15. Intersecting Cylinders. The axes of symmetry of two straight circular cylinders, each with a diameter of 2 cm, intersect at a right angle. What is the volume of the common part of these cylinders?
15. Consider the plane passing through the axes of symmetry of the given cylinders. The cross-section of the common part of the cylinders by this plane is a square. If we draw a plane parallel to this one, it is clear from the figure that the cross-section will also be a square, and the circle inscribed in this square will

![](https://cdn.mathpix.com/cropped/2024_05_21_2b40d984baf9da275c68g-080.jpg?height=451&width=443&top_left_y=894&top_left_x=812)

represent the cross-section of the plane with the sphere inscribed in the common part of the cylinders. Therefore, the volume of the common part of our cylinders is related to the volume of the inscribed sphere as the area of the square is to the area of the inscribed circle.

$$
V=\frac{4 \pi r^{3}}{3} \cdot \frac{4}{\pi}=\frac{16 r^{3}}{3}=\frac{16}{3} \mathrm{~cm}^{3}
$$

[L. Moser, M. M., 25, 290 (May 1952).]

From this, it follows that the common part of the two given cylinders can be divided into infinitely small pyramids, the vertices of which lie at the point of intersection of the axes of the cylinders, and their bases are elements of the cylinders. All such pyramids have a height of one unit. Therefore, the surface area of the common part of the cylinders is $16 \mathrm{~cm}^{2}$.

[J. H. Butchart, M. M., 26, 54 (September 1952).]