The radii of two spheres are 5 and 3 units, and the distance between their centers is 4 units. Let's calculate the volume of the common part of the two spheres.
I. Solution. Consider an arbitrary plane section passing through the centers of the two spheres. Let the centers of the two circles visible in the plane section be $O_{1}$ and $O_{2}$, and the intersection points be $P_{1}$ and $P_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0e9839144bda379391a0g-1.jpg?height=491&width=591&top_left_y=267&top_left_x=756)

$\angle P_{1} O_{1} O_{2} = 90^{\circ}$, since the sides of the triangle $P_{1} O_{1} O_{2}$ form a Pythagorean triple, and the angle in question is opposite the hypotenuse of the triangle.

Let the intersection point of $O_{1} O_{2}$ and $P_{1} P_{2}$ be $M$. We know that the line connecting the intersection points of two circles is perpendicular to the line segment connecting the centers, so $\angle P_{1} M O_{2} = 90^{\circ}$.

From the above, it follows that $M = O_{1}$, i.e., $O_{1}$ lies on the segment $P_{1} P_{2}$.

Returning to the space: the common part of the two spheres consists of two spherical caps (which are separated by a circle in a plane perpendicular to $O_{1} O_{2}$ and centered at $O_{1}$), one of which is equal to half of the smaller sphere. The volume of this spherical cap is $V_{1}$, and the volume of the other is $V_{2}$. Let the height of the spherical cap be $m_{1}$, with its value being: $m_{1} = 5 - 4 = 1$. Using the formula for the volume of a sphere and a spherical cap:

$$
\begin{aligned}
& V_{1} = \frac{1}{2} \cdot \frac{4}{3} \cdot 3^{3} \pi = 18 \pi \\
& V_{2} = 5 \cdot 1^{2} \cdot \pi - \frac{1^{3} \pi}{3} = \frac{14}{3} \pi
\end{aligned}
$$

The volume of the common part of the two spheres is $V_{1} + V_{2} = \frac{68}{3} \pi \approx 71.21$.

II. Solution. We cut the two spheres with a plane passing through their centers. The plane section will be a circle with radius 3 and a circle with radius 5, and the distance between the centers of the two circles is 4.

![](https://cdn.mathpix.com/cropped/2024_05_02_0e9839144bda379391a0g-1.jpg?height=469&width=547&top_left_y=1657&top_left_x=778)

Place the two circles in a coordinate system such that their centers lie on the $x$-axis, and the larger circle touches the $y$-axis. Then the equations of the two circles are:

$$
\begin{array}{r}
(x-9)^{2} + y^{2} - 9 = 0 \\
(x-5)^{2} + y^{2} - 25 = 0
\end{array}
$$

We can calculate the coordinates of the intersection points of the two circles by equating the left-hand sides of the two equations: $x^{2} - 18x + 72 = x^{2} - 10x$, from which $x = 9$ and $y = \pm 3$.

We divide the desired solid into two parts and calculate the volume using integration. One part is formed by the rotation of the smaller circle around the $x$-axis, and the other part is formed by the rotation of the larger circle around the $x$-axis. For the smaller circle, $f_{1}^{2}(x) = -x^{2} + 18x - 72$, and for the larger circle, $f_{2}^{2}(x) = -x^{2} + 10$. Therefore:

$$
\begin{aligned}
V_{1} & = \pi \int_{6}^{9} f_{1}^{2}(x) \, dx = \pi \int_{6}^{9} \left(-x^{2} + 18x - 72\right) \, dx = \pi \left[ \frac{-x^{3}}{3} + \frac{18x^{2}}{2} - 72x \right]_{6}^{9} = \\
& = \pi \left( \frac{-9^{3}}{3} + \frac{18 \cdot 9^{2}}{2} - 72 \cdot 9 - \frac{-6^{3}}{3} - \frac{18 \cdot 6^{2}}{2} + 72 \cdot 6 \right) = 18 \pi \\
V_{2} & = \pi \int_{9}^{10} f_{2}^{2}(x) \, dx = \pi \int_{9}^{10} \left(-x^{2} + 10\right) \, dx = \pi \left[ \frac{-x^{3}}{3} + \frac{10x^{2}}{2} \right]_{9}^{10} = \\
& = \pi \left( \frac{-10^{3}}{3} + \frac{10 \cdot 10^{2}}{2} - \frac{-9^{3}}{3} - \frac{10 \cdot 9^{2}}{2} \right) = 4 \frac{2}{3} \pi
\end{aligned}
$$

The sum of the two volumes is: $V_{1} + V_{2} = \frac{68}{3} \pi$.