AP Calculus BC 8.9 Solids of Revolution Overview
Review solids of revolution by selecting disk or washer radii and integrating cross-sectional areas around an axis, with bounds set by the region.
- Syllabus
- Effective Fall 2025
- Course
- AP Calculus BC
Review solids of revolution by selecting disk or washer radii and integrating cross-sectional areas around an axis, with bounds set by the region.

A company designs spinning toys using the family of functions y=cx4−x2, where c is a positive constant. The figure above shows the region in the first quadrant bounded by the x-axis and the graph of y=cx4−x2, for some c. Each spinning toy is in the shape of the solid generated when such a region is revolved about the x-axis. Both x and y are measured in inches.
For another spinning toy, the volume is 2π cubic inches. What is the value of c for this spinning toy?
\begin{tabular}{|l|l|l|}
\hline \multirow[t]{2}{*}{Volume =∫02π(cx4−x2)2dx=πc2∫02x2(4−x2)dx} & Form of the integrand & 1 point \\
\hline & Limits and constant & 1 point \\
\hline =πc2∫02(4x2−x4)dx=πc2(34x3−51x502) & Antiderivative & 1 point \\
\hline =πc2(332−532)=1564πc21564πc2=2π⇒c2=3215⇒c=3215 & Answer & 1 point \\
\hline
\end{tabular}
Scoring notes:
- The first point is earned for presenting an integrand of the form A(x4−x2)2 in a definite integral
with any limits of integration (numeric or alphanumeric) and any nonzero constant A. Mishandling
the c will result in the response being ineligible for the fourth point.
- The second point can be earned without the first point. The second point is earned for the limits of
integration, x=0 and x=2, and the constant π ( but not for 2π ) as part of an integral with a
correct or incorrect integrand.
- If an indefinite integral is presented with the correct constant π, the second point can be earned if
the antiderivative (correct or incorrect) is evaluated using the correct limits of integration.
- A response that presents 2=∫02(cx4−x2)2dx earns the first and second points.
- The third point is earned for presenting a correct antiderivative of the presented integrand of the
form A(x4−x2)2 for any nonzero A. If there are subsequent errors in simplification of the
antiderivative, linkage errors, or sign errors, the response will not earn the fourth point.
- The fourth point cannot be earned without the third point. The fourth point is earned only for the
correct answer. The expression does not need to be simplified to earn the fourth point.