AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/8/8-10)
Unit 8 · Topic 8.10
8.10 Volume with Disc Method: Revolving Around Other Axes
When a region spins around a line other than an axis, such as y = 3 or x = −1, the disc method still works. The radius is the distance from that line to the curve, so you measure it, square it and integrate.
Key terms
- axis of rotation
- horizontal line
- vertical line
- radius as a distance
Radius as a distance
The radius of each disc is the distance from the axis of rotation to the edge of the region. For a horizontal line y = k and a curve y = f(x), that distance is the bigger y-value minus the smaller: |k − f(x)|. For a vertical line x = h and a curve x = g(y), it's |h − g(y)|. In practice you don't write absolute values; you put the larger value first.
| Axis | Curve | Radius |
|---|---|---|
| y = k, region below the line | y = f(x) | k − f(x) |
| y = k, region above the line | y = f(x) | f(x) − k |
| x = h, region to the left of the line | x = g(y) | h − g(y) |
| x = h, region to the right of the line | x = g(y) | g(y) − h |
The setup
- Horizontal axis y = k: slice vertically, V = π ∫ₐᵇ [radius(x)]² dx.
- Vertical axis x = h: slice horizontally, V = π ∫ from c to d of [radius(y)]² dy.
- Squaring makes the order of subtraction not matter in the end, but getting it right keeps you thinking clearly, especially for washers (8.12).
When it's a disc and when it isn't
This is the disc method only if the region touches the axis of rotation along its whole length. For example, the region between y = x² and y = 4 touches the line y = 4 all along its top edge, so spinning it around y = 4 gives solid discs. Spinning the same region around y = 5 leaves a gap, so you'd get washers (8.12).
Habits that prevent errors
A radius is always a distance, so it's always positive, and it's always measured perpendicular to the axis of rotation. If the axis is horizontal, the radius is a vertical segment, and its length is a difference of y-values.
Mark the axis on your sketch, draw one slice and draw the radius as a segment. Label its two ends with their coordinates, like (x, 4) and (x, x²), and subtract. This is far more reliable than trying to remember a rule. Then square the whole radius expression, not each piece.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Around the line y = 4
The region bounded by y = x² and y = 4 is revolved around the line y = 4. Find the volume.
Show the solutionHide the solution
- Step 1: The region touches y = 4 along its top, so slices are discs.
- Step 2: Radius from the line down to the curve: r = 4 − x².
- Step 3: The region runs from x = −2 to x = 2.
- Step 4: V = π ∫ from −2 to 2 of (4 − x²)² dx = π ∫ from −2 to 2 of (16 − 8x² + x⁴) dx = π(512/15).
Answer: 512π/15 ≈ 107.233 cubic units
- Example 2
Around a vertical line
The region bounded by y = √x, the x-axis and x = 4 is revolved around the line x = 4. Find the volume.
Show the solutionHide the solution
- Step 1: Vertical axis, so slice horizontally and work in y. The curve is x = y², and y runs from 0 to 2.
- Step 2: The region touches x = 4 along its right edge, so slices are discs.
- Step 3: Radius from the line x = 4 to the curve: r = 4 − y².
- Step 4: V = π ∫₀² (4 − y²)² dy = π ∫₀² (16 − 8y² + y⁴) dy = π(32 − 64/3 + 32/5) = 256π/15.
Answer: 256π/15 ≈ 53.617 cubic units
- Example 3
Trap: the radius isn't the x-value
The region bounded by y = x², the x-axis and x = 2 is revolved around the line x = 2. Find the volume.
Show the solutionHide the solution
- Step 1: Slice horizontally. The curve is x = √y, and y runs from 0 to 4.
- Step 2: Wrong radius: √y (the distance to the y-axis, not to the line x = 2).
- Step 3: Right radius: 2 − √y, from the curve over to the line.
- Step 4: V = π ∫₀⁴ (2 − √y)² dy = π ∫₀⁴ (4 − 4√y + y) dy = π(16 − 64/3 + 8) = 8π/3.
Answer: 8π/3 ≈ 8.378 cubic units
Common mistakes
- Using the curve's value, f(x), as the radius when the axis isn't the x-axis.
- Writing (4 − x)² when the radius is 4 − x²: copy the whole expression.
- Integrating in x for a vertical axis of rotation with disc slices.
- Squaring the pieces separately: (4 − x²)² is not 16 − x⁴.
On the exam
- Free-response questions often ask for volume about a line like y = −1 or y = 3 and only the setup. A correct radius expression is the key point.
- Check your radius at one easy point. At x = 0 in the first example, the distance from y = 0 to y = 4 is 4, and 4 − 0² = 4.
Connected topics
Videos
Check yourself
4 questions on 8.10 Volume with Disc Method: Revolving Around Other Axes. Pick an answer to see if you got it, and why.
The region bounded by the graph of y = √x, the line y = 3, and the y-axis is revolved about the line y = 3. What is the volume of the resulting solid?
Let R be the region bounded by the graph of y = x², the line x = 2, and the x-axis. Which of the following gives the volume of the solid generated when R is revolved about the line x = 2?
The region bounded by the graph of y = x² and the line y = 1 is revolved about the line y = 1. What is the volume of the resulting solid?
The region bounded by the line y = x, the x-axis, and the line x = 3 is revolved about the line x = 3. What is the volume of the resulting solid?
0 of 4 answered