AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/8/8-9)
Unit 8 · Topic 8.9
8.9 Volume with Disc Method: Revolving Around the x- or y-Axis
Spinning a region around the x- or y-axis makes a solid of revolution. When the region touches the axis, every slice is a solid disc with area πr², so the volume is π times the integral of the radius squared.
Key terms
- solid of revolution
- disc method
- radius
- axis of rotation
Discs
Take the region under y = f(x) from x = a to x = b and spin it around the x-axis. Each thin vertical strip sweeps out a flat coin, called a disc. Its radius is the distance from the axis to the curve, r = f(x), and its thickness is dx. Its volume is about πr²·dx. So
V = π ∫ₐᵇ [f(x)]² dx (around the x-axis).
For a region bounded by x = g(y) and the y-axis, spun around the y-axis, use horizontal slices: V = π ∫ from c to d of [g(y)]² dy.
Matching the slice to the axis
For the disc method, slices are always perpendicular to the axis of rotation. Around a horizontal axis (like the x-axis), slice vertically and integrate in x. Around a vertical axis (like the y-axis), slice horizontally and integrate in y, so you'll need the curve written as x in terms of y.
| Axis of rotation | Slice | Radius | Integral |
|---|---|---|---|
| x-axis | vertical (dx) | r = f(x) | π ∫ₐᵇ [f(x)]² dx |
| y-axis | horizontal (dy) | r = g(y) | π ∫ from c to d of [g(y)]² dy |
When to use discs vs. washers
Use discs when the region touches the axis of rotation along the whole interval, so each slice is solid with no hole. If there's a gap between the region and the axis, slices have holes, and you need the washer method (8.11).
Note: the shell method (another way to find volumes) isn't part of the AP Calculus AB or BC exam. Every revolution problem can be done with discs or washers.
Exact answers and units
Without a calculator, leave answers in terms of π, like 8π. With one, give a decimal to three places. Volume is in cubic units: if x and y are in centimeters, the volume is in cubic centimeters. A quick reasonableness check is to compare with a cylinder that encloses the solid; your volume must be smaller.
Steps
- Sketch the region and the axis.
- Draw a slice perpendicular to the axis, and its radius from the axis to the edge of the region.
- Write r in terms of the integration variable.
- Set up π ∫ r² and use the region's limits along the axis.
- Evaluate (or leave as an integral if the question says so).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Around the x-axis
The region under y = √x from x = 0 to x = 4 is revolved around the x-axis. Find the volume.
Show the solutionHide the solution
- Step 1: Slices perpendicular to the x-axis, radius r = √x.
- Step 2: V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π(16/2) = 8π.
Answer: 8π ≈ 25.133 cubic units
- Example 2
Around the y-axis
The region bounded by y = x², the y-axis and y = 4 (in the first quadrant) is revolved around the y-axis. Find the volume.
Show the solutionHide the solution
- Step 1: Rotating about a vertical axis, so slice horizontally and integrate in y.
- Step 2: Radius = distance from the y-axis to the curve = x = √y.
- Step 3: y runs from 0 to 4.
- Step 4: V = π ∫₀⁴ (√y)² dy = π ∫₀⁴ y dy = 8π.
Answer: 8π ≈ 25.133 cubic units
- Example 3
Trap: square the radius
The region under y = 1/x from x = 1 to x = 3 is revolved around the x-axis. Find the volume.
Show the solutionHide the solution
- Step 1: Radius r = 1/x, so r² = 1/x².
- Step 2: V = π ∫₁³ x⁻² dx = π[−1/x]₁³ = π(−1/3 + 1) = 2π/3.
- Step 3: Common wrong setup: π ∫₁³ (1/x) dx = π ln 3, which forgets to square the radius.
Answer: 2π/3 ≈ 2.094 cubic units
Common mistakes
- Forgetting to square the radius, or squaring it outside the integral.
- Forgetting the π.
- Integrating with respect to x when rotating around the y-axis with disc slices.
- Using the disc method when the region doesn't touch the axis.
On the exam
- Volume-of-revolution questions often ask you to “write, but do not evaluate, an integral expression.” Then the setup, including π, the squared radius and the limits, is the answer.
- Sketching the radius as a line segment from the axis to the curve is the quickest way to get it right.
Connected topics
Videos
Check yourself
4 questions on 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis. Pick an answer to see if you got it, and why.
The region bounded by the graph of y = 1/x, the x-axis, and the lines x = 1 and x = 3 is revolved about the x-axis. What is the volume of the resulting solid?
The region bounded by the graph of y = x³, the line y = 8, and the y-axis is revolved about the y-axis. What is the volume of the resulting solid?
The region bounded by the graph of y = sec x, the x-axis, the y-axis, and the line x = π/4 is revolved about the x-axis. What is the volume of the resulting solid?
Let R be the region bounded by the graph of y = ln x, the y-axis, the x-axis, and the line y = 2. The region R is revolved about the y-axis. What is the volume of the resulting solid?
0 of 4 answered