Find volumes about the x axis with disks and washers
Use squared distances from the rotation axis, and subtract an inner disk when the region creates a hole.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Section 1.8 includes rational and linear-composite powers except exponent −1, simple improper integrals, areas and volumes about either coordinate axis, including regions away from the rotation axis. Written practice is self-checked.
Before you begin
- Find curve intersections and integrate polynomials.
- Use the area of a circle and identify distances from a coordinate axis.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A washer has outer radius 9 and inner radius 5. What is its area?
What you will learn
- Integrate circular cross-section areas for volumes about the x-axis.
- Use π(R² − r²) for a region away from the rotation axis and choose radii as distances.
Rotating a strip produces a circular cross-section
A region from the x-axis to a curve over a finite x interval produces disks when rotated about the x-axis. The radius is the distance |y| from the axis, so each disk has area πy² and V = π∫y² dx. Square the radius expression before integrating. A curve below the axis still produces a positive volume because its radius is a distance, not a signed coordinate.
If a strip occupies distances from r(x) to R(x), with 0 ≤ r ≤ R, rotation produces a washer: the outer disk with the inner disk removed. Its area is π[R(x)² − r(x)²], not π[R(x) − r(x)]². Locate the actual region and check whether it reaches the axis. If it crosses the axis, rotation fills a disk; subtracting two signed boundary squares blindly can remove material that is present.
A region away from the axis needs its inner radius
The region between y = 9 − x² and y = 5 has intersections at x = −2 and 2. Throughout this interval both boundaries are above the axis, with outer radius R = 9 − x² and inner radius r = 5. Hence V = π∫₋₂²[(9 − x²)² − 25] dx = π∫₋₂²(56 − 18x² + x⁴) dx. The inner cylinder is absent from the solid and must be subtracted.
Use an x differential for slices perpendicular to the x-axis. A cross-sectional area has square length units; multiplying by an axial length gives cubic units. Where boundary distances change order, split the geometry before integrating. Draw a representative slice and identify both radii; the upper y coordinate is not always the farther boundary when a region lies below the axis.
Pause and explain
For the region between y = 9 − x² and y = 5 on −2 ≤ x ≤ 2, which cross-section area is correct about the x-axis?
Worked example
Find the volume formed by rotating the region between y = 9 − x² and y = 5 about the x-axis.
Show the worked solution
- Solve 9 − x² = 5 to get x = −2 and 2; use outer radius 9 − x² and inner radius 5.
- V = π∫₋₂²(56 − 18x² + x⁴) dx, with antiderivative 56x − 6x³ + x⁵/5.
- The endpoint values are 352/5 and −352/5, giving V = 704π/5 cubic units.
Answer Volume 704π/5 cubic units, using washers with inner radius 5.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Rotate the region between y = x and the x-axis for 0 ≤ x ≤ 2 about the x-axis. Find its volume.
Give me a hint
Each perpendicular slice is a disk of radius x.
Compare my reasoning
- The disk area at x is πx².
- V = π∫₀²x² dx = π[x³/3]₀².
- The volume is 8π/3 cubic units, consistent with a cone of radius 2 and height 2.
Volume 8π/3 cubic units.
Look for these in your work
- Use squared radius in the integrand.
- Include π and state cubic units.
Rotate the region between y = 4 and y = x² for −2 ≤ x ≤ 2 about the x-axis. Find its volume.
Give me a hint
The outer radius is 4 and the inner radius is x² on the stated interval.
Compare my reasoning
- The washer area is π[4² − (x²)²] = π(16 − x⁴).
- V = π[16x − x⁵/5]₋₂².
- The endpoint difference is 64 − 64/5 = 256/5, giving 256π/5 cubic units.
Volume 256π/5 cubic units.
Look for these in your work
- Square the inner radius x² to obtain x⁴.
- Use both signed bounds in endpoint subtraction.
A supplied hollow-solid model has axial coordinate 0 ≤ x ≤ 2 cm, outer radius R(x) = 2 − x/2 cm and inner radius 1 cm. Find the material volume.
Give me a hint
Confirm R ≥ 1 on the model interval, then subtract the fixed inner disk from each outer disk.
Compare my reasoning
- For 0 ≤ x ≤ 2, R decreases from 2 to 1 and stays at least the inner radius.
- V = π∫₀²[(2 − x/2)² − 1] dx = π∫₀²(3 − 2x + x²/4) dx.
- Evaluation gives π[3x − x² + x³/12]₀² = π(6 − 4 + 2/3) = 8π/3 cm³.
Material volume 8π/3 cm³.
Look for these in your work
- Check the outer radius is at least the inner radius on the entire interval.
- Subtract the hole and use cubic length units.
Common mistakes
- Square radii before integrating the cross-section area.
- Subtract squared radii, not the square of their difference.
- Use distance from the axis and check whether the rotated strip actually has a hole.
What can you explain now?
Rotate the region between y = −x and the x-axis for 0 ≤ x ≤ 2 about the x-axis. Find its volume.
Compare with the explanation
8π/3 cubic units.
The radius is |−x| = x; V = π∫₀²x² dx = 8π/3, with no negative volume.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Sketch a representative disk or washer, identify the radii as distances and explain why the cross-section area uses a difference of squares.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.