} = R \sin 60^{\circ} = \frac{R \sqrt{3}}{2}; l = R \sqrt{\frac{3}{2}}$. The height of the cone $H = \sqrt{l^2 - R^2} = \sqrt{\frac{3}{2} R^2 - R^2} = \frac{R \sqrt{2}}{2}$. The volume of the cone $V = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi R^2 \cdot \frac{R \sqrt{2}}{2} = \frac{\pi R^3 \sqrt{2}}{6}$.

Answer: $\frac{\pi R^3 \sqrt{2}}{6}$.