Volume of three cylinders

This one is not a M.SE question that I answered, but rather a response to a challenge from another M.SE user. The problem is to find the volume of three orthogonal, intersecting cylinders:

$$\begin{align}x^2+y^2&=1\\x^2+z^2&=1\\ y^2+z^2&=1\end{align}$$

The intersection region is pictured below:

3cyls

Problems like these are notoriously difficult because they are difficult to visualize properly. I will not attempt to do so here; my favored approach of calculating cross-sections will lead to frustration and misery.

Rather, I note that there are two ways to bound the volume over $x$:

$$\begin{align}|x| &\le \sqrt{1-y^2}\\ |x| &\le \sqrt{1-z^2}\end{align}$$

Since we are computing the volume of the interior of the region defined by these bounds, it stands to reason that $|x|$ must be bounded by the smaller of these two bounds:

$$|x| \le \min{\left(\sqrt{1-y^2},\sqrt{1-z^2}\right)}$$

so that the volume integral takes the form

$$\int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: \int_{-m(y,z)}^{m(y,z)} dx = 2 \int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: m(y,z)$$

Below is a representation of the integration region for this integral:

3cylscs

The reason for the lines is because $\sqrt{1-y^2} \lt \sqrt{1-z^2}$ according to whether $|y| \gt |z|$. The integral is then symmetric over the regions bounded by the sloped lines; thus, we need only consider one such region and the others will yield the same result. Let’s then consider the region surrounding the positive $y$ (horizontal) axis in the above figure. In this case, $m(y,z) = \sqrt{1-y^2}$; when we use polar coordinates, the integral becomes

$$\begin{align}8 \int_{-\pi/4}^{\pi/4} d\phi \: \int_0^1 d\rho \, \rho \sqrt{1-\rho^2 \cos^2{\phi}} &= 4 \frac{2}{3} \int_{-\pi/4}^{\pi/4} d\phi \: \left( 1- \left|\sin^3{\phi}\right|\right) \sec^2{\phi}\\ &= \frac{16}{3} – \frac{16}{3} \int_0^{\pi/4} d\phi \: \sin^3{\phi} \, \sec^2{\phi}\\ &= 8 \left (2 – \sqrt{2}\right ) \end{align}$$

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