Which of the following is a correct form of the volume by washers formula for outer radius R(x) and inner radius r(x)?

Which of the following is a correct form of the volume by washers formula for outer radius R(x) and inner radius r(x)?

• A. π ∫_a^b (R)^2 dx − 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: (A)

When rotating a region around the x-axis, each cross-section perpendicular to the axis is a washer with outer radius R(x) and inner radius r(x). The area of that washer is π[R(x)^2 − r(x)^2]. Integrating this area from a to b gives the total volume: V = π ∫_a^b [R(x)^2 − r(x)^2] dx. This is the same as π ∫_a^b R(x)^2 dx − π ∫_a^b r(x)^2 dx, since the limits are the same. The subtraction reflects removing the hole from the outer disk slice by slice. The other forms would misrepresent the cross-sectional area: adding the two areas ignores the hole, omitting π would fail to convert radii to area, and integrating only the outer radius ignores the hollow region.

When rotating a region around the x-axis, each cross-section perpendicular to the axis is a washer with outer radius R(x) and inner radius r(x). The area of that washer is π[R(x)^2 − r(x)^2]. Integrating this area from a to b gives the total volume: V = π ∫_a^b [R(x)^2 − r(x)^2] dx. This is the same as π ∫_a^b R(x)^2 dx − π ∫_a^b r(x)^2 dx, since the limits are the same. The subtraction reflects removing the hole from the outer disk slice by slice. The other forms would misrepresent the cross-sectional area: adding the two areas ignores the hole, omitting π would fail to convert radii to area, and integrating only the outer radius ignores the hollow region.