Volumes of revolution, explained with examples
Rotating a region around a line sweeps out a solid. Its volume is found by slicing it into thin discs or washers and integrating. These questions appear on AP Calculus free-response sections almost every year.
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The idea: slice and add
Imagine rotating the region under y = f(x) from x = a to x = b around the x-axis. Each thin vertical strip of width dx sweeps out a thin disc. Its radius is f(x), so its volume is approximately π[f(x)]² dx. Adding up all the discs and letting their width shrink to zero gives an integral.
The disc method
V = π ∫ab [R(x)]² dx
R(x) is the radius: the distance from the axis of rotation to the edge of the region. Use the disc method when the region touches the axis of rotation, so each slice is a solid disc.
Example 1: a disc
Find the volume when the region under y = √x from x = 0 to x = 4 is rotated about the x-axis.
- R(x) = √x, so R² = x.
- V = π ∫04 x dx = π[x²/2]04 = 8π.
Example 2: a cone
Rotate y = x/2 from x = 0 to x = 6 about the x-axis. Check the answer against the cone formula.
- V = π ∫06 x²/4 dx = π[x³/12]06 = 18π.
- The solid is a cone with radius 3 and height 6: (1/3)π(9)(6) = 18π. It matches.
The washer method
If the region does not touch the axis of rotation, each slice has a hole in the middle: a washer. Subtract the inner disc from the outer one:
V = π ∫ab ([R(x)]² − [r(x)]²) dx
R is the outer radius (furthest edge from the axis) and r is the inner radius (nearest edge). Square each radius separately; (R − r)² is a common and costly mistake.
Example 3: a washer
The region between y = x and y = x² is rotated about the x-axis. Find the volume.
- The curves meet at x = 0 and x = 1. On that interval x ≥ x², so R = x and r = x².
- V = π ∫01 (x² − x4) dx = π(1/3 − 1/5) = 2π/15.
Rotating about other lines
When the axis is a horizontal line y = k, each radius is a vertical distance from that line: radius = |curve − k|. Sketch the region and the axis, then measure from the axis to each edge.
Example 4: about y = −1
Rotate the region between y = x and y = x² about the line y = −1.
- Outer radius: from y = −1 up to y = x, so R = x + 1. Inner radius: up to y = x², so r = x² + 1.
- V = π ∫01 [(x + 1)² − (x² + 1)²] dx = π ∫01 (2x − x4 − x²) dx.
- = π(1 − 1/5 − 1/3) = 7π/15.
Rotating about the y-axis
For rotation about the y-axis (or another vertical line), slice horizontally and integrate with respect to y. Rewrite the curves as x in terms of y, and use y-limits.
Example 5: about the y-axis
The region bounded by y = x², the y-axis and y = 4 (with x ≥ 0) is rotated about the y-axis. Find the volume.
- x = √y, so the radius is √y and R² = y. Limits: y from 0 to 4.
- V = π ∫04 y dy = 8π.
Always sketch first
Before writing any integral, sketch the region, draw the axis of rotation, and draw one typical slice perpendicular to that axis. Mark where the slice starts and ends. This one picture tells you whether you need discs or washers, which variable to integrate with respect to, and what each radius is. Most errors in volume questions come from skipping the sketch, and it takes less than a minute.
Volumes with known cross-sections
Some AP questions describe a solid whose base is a region in the plane and whose cross-sections perpendicular to the x-axis are a given shape. The volume is ∫ A(x) dx, where A(x) is the area of one cross-section, found from the side length s(x) given by the region:
- squares: A = s²;
- semicircles with diameter s: A = πs²/8;
- equilateral triangles: A = (√3/4)s².
Example 6: square cross-sections
The base of a solid is the region between y = √x and the x-axis from x = 0 to x = 4. Cross-sections perpendicular to the x-axis are squares. Find the volume.
- Side s = √x, so A = x.
- V = ∫04 x dx = 8. No π, because the slices are not circles.
Setting up on the free-response section
Many AP volume questions say “write, but do not evaluate, an integral expression”. Full credit needs the constant π (for discs and washers), correct limits, and a correct integrand. Writing the outer and inner radii as labelled expressions first makes your work clear and protects partial credit. When a calculator is allowed, evaluate with it and give three decimal places.
Common mistakes
- Using (R − r)² instead of R² − r².
- Measuring radii from the x-axis when the axis of rotation is a different line.
- Integrating with respect to x when the slices should be horizontal.
- Forgetting π, or adding π for non-circular cross-sections.
Practice questions
Try each one before opening the answer.
Q1. Rotate y = x² from x = 0 to x = 2 about the x-axis.
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Q2. Rotate y = ex from x = 0 to x = 1 about the x-axis.
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Q3. Region between y = 2 and y = x² for −√2 ≤ x ≤ √2, rotated about y = 2. Write the integral.
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Q4. A solid has base y = 1 − x² above the x-axis, with semicircular cross-sections. Write the integral.
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Frequently asked questions
Is the shell method on the AP exam?
No. The current AP Calculus course and exam use the disc and washer methods and cross-sections. The shell method is not required.
Are volumes of revolution in A level Maths?
They are in A level Further Maths rather than A level Maths, mostly with the disc method about the x- or y-axis.
Key takeaways
- Disc: V = π∫ R² dx.
- Washer: V = π∫ (R² − r²) dx.
- Measure radii from the axis of rotation.
- About a vertical axis, integrate with respect to y.
- Known cross-sections: V = ∫ A(x) dx, with no π unless the shape is circular.