Four Intersecting Cylinders

See the solution at the related problem . The only difference is that in this case, for each $z$ , the four long rectangles intersect in an octagon of height $2\sqrt{1-z^2}$ . The area of an octagon of height $h$ is $2(\sqrt{2}-1)h^2$ and so our volume is: $\int_{-1}^{1} 2(\sqrt{2}-1)\cdot 4(1-z^2) dz = 8(\sqrt{2}-1)\bigg[_{-1}^1 z - \frac{z^3}{3} \bigg] = \frac{16}{3}\cdot 2(\sqrt{2}-1) \approx \boxed{4.42}$

Naturally, this is the volume of two unit cylinders intersecting at right angles (16/3) times the ratio of the area of an octagon to that of a square of the same height.