1
$\begingroup$

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.

enter image description here

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.