74 A sector $O A B$ with radius $R$ and central angle $\theta$ is rotated once around one of its sides $O A$. The surface area of the resulting solid of revolution is
A. $\pi R^{2} \sin \theta$
B. $2 \pi R^{2} \sin \frac{\theta}{2}\left(\cos \frac{\theta}{2}+2 \sin \frac{\theta}{2}\right)$
C. $4 \pi R^{2} \sin \frac{\theta}{2}$
D. $\pi R^{2} \sin \theta(1+\sin \theta)$
74 B. The required surface area is the sum of the area of a spherical cap and the lateral area of a cone, with the base radius of the cone and the spherical cap being $R \cdot \sin \theta$, and the height of the spherical cap being $R(1-\cos \theta)$. Therefore,
$$
\begin{aligned}
S & =\pi R \cdot R \sin \theta+\pi\left[(R \sin \theta)^{2}+R^{2}(1-\cos \theta)^{2}\right] \\
& =\pi R^{2}(\sin \theta+2-2 \cos \theta) \\
& =\pi R^{2}\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+4 \sin ^{2} \frac{\theta}{2}\right) \\
& =2 \pi R^{2} \sin \frac{\theta}{2}\left(\cos \frac{\theta}{2}+2 \sin \frac{\theta}{2}\right) .
\end{aligned}
$$