What is the washers formula for volume when there is an outer radius R(x) and an inner radius r(x)?

What is the washers formula for volume when there is an outer radius R(x) and an inner radius r(x)?

• A. π ∫_a^b (R)^2 dx
• B. π ∫_a^b (R)^2 dx − π ∫_a^b (r)^2 dx
• C. ∫_a^b (R^2 − r^2) dx
• D. π ∫_a^b (r)^2 dx

Answer: (B)

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.

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.