12.236. Find the ratio of the volume of a spherical segment to the volume of the entire sphere, if the arc in the axial section of the segment corresponds to a central angle equal to $\alpha$.
## Solution.

Consider the axial section of the given sphere, $O$ - its center, $\angle A O C=\alpha$, $K$ - the vertex of the segment, $D$ - the point of intersection of $A C$ and the radius $O K, K D-$ the height of the segment (Fig. 12.103).

If $R$ is the radius of the given sphere, then its volume $V_{1}=\frac{4}{3} \pi R^{3}$.

In $\triangle A D O\left(\angle A D O=90^{\circ}\right):$

$$
\begin{aligned}
& D O=O A \cos \angle A O D=R \cos \frac{\alpha}{2} \Rightarrow K D=O K-O D=R-R \cos \frac{\alpha}{2}= \\
& =R\left(1-\cos \frac{\alpha}{2}\right)
\end{aligned}
$$

The volume of the segment

$$
\begin{aligned}
& V_{2}=\pi \cdot K D^{2} \cdot\left(R-\frac{1}{3} K D\right)=\pi R^{2}\left(1-\cos \frac{\alpha}{2}\right)^{2} \cdot\left(R-\frac{1}{3} R\left(1-\cos \frac{\alpha}{2}\right)\right)= \\
& =\frac{4}{3} \pi R^{3} \sin ^{4} \frac{\alpha}{4}\left(3-\left(1-\cos \frac{\alpha}{2}\right)\right)=\frac{4}{3} \pi R^{3} \sin ^{4} \frac{\alpha}{4}\left(2+\cos \frac{\alpha}{2}\right)
\end{aligned}
$$

From this, $\frac{V_{2}}{V_{1}}=\sin ^{4} \frac{\alpha}{4}\left(2+\cos \frac{\alpha}{2}\right)$.

Answer: $\sin ^{4} \frac{\alpha}{4}\left(2+\cos \frac{\alpha}{2}\right)$.