Volume Common to Two Perpendicular Cylinders
Two solid circular cylinders, each of radius $1$, intersect at right angles: their axes cross (they meet at a single point) and are perpendicular to each other.
What is the volume of the region that lies inside both cylinders?
For full credit, also give the volume when both cylinders have radius $r$.
Hints
- Set the cylinders up as $x^2 + z^2 \le r^2$ (axis along $y$) and $y^2 + z^2 \le r^2$ (axis along $x$). What does a horizontal slice at height $z$ look like?
- At height $z$ the slice satisfies $|x| \le \sqrt{r^2 - z^2}$ and $|y| \le \sqrt{r^2 - z^2}$ independently: it is a square. Integrate the square's area over $z \in [-r, r]$.
- Shortcut: the sphere of radius $r$ has circular slices of radius $\sqrt{r^2 - z^2}$ at the same heights, and a square of side $2a$ has $4/\pi$ times the area of a circle of radius $a$. So the answer is $\frac{4}{\pi}$ times the sphere's volume.
Worked Solution
How to Think About It: Volume problems become one-dimensional integrals once you find a slicing direction in which every cross-section has a simple shape. Slicing the intersection (a Steinmetz solid) perpendicular to the plane of the two axes turns each slice into a square.
Quick Estimate: The solid contains the unit sphere (volume $4\pi/3 \approx 4.19$) and is contained in the cube of side $2$ (volume $8$). Each horizontal slice is a square rather than the sphere's inscribed circle, and a square is $4/\pi \approx 1.27$ times the area of its inscribed circle, so expect about $1.27 \times 4.19 \approx 5.33$.
Formal Solution:
*Step 1 -- Coordinates.* Let the first cylinder be $\{x^2 + z^2 \le r^2\}$ (axis along the $y$-axis) and the second be $\{y^2 + z^2 \le r^2\}$ (axis along the $x$-axis). The axes are perpendicular and cross at the origin.
*Step 2 -- Cross-sections at height $z$.* Fix $z$ with $|z| \le r$. A point $(x, y, z)$ lies in both cylinders exactly when $$|x| \le \sqrt{r^2 - z^2} \quad \text{and} \quad |y| \le \sqrt{r^2 - z^2}.$$ These are independent constraints on $x$ and $y$, so the slice is a square of side $2\sqrt{r^2 - z^2}$ and area $4(r^2 - z^2)$.
*Step 3 -- Integrate.* $$V = \int_{-r}^{r} 4\,(r^2 - z^2)\,dz = 4\left[r^2 z - \frac{z^3}{3}\right]_{-r}^{r} = 4\left(2r^3 - \frac{2r^3}{3}\right) = \frac{16 r^3}{3}.$$
*Step 4 -- Cavalieri check.* The sphere $x^2 + y^2 + z^2 \le r^2$ has slice area $\pi(r^2 - z^2)$ at the same height, so $V_{\text{solid}} / V_{\text{sphere}} = 4/\pi$ slice by slice, and $V = \frac{4}{\pi}\cdot\frac{4}{3}\pi r^3 = \frac{16r^3}{3}$, confirming Step 3 without any integration.
Answer: With $r = 1$ the common volume is $\dfrac{16}{3} \approx 5.33$; in general it is $\dfrac{16 r^{3}}{3}$.
Intuition
Slicing perpendicular to the plane containing both axes makes every cross-section a square, which is why no $\pi$ appears in the answer $16r^3/3$. Comparing slice by slice with the inscribed sphere (Cavalieri's principle) is the same idea used to price path-dependent claims by matching distributions layer by layer, or to compute a volume of a high-dimensional acceptance region in Monte Carlo by comparing with a known body.