AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/8/8-12)
Unit 8 · Topic 8.12
8.12 Volume with Washer Method: Revolving Around Other Axes
This is the washer method around a line other than an axis. Both radii are distances from the new line, so for an axis below the region, like y = −2, a curve y = f(x) is f(x) + 2 away.
Key terms
- outer radius
- inner radius
- axis of rotation
- distance from a line
Both radii from the new line
Everything from 8.11 still holds: V = π ∫ (R² − r²). The only change is how you measure the radii. Each one is the distance from the axis of rotation to a boundary curve, so each one is (farther value) − (nearer value) relative to the axis.
| Axis | Region position | Outer radius R | Inner radius r |
|---|---|---|---|
| y = k below the region (top f, bottom g) | above the axis | f(x) − k | g(x) − k |
| y = k above the region (top f, bottom g) | below the axis | k − g(x) | k − f(x) |
| x = h left of the region (right p, left q) | right of the axis | p(y) − h | q(y) − h |
| x = h right of the region (right p, left q) | left of the axis | h − q(y) | h − p(y) |
Notice the switch
When the axis is above the region, the bottom curve is farther from the axis, so it gives the outer radius. When the axis is to the right, the left curve gives the outer radius. Don't memorize “top is outer.” Draw the axis, draw a slice, and see which boundary is farther away.
If the axis is y = −2, then k = −2 and the radius to a curve y = f(x) is f(x) − (−2) = f(x) + 2. Subtracting a negative trips people up, so write the subtraction out.
Steps
- Sketch the region and the axis line.
- Draw a slice perpendicular to the axis.
- Draw R (axis to far boundary) and r (axis to near boundary). Label endpoints with coordinates.
- Write R and r as differences of coordinates.
- Set up π ∫ (R² − r²) with the right variable and limits.
Checking your radii
If the axis passes through the region itself, the solid isn't a simple washer stack, and you won't be asked to handle that case. Expect the axis to be outside the region or along its edge.
Pick one value of x (or y) in the interval and compute R and r as numbers. Both must be positive, and R must be bigger than r. If R comes out smaller, you've swapped them. If either is negative, the subtraction is in the wrong order.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Axis below the region
The region between y = x and y = x² (0 ≤ x ≤ 1) is revolved around the line y = −1. Find the volume.
Show the solutionHide the solution
- Step 1: The axis is below the region, so the top curve y = x is farther away.
- Step 2: R = x − (−1) = x + 1 and r = x² − (−1) = x² + 1.
- Step 3: Check at x = 0.5: R = 1.5, r = 1.25. R > r > 0. ✓
- Step 4: V = π ∫₀¹ [(x + 1)² − (x² + 1)²] dx = π ∫₀¹ (−x⁴ − x² + 2x) dx = π(−1/5 − 1/3 + 1) = 7π/15.
Answer: 7π/15 ≈ 1.466 cubic units
- Example 2
Trap: axis above the region
The same region is revolved around the line y = 2. Find the volume.
Show the solutionHide the solution
- Step 1: Now the axis is above the region, so the bottom curve y = x² is farther away. It gives the outer radius.
- Step 2: R = 2 − x² and r = 2 − x.
- Step 3: Check at x = 0.5: R = 1.75, r = 1.5. ✓
- Step 4: V = π ∫₀¹ [(2 − x²)² − (2 − x)²] dx = π ∫₀¹ (x⁴ − 5x² + 4x) dx = π(1/5 − 5/3 + 2) = 8π/15.
- Step 5: If you kept “top curve is outer,” you'd get R = 2 − x and r = 2 − x², and a negative volume.
Answer: 8π/15 ≈ 1.676 cubic units
- Example 3
Vertical axis to the right
The same region is revolved around the line x = 3. Find the volume.
Show the solutionHide the solution
- Step 1: Vertical axis: slice horizontally, 0 ≤ y ≤ 1. The boundaries are x = y (left) and x = √y (right).
- Step 2: The axis x = 3 is to the right, so the left boundary x = y is farther away: R = 3 − y, r = 3 − √y.
- Step 3: V = π ∫₀¹ [(3 − y)² − (3 − √y)²] dy = π ∫₀¹ (6√y − 7y + y²) dy = π(4 − 7/2 + 1/3) = 5π/6.
Answer: 5π/6 ≈ 2.618 cubic units
Common mistakes
- Using f(x) − 2 instead of f(x) + 2 when the axis is y = −2.
- Always taking the top curve as the outer radius, even when the axis is above the region.
- Mixing x and y: for a vertical axis, both radii and the limits must be in y.
- Expanding (R² − r²) incorrectly. Expand each square fully, then subtract.
On the exam
- Rotating about lines like y = −1 or y = 4 is a common final part of a region free-response question, usually “write, but do not evaluate.” The two radii are where the points are.
- Write R and r on their own lines first, then the integral. Scorers can give credit for correct radii even if the final integral has a slip.
Connected topics
Videos
Check yourself
4 questions on 8.12 Volume with Washer Method: Revolving Around Other Axes. Pick an answer to see if you got it, and why.
Let R be the region bounded by the graphs of y = x² and y = 2x. Which of the following gives the volume of the solid generated when R is revolved about the line y = −1?
Let R be the region bounded by the graph of y = √x, the x-axis, and the line x = 4. What is the volume of the solid generated when R is revolved about the line x = −1?
Let R be the region bounded by the graph of y = √x, the x-axis, and the line x = 4. The region R is revolved about the line y = −1. 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 horizontal line y = 5. What is the volume of the resulting solid?
0 of 4 answered