Source: Cal3 PTAC Lecture Practice (Texas A&M University)
Tags: washer method, volume of solids of revolution, disk method, Riemann sum, definite integral, axis of rotation, cross-sectional area, calculus 1, TAMU
When you rotate a region around a horizontal line that sits above the curve (y = M), each cross-sectional slice is a washer, not a simple disk. The volume integral subtracts the inner-radius area from the outer-radius area. The key move is identifying the correct expressions for the outer and inner radii relative to the axis of rotation.
Washer method
A technique for finding the volume of a solid of revolution when each cross-section perpendicular to the axis of integration is a ring (annulus) rather than a full disk. You subtract the volume carved out by the inner radius from the volume enclosed by the outer radius.
Solid of revolution
The three-dimensional shape formed by rotating a two-dimensional region around a given line (the axis of rotation).
Axis of rotation
The line about which the region is revolved. In this problem, it is y = M, a horizontal line sitting at the maximum value of f on [a, b].
Outer radius (r_outer)
The distance from the axis of rotation to the farthest boundary of the region. Here, r_outer = M (the full distance from y = M down to y = 0).
Inner radius (r_inner)
The distance from the axis of rotation to the nearest boundary of the region. Here, r_inner = M − f(x) (the gap between y = M and the curve y = f(x)).
Riemann sum
The finite sum of slice volumes that, in the limit as the partition norm approaches zero, converges to the definite integral giving exact volume.
Given f(x) ≥ 0 on [a, b].
The region is bounded by y = f(x), y = 0, x = a, and x = b.
Rotate this region about the line y = M, where M is the maximum value of f on [a, b].
Because the axis of rotation is above the curve, each cross-section is a washer.
The outer radius is the distance from y = M down to y = 0:
r_outer = M
The inner radius is the distance from y = M down to the curve y = f(x):
r_inner = M − f(x)
Both radii are measured from the axis of rotation, perpendicular to the x-axis.
Partition [a, b] on the x-axis into n subintervals of width Δxᵢ. Each thin washer-shaped slice has:
Volume of one washer = π · r_outer² · h − π · r_inner² · h
Substituting in: π M² Δxᵢ − π (M − f(xᵢ*))² Δxᵢ
The total approximate volume is:
V ≈ Σ (i = 1 to n) [ π M² − π (M − f(xᵢ*))² ] Δxᵢ
As the partition norm ‖P‖ → 0, the Riemann sum converges to a definite integral:
V = ∫ from a to b [ π M² − π (M − f(x))² ] dx
This can also be written as:
V = π ∫ from a to b [ M² − (M − f(x))² ] dx
Single washer volume:
V_slice = π r_outer² h − π r_inner² h
Washer method integral (rotation about y = M):
V = ∫ from a to b [ π M² − π (M − f(x))² ] dx
where M is the maximum value of f on [a, b], r_outer = M, and r_inner = M − f(x).
General washer method (for any outer/inner radii R(x) and r(x)):
V = π ∫ from a to b [ R(x)² − r(x)² ] dx
⚠️ The most common mistake is mixing up which radius is outer and which is inner. Always measure from the axis of rotation outward. When the axis is above the curve, the outer radius goes to y = 0 (not to f(x)).
⚠️ If you forget to subtract the inner radius term, you get the volume of a solid cylinder, not the washer-shaped solid. The whole point of the washer method is the subtraction.
⚠️ When the axis of rotation is not the x-axis or y-axis, the radius expressions change. The distance from a point on the curve to the axis y = M is |M − f(x)|, not just f(x).
⚠️ Expanding (M − f(x))² before integrating can simplify computation: M² − 2M f(x) + [f(x)]². This means the integrand simplifies to 2M f(x) − [f(x)]² after cancelling the M² terms.
Q: When rotating about y = M (the max of f on [a, b]), why is the outer radius simply M and not M − f(x)?
A: The outer radius is the distance from the axis of rotation (y = M) to the farthest edge of the region, which is y = 0. That distance is M − 0 = M. The curve y = f(x) is closer to the axis, so it forms the inner boundary.
Q: What does the integrand π M² − π (M − f(x))² simplify to after expanding?
A: Expanding gives π M² − π [M² − 2Mf(x) + f(x)²] = π [2Mf(x) − f(x)²]. The M² terms cancel.
Q: How would the formula change if the region were rotated about the x-axis (y = 0) instead of y = M?
A: With the axis at y = 0, there is no hole in the middle (since the bottom boundary is the axis itself). Each cross-section is a disk, not a washer, and the volume becomes V = π ∫ from a to b [f(x)]² dx.
Q: Why do we partition on the x-axis rather than the y-axis for this problem?
A: Because the region is described with y as a function of x (y = f(x)) and the boundaries are vertical lines x = a and x = b. Slicing perpendicular to the x-axis produces clean washer cross-sections whose radii are straightforward functions of x.
Q: If f(x) = x² on [0, 2], and M = 4, what is the volume integral for rotation about y = 4?
A: V = π ∫ from 0 to 2 [4² − (4 − x²)²] dx = π ∫ from 0 to 2 [16 − (4 − x²)²] dx. Expanding the inner square: (4 − x²)² = 16 − 8x² + x⁴, so the integrand becomes π(8x² − x⁴).
washer method, disk method, shell method, volume of revolution, solid of revolution, cross-sectional area, Riemann sum for volume, axis of rotation above curve, rotating about y = M, calculus 1 volume problems, TAMU calculus, MATH 151, MATH 152, integral applications, area between curves