The center of a square moves along a diameter of a circle

The center of a square moves along a diameter of a circle of radius $a$, with the plane of the square remaining perpendicular to the plane of the circle, and two opposite vertices of the square moving on the circle. Find the volume of the solid generated by the moving square.

We express the square's cross-sectional area at each point along the diameter, then integrate.

Step 1: Set up coordinates

Let the circle $x^2+y^2=a^2$ lie in the $xy$-plane, and let the square's center move along the $x$-axis. At position $x\in[-a,a]$, the two vertices on the circle are $(x,\pm\sqrt{a^2-x^2},0)$.

Step 2: Relate the diagonal to the side length

These two vertices are opposite corners of the square, so they lie at the ends of one diagonal, of length $d=2\sqrt{a^2-x^2}$. For a square, diagonal and side are related by $s\sqrt2=d$, so:

\[s = \sqrt2\sqrt{a^2-x^2}\]

Step 3: Area of the square cross-section

\[A(x) = s^2 = 2(a^2-x^2)\]

Step 4: Integrate along the diameter

\[V = \int_{-a}^{a}2(a^2-x^2)\,dx = 4\int_0^a(a^2-x^2)\,dx = 4\left[a^2x-\frac{x^3}{3}\right]_0^a = 4\cdot\frac{2a^3}{3} = \frac{8a^3}{3}\]

\[\boxed{V = \frac{8a^3}{3} \text{ cubic units}}\]

Step 5: Sanity checks

At $x=0$ (center of the circle): $s=a\sqrt2$, area $=2a^2$ (the maximum cross-section). At $x=\pm a$ (the circle's endpoints): $s=0$, area $=0$ — the solid tapers to a point at both ends, matching the picture of a square shrinking to nothing as its center approaches the circle's edge.

Summary

\[\boxed{V = \frac{8a^3}{3} \text{ cubic units}}\]

Key insight: Because the constraint is on the square's diagonal (not its side) touching the circle, the cross-sectional area $2(a^2-x^2)$ is exactly twice what a circle of the same radius would contribute at that $x$ — giving $\frac{8a^3}{3}$ instead of the sphere's $\frac{4\pi a^3}{3}$, with $\pi$ replaced by the constant $2$.