127. A truncated cone is described around a sphere. The total surface area of this cone is $S$. A second sphere touches the lateral surface of the cone along the circumference of the cone's base. Find the volume of the truncated cone, given that the part of the surface of the second sphere that is inside the first has an area of $Q$.
127. When solving the problem, the following facts are used:
1) The center of the sphere inscribed in the cone lies on the surface of the second sphere (consider the corresponding statement from planimetry);
2) From the fact that the center of the inscribed sphere lies on the surface of the second, it follows that the surface area of the inscribed sphere is $4 Q$, and its radius will be $\sqrt{Q / \pi}$;
3) The volume of the frustum of the cone, in which the sphere is inscribed, is also expressed through the total surface area of the frustum and the radius of the sphere, as is the volume of the circumscribed polyhedron, i.e., $V=\frac{1}{3} S \sqrt{\frac{Q}{\pi}}$. 128. Prove that if $R$ and $r$ are the radii of the circles of the bases of the frustum, then the radius of the inscribed sphere will be $\sqrt{R r}$. Answer: $\frac{S}{2}$.