When a region is rotated about an axis, it sweeps out a solid, and its volume comes from adding up thin discs. About the x-axis, V = π ∫ y² dx, so the key step is to square y before integrating.
This lesson is part of SPM Additional Mathematics integration. It uses evaluating definite integrals and expands brackets, so practise those first.
Worked example: rotation about the x-axis
The region under y = x + 1, between x = 0 and x = 2, is rotated through 360° about the x-axis. Find the volume generated.
Each disc has radius y = x + 1, so its area is π(x + 1)².
V = π ∫ (x + 1)² dx from 0 to 2 = π ∫ (x² + 2x + 1) dx = π [x³/3 + x² + x] from 0 to 2.
At x = 2: 8/3 + 4 + 2 = 26/3. At x = 0 the value is 0.
V = 26π/3 cubic units, about 27.2.
As a check, this solid is a truncated cone with end radii 1 and 3 and height 2. The cone formula (πh/3)(r² + rR + R²) = (2π/3)(1 + 3 + 9) = 26π/3.
The mistake that costs marks
The common slip is to integrate y first and square afterwards, or to forget the square completely. Both keep the working short and both are wrong.
| Step | Wrong | Right |
|---|---|---|
| Expression | ∫ (x + 1) dx | ∫ (x + 1)² dx |
| Order | (∫ y dx)² | π ∫ y² dx |
| Expand | (skipped) | x² + 2x + 1 |
| Result | 16π (square after) | 26π/3 |
For the wrong order, ∫ (x + 1) dx from 0 to 2 is 4, and squaring gives 16, then 16π. The volume is not the square of an area.
Rotation about the y-axis
When the region is rotated about the y-axis, use V = π ∫ x² dy, with y-limits. Rotate the region bounded by y = x², the y-axis and y = 4 about the y-axis.
Since y = x², we have x² = y directly. The limits are y = 0 to y = 4.
V = π ∫ y dy from 0 to 4 = π [y²/2] from 0 to 4 = π(8) = 8π cubic units.
If the curve were y = √x, you would rewrite x = y² and so x² = y⁴ before integrating.
A method you can reuse
- Sketch the region and mark the axis of rotation.
- Choose V = π ∫ y² dx for the x-axis, or V = π ∫ x² dy for the y-axis.
- Write y² or x² in terms of the variable you integrate, and expand any bracket.
- Integrate, substitute the limits, then multiply by π.
- State the answer in terms of π with the units.
Check yourself
The region under y = √x, between x = 1 and x = 4, is rotated about the x-axis. Find the volume in terms of π.
Answer
y² = x, so V = π ∫ x dx from 1 to 4 = π [x²/2] from 1 to 4.
At x = 4: 8. At x = 1: 1/2. So V = π(8 − 1/2) = 15π/2 cubic units.
Check the size: the radius grows from 1 to 2, and the length is 3. A cylinder of radius 1 would have volume 3π, and one of radius 2 would have 12π. Since 7.5π lies between them, the answer is reasonable.
What to study next
Link integration to motion in connecting integration to displacement and distance. Then try the mixed integration practice set.
For a teacher to go through solids with you, see online one-to-one Additional Mathematics tuition. The word problem structure worksheet helps with setting out.