12.052. Triangle $ABC$ rotates around a line lying in the plane of this triangle, passing outside it through vertex $A$ and equally inclined to sides $AB$ and $AC$. Find the volume of the solid of revolution if $AB=a$, $AC=b$, and $\angle BAC=\alpha$.
## Solution.

The volume of the given body can be obtained by subtracting the volumes of the cones with vertex at $A$ and diameters $B B_{1}$ and $C C_{1}$ in the bases from the volume of the frustum of a cone with diameters $B B_{1}$ and $C C_{1}$ in the bases (Fig. 12.54). Drop perpendiculars $B K$ and $C L$ to the line $l$. Then $\angle K A B=\angle C A L=\frac{180^{\circ}-\alpha}{2}=90^{\circ}-\frac{\alpha}{2}$. Therefore, $K B=A B \sin \left(90^{\circ}-\frac{\alpha}{2}\right)=a \cos \frac{\alpha}{2}$, and $C L=A C \sin \left(90^{\circ}-\frac{\alpha}{2}\right)=b \cos \frac{\alpha}{2}$; $A K=A B \cos \left(90^{\circ}-\frac{\alpha}{2}\right)=a \sin \frac{\alpha}{2}$, $A L=b \sin \frac{\alpha}{2}$. The height of the frustum of the cone $K L=A K+A L=\sin \frac{\alpha}{2}(a+b)$. The volume of the frustum of the cone

$$
V^{\prime}=\frac{1}{3} \pi \cdot K L \cdot\left(K B^{2}+K B \cdot L C+L C^{2}\right)=\frac{1}{3} \pi \sin \frac{\alpha}{2}(a+b) \cos ^{2} \frac{\alpha}{2}\left(a^{2}+a b+b^{2}\right)
$$

The volume $V_{1}=\frac{1}{3} \pi A K \cdot K B^{2}=\frac{1}{3} \pi a^{3} \cos ^{2} \frac{\alpha}{2} \sin \frac{\alpha}{2}$. The volume $V_{2}=\frac{1}{3} \pi \cdot A L \cdot L C^{2}=$ $=\frac{1}{3} \pi b^{3} \cos ^{2} \frac{\alpha}{2} \sin \frac{\alpha}{2}$. Then the volume of the obtained solid of revolution

$$
\begin{aligned}
& V=V^{\prime}-V_{1}-V_{2}=\frac{1}{3} \pi \cos ^{2} \frac{\alpha}{2} \sin \frac{\alpha}{2}\left(a^{3}+2 a^{2} b+2 a b^{2}+b^{3}-a^{3}-b^{3}\right)= \\
& =\frac{2}{3} \pi a b(a+b) \cos ^{2} \frac{\alpha}{2} \sin \frac{\alpha}{2}=\frac{\pi}{3} a b(a+b) \sin \alpha \cos \frac{\alpha}{2}
\end{aligned}
$$

Answer: $\quad \frac{\pi}{3} a b(a+b) \sin \alpha \cos \frac{\alpha}{2}$.