12.329. The lateral side of an isosceles triangle is equal to $a$, and the angle at the base is $\alpha$. This triangle rotates around a line passing through the vertex opposite the base, parallel to the bisector of angle $\alpha$. Find the surface area of the solid of revolution.
Solution.

In $\triangle BAC, BA = AC = a, \angle ACB = \alpha, CD$ is the bisector of $\angle ACB$. Consider the axial section of the resulting solid of revolution (Fig. 12.190). Its surface area $S$ is the sum of the lateral surface area $S_{1}$ of the cone with axial section

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1086.jpg?height=642&width=652&top_left_y=79&top_left_x=368)

Fig. 12.190

$BAB_{1}$, the lateral surface area $S_{2}$ of the cone with axial section $CAC_{1}$, and the lateral surface area $S_{3}$ of the frustum of a cone with axial section $BCC_{1}B_{1}$.

$O_{1}$ is the midpoint of $BB_{1}, O_{2}$ is the midpoint of $CC_{1} \cdot$ Let $O_{1}B = R, O_{2}C = r, BC = b$.

In $\triangle BAC: b = 2a \cos \alpha, \angle BAC = \pi - 2\alpha$.

Since the axis of rotation $O_{1}O_{2}$ is parallel to $CD$, then $\angle O_{2}AC = \angle ACD = \frac{\alpha}{2} \Rightarrow \angle BAO_{1} = \pi - \left(\angle O_{2}AC + \angle BAC\right) = \frac{3\alpha}{2}$.

In $\triangle AO_{2}C \left(\angle AO_{2}C = \frac{\pi}{2}\right): r = a \sin \frac{\alpha}{2}$.

In $\triangle AO_{1}B \left(\angle AO_{1}B = \frac{\pi}{2}\right): R = a \sin \frac{3\alpha}{2}$.

$S = S_{1} + S_{2} + S_{3} = \pi R a + \pi r a + \pi b(R + r) = \pi a(R + r) + \pi b(R + r) = \pi(R + r)(a + b) = \pi \left(a \sin \frac{3\alpha}{2} + a \sin \frac{\alpha}{2}\right)(a + 2a \cos \alpha) = \pi a^{2} \cdot 2 \sin \alpha \cos \frac{\alpha}{2} \times$

$\times (1 + 2 \cos \alpha) = 4 \pi a^{2} \sin \alpha \cos \frac{\alpha}{2} \left(\frac{1}{2} + \cos \alpha\right) = 4 \pi a^{2} \sin \alpha \cos \frac{\alpha}{2} \times$

$\times \left(\cos \frac{\pi}{3} + \cos \alpha\right) = 8 \pi a^{2} \sin \alpha \cos \frac{\alpha}{2} \cos \left(\frac{\pi}{6} + \frac{\alpha}{2}\right) \cos \left(\frac{\pi}{6} - \frac{\alpha}{2}\right)$

Answer: $8 \pi a^{2} \sin \alpha \cos \frac{\alpha}{2} \cos \left(\frac{\pi}{6} + \frac{\alpha}{2}\right) \cos \left(\frac{\pi}{6} - \frac{\alpha}{2}\right)$.