AP Calculus BC Practice Test

When a region is revolved around an axis and there is a hollow inside, each cross-section perpendicular to the axis is a washer: an outer disk of radius R(x) with a hole of radius r(x). The area of that washer is π[R(x)]^2 − π[r(x)]^2. To get the volume, you sum these cross-sectional areas from a to b, which gives V = ∫_a^b [π(R(x)^2) − π(r(x)^2)] dx. This is the same as V = π ∫_a^b [R(x)^2 − r(x)^2] dx, i.e., the outer disk’s volume minus the inner hole’s volume. Expressed differently, you can see it as the difference of two separate disks: π ∫_a^b R(x)^2 dx minus π ∫_a^b r(x)^2 dx. This matches the correct idea: the outer radius contributes the full disk area, the inner radius subtracts the hollow part. The other forms miss a piece: dropping the inner subtraction inside the integral omits the hole, and dropping the π factor would misscale the volume. If the inner radius were zero, it would reduce to the disk method.