Consider three cylinders intersecting with a unit cube. Their intersection within the unit cube produces a 3-sided solid with a volume of about .386.

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One cylinder has center axis (0,0,1) to (0,1,1) with unit radius, the others are rotations.

What is an exact solution for the volume?

  • $\begingroup$ physicsforums.com/threads/… but the result for $R=1$ doesn't coincide with yours $\endgroup$ – Jean Marie Feb 11 at 17:00
  • $\begingroup$ The structure Jean Marie points to has three intersecting axes. In the figure I'm asking about, the axes do not intersect. $\endgroup$ – Ed Pegg Feb 11 at 17:06
  • 1
    $\begingroup$ I should have paid more attention... $\endgroup$ – Jean Marie Feb 11 at 17:11

Convert the volume integral to a surface integral: $$\int_D dV = \frac 1 3 \int_{\partial D} \mathbf r \cdot d\mathbf S.$$ Parametrize one piece of the surface as $(x, y, z) = (1 + \cos t, \sin t, z)$. The $(t, z)$ domain will be $\pi/2 < t < \pi, \,f(t) < z < g(t)$, where $f$ and $g$ are found from the equations of the other two cylinders. The integrals over the other two pieces are the same due to symmetry. This gives $$V = \int_{\pi/2}^\pi \left( \sqrt {(2 - \sin t) \sin t} + \sqrt {-(2 + \cos t) \cos t} - 1 \right) (1 + \cos t) \,dt = \\ \frac {15} 2 E {\left( \frac 1 9 \right)} - 6 K {\left( \frac 1 9 \right)} -\frac {3 \pi} 4 + 1,$$ with the elliptic integrals given in the parameter notation.


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