3.92. On the segment $A B$, equal to $2 R$, a semicircle is constructed as if on a diameter and a chord $C D$ is drawn parallel to $A B$. Find the volume of the body formed by the rotation of triangle $A C D$ around the diameter $A B$, if the inscribed angle subtending the arc $A C$ is $\alpha$ $(A C<A D)$.
3.92. By the condition, $A B=2 R, C D \| A B, O$ - the center of the semicircle, $\angle A D C=\alpha, A C<A D$ (Fig. 3.92). The volume of the body of revolution of $\triangle A C D$ around $A B$ will be found using the formula

$$
\begin{aligned}
& V_{\text {T. } \mathrm{pp}}=V_{A O_{2} C}+V_{O_{2} C D O_{1}}-V_{A O_{1} D}=\frac{1}{3} \pi O_{2} C^{2} \cdot A O_{2}+\pi O_{2} C^{2} \cdot O_{1} O_{2}- \\
& -\frac{1}{3} \pi O_{1} D^{2} \cdot A O_{1}=\frac{1}{3} \pi O_{2} C^{2}\left(A O_{2}+3 O_{1} O_{2}-A O_{1}\right)=\frac{2}{3} \pi O_{2} C^{2} \cdot O_{1} O_{2}
\end{aligned}
$$

since $O_{1} D=O_{2} C$ and $A O_{1}-A O_{2}=O_{1} O_{2}$. Considering that $\angle A B C=$ $=\angle A D C=\alpha$ as inscribed angles subtending $\cup A C$, from $\triangle A C B$, where $\angle A C B=90^{\circ}$ (an inscribed angle subtending the diameter $A B$), we find $B C=2 R \cos \alpha, A C=2 R \sin \alpha$. In $\triangle A O_{2} C$ we have $\angle O_{2} A C=$ $=90^{\circ}-\alpha, \angle O_{2} C A=\alpha, O_{2} C=A C \cos \alpha=2 R \sin \alpha \cos \alpha=R \sin 2 \alpha, O_{2} A=$ $=A C \sin \alpha=2 R \sin ^{2} \alpha$; then $O_{1} O_{2}=2 R-2 O_{2} A=2 R\left(1-2 \sin ^{2} \alpha\right)=$ $=2 R \cos 2 \alpha$. Thus,

$$
V_{\mathrm{T} . \mathrm{Bp}}=\frac{2}{3} \pi R^{2} \sin ^{2} 2 \alpha \cdot 2 R \cos 2 \alpha=\frac{2}{3} \pi R^{3} \sin 2 \alpha \sin 4 \alpha
$$