Cylinders Intersecting In 3 Dimensions

The intersection of any two of these cylinders is two base-to-base "vaults", each of which has square cross sections. Here is the final intersection of all three cylinders: The easiest way to think of this shape is as a cube plus six partial vaults. The cube has face diagonal 2 (two radii of a unit cylinder), side length $\sqrt{2}$ , and volume $2\sqrt{2}$ .

Each partial vault is bounded by the cube and by the surfaces of two of the cylinders. Its base (at the cube) is half the cube side length, or $\sqrt{2}/2$ , from the origin along the axis (call it $z$ ) of the other cylinder. We'll slice the partial vault, so that for each value of $z$ from $\sqrt{2}/2$ to 1 we have a square of side $2\sqrt{1-z^2}$ and area $4(1-z^2)$ . The volume of each partial vault, then, is: $\int_{\sqrt{2}/2}^{1} 4(1-z^2) dz = 4\bigg[_{\sqrt{2}/2}^1 z - \frac{z^3}{3} \bigg] =\frac{8-5\sqrt{2}}{3}$

So the total volume is: $2\sqrt{2} + 6\cdot\frac{8-5\sqrt{2}}{3} = 8\cdot(2-\sqrt{2}) \approx \boxed{4.69}$