4. Given a circle (c) with center $O$ and radius $R$. The point $O_{1}$, which lies on the circle (c), is the center of another circle with radius $\frac{R}{2}$.

If $A$ and $B$ are the intersection points of these two circles, determine the approximate value of $\measuredangle A O B$ (in radians), and then the volume of the intersection of the spheres obtained by rotating the two circles around the line $O O_{1}$.
Solution. Let $\alpha=\measuredangle A O B$. Then, in the isosceles triangle $O_{1} A O$ ($A O=O O_{1}=R, O_{1} A = \frac{R}{2}$), the angle at the vertex is $\frac{\alpha}{2}$, so we have

$$
\sin \frac{\alpha}{4}=\frac{\frac{R}{4}}{R}=\frac{1}{4}
$$

Therefore,

$$
\measuredangle A O B=\alpha=4 \arcsin \frac{1}{4} \approx 1.01072102057
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_c53e313ed51c88c0c5c5g-2.jpg?height=325&width=454&top_left_y=1097&top_left_x=1030)

Let $C$ be the intersection of the lines $A B$ and $O O_{1}$. Since $\measuredangle C A O_{1}=\frac{\alpha}{4}$, we get

$$
O_{1} C=O_{1} A \sin \frac{\alpha}{4}=\frac{R}{8} \text { and } O C=\frac{7 R}{8} .
$$

The desired volume is equal to the sum of the volumes of two spherical segments, the first with radius $O_{1} A=\frac{R}{2}$ and distance from the center of the sphere $O_{1} C=\frac{R}{8}$, and the second with radius $A O=R$ and distance from the center of the sphere $O C=\frac{7 R}{8}$. Using the formula

$$
V_{0}=\pi\left(\frac{2}{3} R^{3}-R^{2} x+\frac{1}{3} x^{3}\right)
$$

for the volume of a spherical segment with radius $R$ and distance from the center of the sphere $x$, we get for the desired volume

$$
V=\frac{13}{192} \pi R^{3}
$$