This whole subtopic is higher level. Nothing in it is on an SL paper.
Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 5 Calculus · 5.17 Areas against the y-axis and volumes of revolution
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Rewrite a curve as x = g(y) and find the area between it and the y-axis for c ≤ y ≤ d | HL only | "Find the area of the region enclosed by the curve, the y-axis and the line y = 4" (4 marks) |
| Deal with parts of the curve left of the y-axis, where x < 0 | HL only | Split the integral at the crossing, or use the modulus on the GDC (Paper 2) |
| Find a volume of revolution about the x-axis with V = ∫ πy² dx | HL only | "Find the exact volume when R is rotated through 2π about the x-axis" (5 to 6 marks) |
| Find a volume of revolution about the y-axis with V = ∫ πx² dy | HL only | Same question, y-axis: x² must first be written in terms of y (5 to 7 marks) |
| Find the volume from a region between two curves, as a difference of two volumes | HL only | "R lies between y = x and y = x². Find the volume …" (5 marks) |
| Model a real object as a solid of revolution and use the GDC | HL only | A vase, bowl or lathe-turned part: capacity, or the depth for a given volume (Paper 2, 4 to 6 marks) |
Before you start
You need SL areas from 5.5 and 5.11: A = ∫ |y| dx, and the rule that you write the integral before you evaluate it. You need to rearrange y = f(x) into x = g(y), which is finding an inverse function (2.2), and the integration techniques of 5.10, 5.15 and 5.16. The formula booklet gives all three formulas on this page: A = ∫ |x| dy, V = ∫ πy² dx and V = ∫ πx² dy.
1The idea in one paragraph
At SL an area is a pile of thin vertical strips, each of height y and width δx, added up by ∫ y dx. Turn the strips on their side and you have the first idea here: horizontal strips of length x and height δy, added up by ∫ x dy, give the area between a curve and the y-axis. The second idea spins the region instead of slicing it flat. Rotate a region through a full turn (2π) about the x-axis and it sweeps out a solid of revolution. Every vertical strip becomes a thin disc of radius y and thickness δx, with volume πy²δx, and adding the discs gives V = ∫ πy² dx. Spin about the y-axis instead and the discs are horizontal, of radius x and thickness δy, so V = ∫ πx² dy. In every case you are adding slices; the only decisions are which way the slices lie and what their radius is.
2Area between a curve and the y-axis
Figure 1 shows the same curve, y = x², with its region sliced two ways. On the left, the familiar region against the x-axis: vertical strips. On the right, the region between the curve and the y-axis from y = 0 to y = 4: horizontal strips. Each horizontal strip has length x (its distance from the y-axis) and height δy, so its area is about x δy.
Area against the y-axis: A = ∫ |x| dy, from y = c to y = d, with x written as a function of y and the limits taken on the y-axis.
The method has three steps: rearrange to x = g(y), find the y-limits, integrate with respect to y.
Example 1. Find the area of the region enclosed by y = x² (x ≥ 0), the y-axis and the line y = 4 (Figure 1b).
Check it another way. The rectangle from the origin to (2, 4) has area 8. The area under the curve against the x-axis is ∫₀² x² dx = 8/3. What is left over is the region against the y-axis: 8 − 8/3 = 16/3. That rectangle-subtraction method always works too, and it is a good check when the rearrangement is awkward.
Example 2. Find the exact area of the region bounded by y = ln x, the y-axis, the x-axis and the line y = 2.
Figure 2 shows the region. Measured against the x-axis it would be awkward, because the curve starts at x = 1, not at the y-axis. Measured against the y-axis it is easy.
The rectangle check needs the parts result from 5.16: the rectangle to (e², 2) has area 2e², and ∫₁e² ln x dx = [x ln x − x] from 1 to e², which is (2e² − e²) − (0 − 1) = e² + 1. So the area is 2e² − (e² + 1) = e² − 1, as before. Working against the y-axis saved a whole integration by parts.
3When the curve crosses the y-axis
The |x| in the formula matters for the same reason |y| mattered at SL. Where the curve is to the left of the y-axis, x is negative, so ∫ x dy counts that area as negative. If a region lies partly on each side, a single integral lets the two parts cancel.
Example 3. Find the total area enclosed by the curve x = y² − 1, the y-axis, and the lines y = 0 and y = 2.
Figure 3 shows the problem. The curve crosses the y-axis where y² − 1 = 0, at y = 1. Below that, x < 0.
The single integral ∫₀² (y² − 1) dy gives −2/3 + 4/3 = 2/3, which is not the area. On Paper 1 you split at the crossing, as above. On Paper 2 you can type ∫₀² |y² − 1| dy into the GDC, which returns 2 directly, but write the integral with the modulus on the page first: the M1 is for the correct expression.
4Volumes of revolution about the x-axis
Take the region under y = √x from x = 0 to x = 4 and spin it through a full turn about the x-axis. It sweeps out a solid shaped like the nose of a rocket, drawn in Figure 4. Slice the solid at right angles to the x-axis. Each slice is a thin disc: a circle of radius y (the height of the curve at that x), with thickness δx. Its volume is area × thickness = πy² δx. Adding the discs and letting δx → 0 gives the formula.
Volume about the x-axis: V = ∫ₐᵇ πy² dx, with x-limits. About the y-axis: V = ∫ πx² dy, from y = c to y = d. Both are in the formula booklet; the rotation is always through 2π.
For the solid in Figure 4:
Because y is squared, you never need to worry about the sign of y here: a region below the x-axis sweeps out a positive volume just the same.
The formula knows the cone. Rotate the line y = (r/h)x from x = 0 to x = h. You get a cone of base radius r and height h, and
which is the cone formula you already know. Rotating y = √(r² − x²) from −r to r gives (4/3)πr³, the sphere. The volume formula is not a new fact; it is the same slicing idea behind the formulas you learnt at school.
Example 4 (Paper 1). The region under y = sin x from x = 0 to x = π is rotated through 2π about the x-axis. Find the exact volume.
Squaring first often creates the integral you have to work at: sin²x needs the identity, and in Try it Q3 squaring √(ln x) creates ln x, which needs parts. Expect the volume questions on Paper 1 to be integration-technique questions in disguise.
5Volumes of revolution about the y-axis
Now spin a region about the y-axis. The discs are horizontal, each of radius x and thickness δy, so V = ∫ πx² dy. Two things change from section 4, and both are the same two changes as in section 2: write x² in terms of y, and use y-limits.
Example 5. The region between y = x², the y-axis and the line y = 4 is rotated through 2π about the y-axis to make a bowl (Figure 5). Find its volume.
That shape, a paraboloid, has a long history: the eleventh-century scholar Ibn al-Haytham worked out the volume of a paraboloid by summing slices, centuries before calculus had a name.
Example 6. The region between y = ln x, the y-axis, the x-axis and y = 1 is rotated through 2π about the y-axis. Find the exact volume.
A common slip is to rotate about the y-axis but integrate πy² dx, which gives the volume about the x-axis instead. Say which axis out loud before you write: x-axis, πy² dx; y-axis, πx² dy.
6A region that does not touch the axis: subtract two volumes
When the region lies between two curves and not against the axis, the solid has a hole down its middle. Each slice is a washer (a ring): a disc of the outer radius with a disc of the inner radius removed. Figure 6 shows the region between y = x and y = x² for 0 ≤ x ≤ 1, and one of its slices.
The volume is the volume from the outer curve minus the volume from the inner one:
Square each radius separately, then subtract. Writing π ∫ (x − x²)² dx squares the gap between the curves instead. It gives π/30, a quarter of the true answer, and it scores nothing after the first line.
7A design problem on Paper 2
The guide links this subtopic to industrial design. Designers draw the profile of a vase, a lamp base or a bottle as a curve and let the computer spin it. On Paper 2 you do the same with the GDC, because the profile is chosen for its shape, not for being easy to integrate.
Example 7 (Paper 2). The inside wall of a glass vase is modelled by rotating the curve x = 4 + 1.5 sin(y/3), 0 ≤ y ≤ 20, through 2π about the y-axis, with x and y in centimetres (Figure 7). (a) Find the capacity of the vase. (b) 500 cm³ of water is poured in. Find the depth of the water.
For (b), enter the left side as a function of h and intersect it with the line 500, or use the solver. The mark scheme gives the M1 for the equation with h as the upper limit, so that line must be on your page.
8Where marks are lost
Using x-limits against the y-axis. For ∫ x dy and ∫ πx² dy the limits are y-values. In Example 1 the limits are 0 and 4 on the y-axis, not 0 and 2.
Integrating the wrong variable. ∫ x dx is not an area against the y-axis. Rewrite x in terms of y first, and check that the integral has dy in it.
Letting parts cancel. Where the curve is left of the y-axis, ∫ x dy is negative. Split at the crossing or use the modulus.
Mixing up the axes. About the x-axis the radius is y, so V = ∫ πy² dx; about the y-axis the radius is x, so V = ∫ πx² dy.
Forgetting to square, or forgetting π. V = π ∫ y dx is not a volume; nor is ∫ y² dx.
Squaring the difference for a washer. π ∫ (f − g)² dx is wrong. Use π ∫ (f² − g²) dx, with f the outer curve.
Pressing the GDC before writing the integral. On Paper 2, a correct number with no integral written often loses the method mark. Write V = π ∫ … with its limits, then evaluate.
9Work it right
- Sketch the region and mark which axis it is against, or which axis it spins about.
- Area against the y-axis, or volume about the y-axis: rearrange to x = g(y) (or x² in terms of y) and find the y-limits.
- Write the full integral, with its limits and its π if it is a volume, before evaluating anything.
- Split any area integral where the curve crosses the axis; add the sizes of the parts.
- For a region between two curves, subtract the squares of the two radii, outer minus inner.
- Paper 1: exact answers, in terms of π and e. Paper 2: three significant figures with units (cm², cm³).
- Check with a rectangle subtraction (areas) or a known shape such as a cone (volumes) when you can.
10Try it
Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style, with a GDC.
Q1. The region R is enclosed by the curve y = x³, the y-axis and the line y = 8.
(a) Find the area of R. 3 marks
(b) Find the exact volume of the solid formed when R is rotated through 2π about the y-axis. 4 marks
Q2. Find the area of the region enclosed by the curve x = y² − 2y and the y-axis. 4 marks
Q3. The region between the curve y = √(ln x), the x-axis and the line x = e is rotated through 2π about the x-axis. Show that the volume of the solid formed is π. 5 marks
Q4. The region between the curves y = 2x and y = x² is rotated through 2π about the x-axis. Find the exact volume of the solid formed. 6 marks
Q5. A wooden candlestick is made on a lathe. Its profile is the curve y = 2 + 0.8 cos(x/2), for 0 ≤ x ≤ 15, rotated through 2π about the x-axis, with x and y in cm.
(a) Write down an integral for the volume of the candlestick. 2 marks
(b) Find the volume of wood. 1 mark
(c) A hole of radius 1 cm and depth 4 cm is drilled along the axis for the candle. Find the volume of wood that remains. 2 marks
11In one breath
Every area and volume here is a pile of thin slices added by an integral. Against the y-axis the slices are horizontal, so rewrite the curve as x = g(y), take y-limits, and use A = ∫ |x| dy, splitting where the curve crosses to the left of the axis, where x is negative. Spin a region about the x-axis and each slice is a disc of radius y, so V = ∫ πy² dx; spin it about the y-axis and each disc has radius x, so V = ∫ πx² dy, with x² in terms of y and y-limits. A region that does not touch the axis gives washers: square the outer and inner radii separately and subtract, never square the difference. Squaring often creates a harder integral, so expect the techniques of 5.15 and 5.16. On Paper 1 give exact answers in π and e; on Paper 2 write the integral first, then let the GDC evaluate it, or solve for an unknown limit.
Answers
Q1. (a) x = y1/3, so A = ∫₀⁸ y1/3 dy = [(3/4)y4/3]₀⁸ = (3/4)(16) = 12. M1 for x = y1/3 and an integral with respect to y, A1 for correct limits 0 and 8, A1 for 12. Limits 0 and 2 score M1 A0 A0. (b) x² = y2/3, so V = π ∫₀⁸ y2/3 dy = π[(3/5)y5/3]₀⁸ = π(3/5)(32) = 96π/5. M1 for π ∫ x² dy, A1 for y2/3 with limits 0 and 8, A1 for (3/5)y5/3, A1 for 96π/5.
Q2. x = y² − 2y = y(y − 2) meets the y-axis at y = 0 and y = 2, and between them x < 0. ∫₀² (y² − 2y) dy = [y³/3 − y²]₀² = 8/3 − 4 = −4/3, so the area is 4/3. A1 for the limits y = 0 and y = 2, M1 for integrating x with respect to y between their limits, A1 for −4/3, A1 for the area 4/3 stated as positive.
Q3. y = √(ln x) meets the x-axis at x = 1. V = π ∫₁e (√(ln x))² dx = π ∫₁e ln x dx. By parts, ∫ ln x dx = x ln x − x, so V = π[(e − e) − (0 − 1)] = π, as required. M1 for π ∫ y² dx, A1 for the lower limit 1, M1 for integration by parts on ln x, A1 for x ln x − x, A1 for substituting the limits to reach π, AG.
Q4. The curves meet where 2x = x², at x = 0 and x = 2, and 2x ≥ x² between them. V = π ∫₀² ((2x)² − (x²)²) dx = π ∫₀² (4x² − x⁴) dx = π[4x³/3 − x⁵/5]₀² = π(32/3 − 32/5) = 64π/15. A1 for the limits 0 and 2, M1 for the difference of two volumes, A1 for 4x² − x⁴, A1 for the integrated expression, M1 for substituting the limits, A1 for 64π/15. π ∫ (2x − x²)² dx scores the limits mark only.
Q5. (a) V = π ∫₀15 (2 + 0.8 cos(x/2))² dx. M1 for π ∫ y² dx, A1 for the correct function and limits. (b) V = 223 cm³ (3 s.f.), from the GDC. A1. (c) The hole is a cylinder of volume π × 1² × 4 = 4π ≈ 12.6 cm³, so the wood left is 223.09 − 12.57 = 211 cm³ (3 s.f.). M1 for subtracting π × 1² × 4 from their (b), A1 for 211. Subtracting the rounded 223 gives 210, which loses the A1: carry full accuracy until the end.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section AHL 5.17 Areas against the y-axis and volumes of revolution. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.