Same area, same volume for the solid of revolution?

Problem

For $-\frac{1}{2}\le a\le 1$,

stays at least $0$ on $[-1,1]$, and the area between the graph and the $x$-axis is always $2$. The region is rotated about the $x$-axis. The question asks whether the volumes are still all equal and, if not, what the maximum is. The full item is on the CEEC 114 AST Mathematics I paper.

Where it goes wrong

The easy mistake is to say that equal areas give equal volumes. Area adds the height $f(x)$. A disk adds the square of the radius, $\pi f(x{)}^2$. With the area fixed, the more unevenly $f$ spreads its height, the larger $\int f^2$ and the volume (it is smallest when $f$ is constant, at $a=0$). The CEEC scoring guidelines split the work into three required steps: write the disk integral, compute the volume in terms of $a$, and find the maximum on the given interval.

Idea

A thin disk of thickness $dx$ has radius about $f(x)$, the distance from the $x$-axis, and volume about

Adding the disks and taking the limit gives

The volume changes with $a^2$, so the solids are not all the same size. On $-\frac{1}{2}\le a\le 1$, the largest $a^2$ is at $a=1$, and

What the figure shows

• Move Parameter a. The area stays $2$, while the height shifts and the solid, rotated about the $x$-axis, gets thicker or thinner.
• The gold section curves are the disk edges. Their radius is $f(x)$. Raise Number of slices n and the midpoint Riemann sum moves toward the exact volume.
• Press Replay the sweep to watch the graph turn about the $x$-axis into the wireframe solid.