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Problem

Tricylinder

Consider the solid formed by intersecting three perpendicular cylinders:

$ x^2 + y^2 = r^2 $
$ x^2 + z^2 = r^2 $
$ y^2 + z^2 = r^2 $.

This solid is called the tricylinder. Determine the volume of the tricylinder. Express the volume in terms of the radius of the cylinder. Then express the volume in terms of the diameter of the cylinder.

Solution

Split the solid into 48 congruent segments. The segments are bounded by the circular arc $ \sqrt{r^2 - {(r\cos{\phi})}^{2}} = r \sin{\phi} $ swept across the azimuthal angle $ 0 \le \phi \le \frac{\pi}{4} $.

$ \left\{ (\rho, \phi, z)| 0 \le \rho \le r, \: 0 \le \phi \le \frac{\pi}{4}, \: 0 \le z \le r\sin{\phi} \right\} $
$ {V}_{segment} = \int_{0}^{\pi/4} \int_{0}^{r} \int_{0}^{r\sin \phi} \rho \: dz d\rho d\phi $
$ {V}_{segment} = \int_{0}^{\pi/4} \int_{0}^{r} \rho \left[z \Biggr|_{0}^{r\sin{\phi}}\right] d\rho d\phi $
$ {V}_{segment} = \int_{0}^{\pi/4} \int_{0}^{r} {\rho}^2 \sin{\phi} \: d\rho d\phi $
$ {V}_{segment} = \int_{0}^{\pi/4} \left[\frac{{\rho}^3}{3} \sin{\phi} \Biggr|_{0}^{r} \right] d\phi $
$ {V}_{segment} = \frac{r^3}{3} \int_{0}^{\pi/4} \sin{\phi} \: d\phi $
$ {V}_{segment} = \frac{r^3}{3} \left[-\cos{\phi} \Biggr|_{0}^{\pi/4} \right] $
$ {V}_{segment} = \frac{\left(2-\sqrt{2} \right)r^3}{6} $

Thus the volume of the tricylinder is

$ V = 48 {V}_{segment} $
$ V = 48 \left(\frac{\left(2-\sqrt{2} \right)r^3}{6} \right) $
$ V = 8 \left(2-\sqrt{2} \right)r^3 $.

Since diameter is twice the radius, $ d = 2r $,

$ V = \left(2-\sqrt{2} \right)d^3 $.
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