Chapter 6.2, Problem 48E

To determine

Find the volume common to two circular cylinders.

Answer to Problem 48E

The volume common to two circular cylinders is $\frac{16r^3}{3}$

Explanation of Solution

Given information: Given two circular cylinders.

Calculation:

To find the volume by summing cross-sections parallel to $ABCD$ shown in the diagram.

Note that points $B\text{ and }C$ are on the yellow cylinder, points $C\text{ and D}$ are on the blue cylinder. Here $ABCD$ is actually $1/4$ of one whole cross-section at that level so will need to multiply by 4 somewhere.

  

We would integrate from $z=-r$ to $r$ but we can use symmetry by integrate from $z=0\text{ to }r$ and multiply by $2$ .

The length $AD\text{ and }BC$ at $z$ would be $\sqrt{r^2-z^2}$ ,from the circle equation $y^2+z^2=r^2$ ,the lengths $AB\text{ and }CD$ would be the same since the cylinders have the same radius $\left(x^2+z^2=r^2\right)$

Whole cross-section at $z$ would be

  $\begin{array}{l}A\left(x\right)=4{\left(\sqrt{r^2-z^2}\right)}^2=4\left(r^2-z^2\right)\\ V=2{\displaystyle \underset{0}{\overset{r}{\int }}4\left(r^2-z^2\right)}dz\\ =8{\left[r^{2}z-\frac{1}{3}{z}^3\right]}_0^r\\ =8\left[r^2\left(r\right)\text{-}\frac{1}{3}{r}^3\text{-}\left(0-0\right)\right]\\ =8\left[r^3-\frac{1}{3}{r}^3\right]\\ =8\left[\frac{3r^3-r^3}{3}\right]\\ =8\left(\frac{2r^3}{3}\right)\\ =\frac{16r^3}{3}\end{array}$

Thus, the volume of the solid is $\frac{16r^3}{3}$