12.091. The side of the rhombus is $a$, and its acute angle is $\alpha$. The rhombus rotates around a line passing through its vertex and parallel to the longer diagonal. Find the volume of the solid of revolution.
## Solution.

The volume of the solid of revolution is the sum of the volumes of two identical figures $V_{\text {s.r }}=2 V$ (Fig. 12.92). The volume of such a figure is: $V=V_{\text {tr.c }}-V_{\text {c }}$. The volume of the frustum of a cone is: $V_{\text {tr.c }}=\frac{\pi H}{3}\left(R^{2}+r^{2}+R r\right)$. The radius of the larger base $R=O D=2 a \sin \frac{\alpha}{2}$. The radius of the smaller base $r=B C=a \sin \frac{\alpha}{2}$. The height of the frustum of the cone: $H=B A=a \cos \frac{\alpha}{2}$. Substituting, we get:

$$
V_{\mathrm{tr.c}}=\frac{\pi}{3} a \cos \frac{\alpha}{2}\left(4 a^{2} \sin ^{2} \frac{\alpha}{2}+a^{2} \sin ^{2} \frac{\alpha}{2}+2 a^{2} \sin ^{2} \frac{\alpha}{2}\right)=\frac{7}{6} \pi a^{3} \sin \frac{\alpha}{2} \sin \alpha
$$

The volume of the cone with the smaller base:

$V_{\mathrm{c}}=\frac{1}{3} \pi r^{2} H=\frac{1}{3} \pi a^{3} \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}=\frac{1}{6} \pi a^{3} \sin \frac{\alpha}{2} \sin \alpha$. Then the volume of the solid of revolution is:

$$
V_{\text {s.r }}=2\left(V_{\text {tr. } \mathrm{c}}-V_{\mathrm{c}}\right)=2\left(\frac{7}{6} \pi a^{3} \sin \frac{\alpha}{2} \sin \alpha-\frac{1}{6} \pi a^{3} \sin \frac{\alpha}{2} \sin \alpha\right)=2 \pi a^{3} \sin \frac{\alpha}{2} \sin \alpha
$$

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