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

To find the ratio of the volume of a spherical segment to the volume of the whole sphere, where the arc in the axial section of the segment corresponds to a central angle equal to $\alpha$, we can proceed as follows:

1. **Volume of the Spherical Segment:**
   The volume \( V_{\text{segment}} \) of a spherical segment can be given by the formula:
   \[
   V_{\text{segment}} = \frac{\pi h^2 (3R - h)}{3}
   \]
   where \( R \) is the radius of the sphere, and \( h \) is the height of the segment.

2. **Height of the Spherical Segment:**
   The height \( h \) of the spherical segment can be related to the central angle \( \alpha \) and the radius \( R \) of the sphere. The height \( h \) is given by:
   \[
   h = R (1 - \cos(\alpha/2))
   \]

3. **Volume of the Whole Sphere:**
   The volume \( V_{\text{sphere}} \) of the whole sphere is:
   \[
   V_{\text{sphere}} = \frac{4}{3} \pi R^3
   \]

4. **Ratio of Volumes:**
   The ratio of the volume of the spherical segment to the volume of the whole sphere is:
   \[
   \text{Ratio} = \frac{V_{\text{segment}}}{V_{\text{sphere}}} = \frac{\frac{\pi h^2 (3R - h)}{3}}{\frac{4}{3} \pi R^3} = \frac{h^2 (3R - h)}{4R^3}
   \]

5. **Substitute \( h \) into the Ratio:**
   Substitute \( h = R (1 - \cos(\alpha/2)) \) into the ratio:
   \[
   \text{Ratio} = \frac{[R (1 - \cos(\alpha/2))]^2 [3R - R (1 - \cos(\alpha/2))]}{4R^3}
   \]
   Simplify the expression:
   \[
   \text{Ratio} = \frac{R^2 (1 - \cos(\alpha/2))^2 [3R - R (1 - \cos(\alpha/2))]}{4R^3}
   \]
   \[
   \text{Ratio} = \frac{(1 - \cos(\alpha/2))^2 [3 - (1 - \cos(\alpha/2))]}{4}
   \]
   \[
   \text{Ratio} = \frac{(1 - \cos(\alpha/2))^2 (2 + \cos(\alpha/2))}{4}
   \]

Thus, the ratio of the volume of the spherical segment to the volume of the whole sphere is:
\[
\boxed{\frac{(1 - \cos(\alpha/2))^2 (2 + \cos(\alpha/2))}{4}}
\]
3.105. By the condition, \(A \cap B\) is a spherical segment, \(O\) is the center of the sphere, \(\cup A \cap B = \angle AOB = \alpha\) (Fig. 3.105). Let \(R\) be the radius of the sphere, \(OC\) be perpendicular to the base of the segment, \(O_1\) be the center of the base of the segment, \(O_1C = H\); then \(H = R - OO_1\). From \(\triangle OO_1A\) we find \(OO_1 = R \cos \frac{\alpha}{2}\), hence

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-570.jpg?height=488&width=464&top_left_y=660&top_left_x=687)

Fig. 3.105 \(H = R - R \cos \frac{\alpha}{2} = 2R \sin^2 \frac{\alpha}{4}\). Therefore,

\[
\begin{gathered}
V_{\text{map. segm}} = \pi H^2 \left(R - \frac{H}{3}\right) = \pi \cdot 4R^2 \sin^4 \frac{\alpha}{4} \left(R - \frac{2R}{3} \sin^2 \frac{\alpha}{4}\right) = \\
= \frac{4\pi R^3 \sin^4 \frac{\alpha}{4}}{3} \left(3 - 2 \sin^2 \frac{\alpha}{4}\right) = \frac{4\pi R^3}{3} \sin^4 \frac{\alpha}{4} \left(2 + \cos \frac{\alpha}{2}\right) \\
V_{\text{map}} = \frac{4\pi R^3}{3}.
\end{gathered}
\]

Thus,

\[
\frac{V_{\text{map. segm}}}{V_{\text{map}}} = \sin^4 \frac{\alpha}{4} \left(2 + \cos \frac{\alpha}{2}\right)
\]