3.50. The radius of the circle inscribed in a right trapezoid is $r$, and the acute angle of the trapezoid is $\alpha$. This trapezoid rotates around the smaller lateral side. Find the lateral surface area of the solid of revolution.
3.50. By the condition, $ABCD$ is a trapezoid, $CD \perp BC, CD \perp AD, CD = 2r$ and $\angle BAD = \alpha$ (Fig. 3.52). When the trapezoid is rotated around the side $CD$, a truncated cone $ABB_1A_1$ is obtained, the lateral surface of which is equal to the difference between the lateral surfaces of the complete cone $AEA_1$ and the additional cone $BEB_1$, i.e.,

$$
S_{\text{lat}} = \pi AD \cdot AE - \pi BC \cdot BE
$$

Draw $BF \perp AD$. Since $AE = \frac{AD}{\cos \alpha}$ (from $\triangle ADE$), $BE = \frac{BC}{\cos \alpha}$ (from $\triangle BCE$), $AF = 2r \operatorname{ctg} \alpha = 2r \frac{\cos \alpha}{\sin \alpha}$ (from $\triangle AFB$), then

$$
S_{\text{lat}} = \pi \left( \frac{AD^2 - BC^2}{\cos \alpha} \right) = \frac{\pi}{\cos \alpha} (AD - BC)(AD + BC)
$$

Thus,

$$
S_{\text{lat}} = \frac{\pi}{\cos \alpha} AF (AD + BC)
$$

Considering that the center $O$ of the inscribed circle is the intersection point of the angle bisectors $AO$ and $OD$, we get $\angle OAK = \frac{\alpha}{2}, OK = KD = r$.

$$
\begin{gathered}
AK = OK \operatorname{ctg} \frac{\alpha}{2} = \frac{r \cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} \text{ and, consequently, } \\
AD = AK + KD = \frac{r \cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} + r = \frac{r \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)}{\sin \frac{\alpha}{2}} \\
BC = AD - AF = \frac{r \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)}{\sin \frac{\alpha}{2}} - \frac{2r \cos \alpha}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \\
= \frac{r \left( \cos^2 \frac{\alpha}{2} + \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} - \cos^2 \frac{\alpha}{2} + \sin^2 \frac{\alpha}{2} \right)}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{r \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)}{\cos \frac{\alpha}{2}} \\
AD + BC = \frac{r \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)}{\sin \frac{\alpha}{2}} + \frac{r \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)}{\cos \frac{\alpha}{2}} = \frac{r \left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)^2}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}
\end{gathered}
$$

Substituting the found expressions into formula (1), we finally get

$$
\begin{aligned}
& S_{\text{lat}} = \frac{\pi}{\cos \alpha} \cdot \frac{2r \cos \alpha}{\sin \alpha} \cdot r \frac{\left( \cos \frac{\alpha}{2} + \sin \frac{\alpha}{2} \right)^2}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \\
& \quad = \frac{2\pi r^2}{\sin \alpha} \cdot \frac{1 + \sin \alpha}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{4\pi r^2 (1 + \sin \alpha)}{\sin^2 \alpha}
\end{aligned}
$$