12.379. On the segment $A B$, equal to $2 R$, a semicircle is constructed with it as the diameter, and a chord $C D$ parallel to $A B$ is drawn. Find the volume of the solid formed by the rotation of triangle $A C D$ around the diameter $A B$, if the inscribed angle subtending arc $A C$ is $\alpha (A C < A D)$.
## Solution.

Let $C_{1}$ and $D_{1}$ be the points symmetric to $C$ and $D$ with respect to the diameter $AB$ (Fig. 12.239), $E$ and $F$ be the points of intersection of $CC_{1}$ and $DD_{1}$ with $AB$, $O$ be the center of the semicircle, $V_{1}$ be the volume of the cone with the axial section $\triangle CAC_{1}$, $V_{2}$ be the volume of the cylinder with the axial section $CDD_{1}C_{1}$, and $V_{3}$ be the volume of the cone with the axial section $DAD_{1}$.

The volume of the solid of revolution is

\[
\begin{aligned}
& V=V_{1}+V_{2}-V_{3}=\frac{1}{3} \pi \cdot CE^{2} \cdot AE + \pi \cdot CE^{2} \cdot EF - \frac{1}{3} \pi \cdot CE^{2} \cdot AF = \\
& =\frac{1}{3} \pi \cdot CE^{2}(AE + 3EF - (AE + EF)) = \frac{2}{3} \pi \cdot CE^{2} \cdot EF.
\end{aligned}
\]

The inscribed angle $\angle ADC$ and the central angle $\angle AOC$ subtend the same arc, hence, $\angle AOC = 2 \angle ADC = 2\alpha$.

In $\triangle CEO (\angle CEO = 90^\circ): CE = CO \sin \angle AOC = R \sin 2\alpha$;

$OE = CO \cos \angle AOC = R \cos 2\alpha$. $EF = 2 \cdot OE = 2R \cos 2\alpha$.

$V = \frac{2}{3} \pi \cdot CE^{2} \cdot EF = \frac{2}{3} \pi \cdot R^{2} \sin^{2} 2\alpha \cdot 2R \cos 2\alpha = \frac{2}{3} \pi R^{3} \sin 4\alpha \sin 2\alpha$.

Answer: $\frac{2}{3} \pi R^{3} \sin 4\alpha \sin 2\alpha$.