The following problem has been solved by Archimedes:

Two distinct circular cylinders are inscribed in a cube; the axes thus intersect each other perpendicularly. Determine the volume common to both cylinders, when the radius of the base of the cylinders is $r$ .

If the solid common to both cylinders is cut with a plane parallel to the axes of both cylinders, the figure of intersection is a square. Denote the distance of the plane from the center of the cube be $x$ . By the Pythagorean theorem, half of the side of the square is $\sqrt{r^2\!-\!x^2}$ and the area of the square is $4(\sqrt{r^2\!-\!x^2})^2$ . Accordingly, we have the function $$A(x) \;:=\; 4(r^2\!-\!x^2)$$ for the area of the intersection square. If we let $x$ here to grow from $0$ to $r$ , then half of the given solid is got. By the volume formula of the parent entry, the half volume of the solid is $$\frac{1}{2}V \;=\; \int_0^r\!4(r^2\!-\!x^2)\,dx \;=\; 4\!\sijoitus{x=0}{\quad r}\!\left(r^2x-\frac{x^3}{3}\right) \;=\; \frac{8}{3}r^3.$$ So the volume in the question is $\frac{16}{3}r^3$ . It is $\displaystyle\frac{2}{3}$ of the volume of the cube.