Archimedes’ cylinders in cube

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

$$A(x):=\mathrm{\hspace{0.33em}4}(r^2-x^2)$$

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 of the parent entry (http://planetmath.org/VolumeAsIntegral), the half volume of the solid is

$$\frac{1}{2}V={\int }_0^{r}4(r^2-x^2)𝑑x=\mathrm{\hspace{0.33em}4}\underset{x=0}{\overset{r}{/}}\left(r^{2}x-\frac{x^3}{3}\right)=\frac{8}{3}{r}^3.$$

So the volume in the question is $\frac{16}{3}{r}^3$.  It is ${\displaystyle \frac{2}{3}}$ of the volume of the cube.