AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/8/8-11)
Unit 8 · Topic 8.11
8.11 Volume with Washer Method: Revolving Around the x- or y-Axis
When the region doesn't touch the axis of rotation, each slice is a washer: a disc with a hole. Its area is π(R² − r²), where R is the outer radius and r is the inner radius, so you subtract the squares, not the radii.
Key terms
- washer method
- outer radius
- inner radius
- solid of revolution
Washers
Spin the region between two curves around the x-axis. A thin vertical slice now sweeps out a flat ring, like a metal washer. The outer edge comes from the curve farther from the axis, and the hole comes from the curve closer to the axis. Its area is the big disc minus the hole:
A = πR² − πr² = π(R² − r²).
So V = π ∫ₐᵇ [R(x)² − r(x)²] dx.
Finding R and r
- Outer radius R: distance from the axis to the farther boundary.
- Inner radius r: distance from the axis to the nearer boundary.
- Around the x-axis with y = f(x) on top and y = g(x) below (both ≥ 0): R = f(x), r = g(x).
- Around the y-axis with x = p(y) on the right and x = q(y) on the left (both ≥ 0): R = p(y), r = q(y). Slice horizontally and integrate in y.
Why you can't subtract first
π(R − r)² is not π(R² − r²). For R = 3 and r = 2, (R − r)² = 1 but R² − r² = 5. The washer's area depends on both the outer size and the hole, not just the ring's thickness. A thin ring far from the axis has much more area than a thin ring close to it. Always square each radius, then subtract.
When one boundary is a line
If one boundary is a horizontal line, one radius may be constant. For the region between y = √x and y = 1 for 1 ≤ x ≤ 4, spun around the x-axis, the outer radius is √x and the inner radius is the constant 1, so V = π ∫₁⁴ (x − 1) dx = 9π/2. A constant radius still gets squared.
Disc or washer?
Look at a slice: if it reaches all the way to the axis, it's a disc (inner radius 0). If there's a gap between the region and the axis, it's a washer. The disc method is just the washer method with r = 0. Region questions sometimes switch between the two within the same problem, depending on which line is the axis.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Washers around the x-axis
The region between y = x and y = x² for 0 ≤ x ≤ 1 is revolved around the x-axis. Find the volume.
Show the solutionHide the solution
- Step 1: On (0, 1), x > x², so y = x is farther from the x-axis.
- Step 2: R = x and r = x².
- Step 3: V = π ∫₀¹ (x² − x⁴) dx = π(1/3 − 1/5) = 2π/15.
Answer: 2π/15 ≈ 0.419 cubic units
- Example 2
Washers around the y-axis
The same region (between y = x and y = x², 0 ≤ x ≤ 1) is revolved around the y-axis. Find the volume.
Show the solutionHide the solution
- Step 1: Vertical axis, so slice horizontally and work in y from 0 to 1.
- Step 2: Rewrite: y = x becomes x = y; y = x² becomes x = √y.
- Step 3: For 0 < y < 1, √y > y, so x = √y is farther from the y-axis. R = √y, r = y.
- Step 4: V = π ∫₀¹ (y − y²) dy = π(1/2 − 1/3) = π/6.
Answer: π/6 ≈ 0.524 cubic units
- Example 3
Trap: (R − r)² instead of R² − r²
A student finds the volume in the first example as π ∫₀¹ (x − x²)² dx. What answer do they get, and why is it wrong?
Show the solutionHide the solution
- Step 1: π ∫₀¹ (x² − 2x³ + x⁴) dx = π(1/3 − 1/2 + 1/5) = π/30.
- Step 2: This treats each slice as a disc with radius equal to the ring's thickness, x − x². That would be the solid you'd get if the ring were pushed down to touch the axis.
- Step 3: The correct volume is π ∫₀¹ (x² − x⁴) dx = 2π/15, which is four times as big.
Answer: The student gets π/30, but the correct volume is 2π/15.
Common mistakes
- Writing π(R − r)² instead of π(R² − r²).
- Swapping R and r, which gives a negative volume.
- Using x-limits when slicing in y around the y-axis.
- Forgetting π, especially when the integral is set up as two separate pieces.
On the exam
- Write R and r explicitly before setting up the integral. It's worth the extra line.
- On calculator free-response questions, a typical part is “find the volume of the solid when R is rotated about the x-axis.” Store the intersection points and enter π ∫ (R² − r²) dx in one go.
Connected topics
Videos
Check yourself
4 questions on 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis. Pick an answer to see if you got it, and why.
The region bounded by the graphs of y = x and y = x² is revolved about the x-axis. What is the volume of the resulting solid?
The region bounded by the graphs of y = 2x and y = x² is revolved about the y-axis. What is the volume of the resulting solid?
Let R be the region bounded by the graph of y = 2cos x and the line y = 1, for −π/2 ≤ x ≤ π/2. The region R is revolved about the x-axis. What is the volume of the resulting solid?
Let R be the region in the first quadrant bounded by the graph of f(x) = 4 − x², the graph of g(x) = eˣ, and the y-axis.
Described region
The region R is revolved about the x-axis. What is the volume of the resulting solid?
0 of 4 answered