Set up volumes about the y axis using horizontal slices
Express boundary distances in terms of y, select the relevant branch and use y bounds with disk or washer areas.
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
- Rearrange a curve to express x in terms of y on a stated branch.
- Use disk and washer cross-section areas and evaluate definite integrals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For y = x² in the first quadrant, what is the disk radius about the y-axis at height y?
What you will learn
- Use V = π∫(R(y)² − r(y)²) dy for rotation about the y-axis.
- Choose a valid inverse branch and convert the limits to y coordinates.
Change both the radius variable and the limits
For rotation about the y-axis, horizontal slices are perpendicular to that axis. The radius is a horizontal distance |x|, expressed as a function of y. A disk volume is V = π∫x(y)² dy; a washer volume is π∫[R(y)² − r(y)²] dy. Use y coordinates at the lower and upper boundary, not old x bounds beside a dy differential. The integration variable, radius formula and limits must describe the same slices.
For the first-quadrant region bounded by y = x², x = 0 and y = 4, the relevant boundary is x = √y. Each horizontal strip runs from the y-axis to √y, so its disk area is πy. The y bounds are 0 and 4. Thus V = π∫₀⁴ y dy = 8π cubic units. Using π∫₀²x² dx would describe a different orientation and a different region.
An inverse branch locates a boundary; distance gives the radius
The equation y = x² also has a second-quadrant branch x = −√y. For that region between the branch and the y-axis, the disk radius is still √y, the positive distance from the axis. A negative x boundary does not produce a negative radius or volume. For a region away from the y-axis, identify outer and inner distances and use their squared difference; do not select the positive inverse automatically when the region is specified elsewhere.
For curves x = y and x = y² on 0 ≤ y ≤ 1, both boundaries are on the same side of the axis and y ≥ y². The outer radius is y and inner radius is y², giving washer area π(y² − y⁴). Their intersections give the y bounds 0 and 1. For more than one interval, check which boundary is farther from the axis on every part and split whenever the geometry changes.
Pause and explain
For x = y and x = y² on 0 ≤ y ≤ 1, which is the washer area about the y-axis?
Worked example
Rotate the first-quadrant region bounded by y = x², x = 0 and y = 4 about the y-axis. Find its volume.
Show the worked solution
- Choose the first-quadrant inverse branch x = √y and bounds 0 ≤ y ≤ 4.
- The disk radius is √y, so the cross-section area is π(√y)² = πy.
- V = π[y²/2]₀⁴ = 8π cubic units.
Answer Volume 8π cubic units, using y bounds 0 and 4.
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 first-quadrant region bounded by y = 2x, x = 0 and y = 6 about the y-axis. Find its volume.
Give me a hint
Rearrange to x = y/2 and use y bounds 0 and 6.
Compare my reasoning
- The disk radius is y/2, giving cross-section area πy²/4.
- V = (π/4)∫₀⁶y² dy = (π/4)[y³/3]₀⁶.
- The volume is 18π cubic units; the corresponding cone has height 6 and radius 3.
Volume 18π cubic units.
Look for these in your work
- Convert the boundary and limits into y coordinates.
- Square the radius y/2 before integrating.
Rotate the region between x = √y and x = 2 for 0 ≤ y ≤ 4 about the y-axis. Find its volume.
Give me a hint
Use outer radius 2 and inner radius √y throughout the stated interval.
Compare my reasoning
- The cross-section area is π[2² − (√y)²] = π(4 − y).
- V = π[4y − y²/2]₀⁴.
- The result is π(16 − 8) = 8π cubic units, with the hole removed.
Volume 8π cubic units.
Look for these in your work
- Identify the farther boundary as the outer radius.
- Remove the inner disk and use the stated y bounds.
Rotate the first-quadrant region between x = y and x = y² for 0 ≤ y ≤ 1 about the y-axis. Find the volume and explain why a squared strip width is wrong.
Give me a hint
Check y ≥ y², then subtract the two disk areas at each height.
Compare my reasoning
- On [0, 1], the outer radius is y and the inner radius is y².
- V = π∫₀¹(y² − y⁴) dy = π[y³/3 − y⁵/5]₀¹ = 2π/15.
- The squared width (y − y²)² would describe a disk of the gap's width, not the actual washer formed by rotation.
Volume 2π/15 cubic units; use a difference of squared radii, not a squared width.
Look for these in your work
- Check radius ordering on the interval.
- Explain the cross-section geometry as well as evaluate the integral.
Common mistakes
- Use y bounds with a dy integral and radius expressed in y.
- Choose the inverse branch belonging to the requested region.
- Use positive distances and subtract an inner disk only where the geometry has a hole.
What can you explain now?
Rotate the second-quadrant region bounded by x = −√y, x = 0, y = 0 and y = 4 about the y-axis. Find its volume.
Compare with the explanation
8π cubic units.
The distance to the axis is √y, so π∫₀⁴y dy = 8π just as for the reflected first-quadrant region.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain how rotation about the y-axis changes the strip direction, radius formula and bounds. Show how the chosen inverse branch affects location but not the sign of a distance.
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.