IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Volumes of Revolution

Spin a 2D region around an axis and you get a 3D solid; integrating π·(radius)² gives its volume. The radius of each thin disk at position x is y = f(x), so each disk has area πy² and thickness dx. Stack the disks via integration: V = π ∫ y² dx. For revolution about the y-axis, the roles swap: V = π ∫ x² dy. Both formulas are in the booklet. Two strategic observations: (1) square roots in the function vanish when you square them, so √-functions are easier than they look; (2) keep your answer in exact form (with π) unless decimals are explicitly requested.

📘 What you need to know

Formula booklet — x-axis revolution: V = π ∫_a^b y² dx, where y = f(x) and x ranges over [a, b].
Formula booklet — y-axis revolution: V = π ∫_a^b x² dy, where x = g(y) and y ranges over [a, b] — REARRANGE first.
The π in front is non-negotiable — every disk has area π·(radius)², so the factor of π appears outside the integral, not inside.
Squaring kills square roots: if y = √(x² + 4), then y² = x² + 4 — much easier to integrate. This is why “square root” functions appear so often in textbook problems.
The solid is 3D but the integration is 1D: you’re stacking circular disks (or annuli) along the axis of revolution. The cross-section perpendicular to the axis is a disk of radius |y| (or |x|).
For y-axis revolution, REARRANGE y = f(x) into x = g(y) — exactly the inverse-function step from the previous note on area between curve and y-axis.
Limits in the right variable: x-axis revolution uses x-limits; y-axis revolution uses y-limits. Don’t mix them up — the limits come from the FORMULA’s differential.
Exact form preserves π: 8π, 32π/3, π²/2, π(e² − 1)/2. Approximate only when the question says “to 3 s.f.” or similar.

The disk method visualized

Volumes of revolution — formula booklet V = π ∫_a^b y² dx (about x-axis) | V = π ∫_a^b x² dy (about y-axis)

Imagine slicing the solid into very thin disks perpendicular to the axis of revolution. Each disk has radius equal to the function value at that point. Disk area = π·(radius)²; disk volume = π·(radius)²·(thickness). Sum infinitely many infinitesimally-thin disks via integration.

The 2D region under y = √x from x = 0 to x = 4 (left, green) is rotated 2π about the x-axis to form a paraboloid (right, also green). A representative disk cross-section (orange) at x = 2 has radius y and thickness dx, contributing volume πy² dx. Integrating over x ∈ [0, 4]: V = π∫y² dx = π∫(√x)² dx = π∫x dx = π · 8 = 8π. Note how the square in y² eliminates the square root in y = √x.

Why squaring eliminates square roots: the formula V = π ∫ y² dx always SQUARES the function before integrating. So if y = √g(x), then y² = g(x) — the square root disappears. This is why textbook problems so often feature y = √(polynomial): the inside integrand is just a polynomial, no calculus tricks needed.

x-axis vs y-axis revolution

Rotation about x-axis

V = π ∫_a^b y² dx

Disks perpendicular to x-axis, radius |y|, thickness dx. Use the function in original form y = f(x); limits are x-values.

Rotation about y-axis

V = π ∫_a^b x² dy

Disks perpendicular to y-axis, radius |x|, thickness dy. Rearrange to x = g(y) first; limits are y-values.

🧭 Recipe — volumes of revolution

Identify the axis of revolution (x-axis or y-axis) and the bounded region. Sketch on a GDC if not given. Find the limits a, b — they must be in the variable matching the integration differential.
For y-axis revolution, rearrange y = f(x) into x = g(y). This is the same inverse-function step as area-between-curve-and-y-axis. For x-axis revolution, keep y = f(x) as given.
Square the function: compute y² (for x-axis) or x² (for y-axis). Square roots in the function disappear; trig functions may need identities like sin²x = (1 − cos 2x)/2.
Apply the formula: V = π ∫ (squared function) d(variable) with the limits from step 1. The π lives OUTSIDE the integral.
Evaluate and keep exact form: preserve π, e, ln, √ in the answer. Only convert to decimal if the question specifically asks for 3 s.f. or similar.

Worked examples

WE 1

Basic x-axis revolution — polynomial

The region bounded by y = x², the x-axis, and the line x = 2 is rotated 2π about the x-axis. Find the volume of the solid of revolution, giving your answer in exact form.

Step 1 — identify limits and axis x-axis revolution, x from 0 to 2 (region starts at origin) Step 2 — already in y = f(x) form y = x² Step 3 — square the function y² = (x²)² = x⁴ Step 4 — apply V = π ∫ y² dx V = π ∫ from 0 to 2 of x⁴ dx = π [x⁵/5] from 0 to 2 = π (32/5 – 0) Step 5 — exact form = 32π/5 V = 32π5 cubic units the 3D solid is a “horn” shape: a parabola rotated about its axis, narrowing to a point at the origin and flaring out to a circular cap of radius 4 at x = 2.

WE 2

y-axis revolution — same parabola, different axis

The region bounded by y = x², the y-axis, and the line y = 4 is rotated 2π about the y-axis. Find the volume.

Step 1 — identify limits and axis y-axis revolution, y from 0 to 4 Step 2 — rearrange y = x² → x = sqrt(y) y = x² ⟹ x = √y (positive root, since x ≥ 0 in the bounded region) Step 3 — square the function x² = (√y)² = y ← square root disappears! Step 4 — apply V = π ∫ x² dy V = π ∫ from 0 to 4 of y dy = π [y²/2] from 0 to 4 = π (8 – 0) Step 5 V = 8π V = 8π cubic units classic example of “square eliminates square root”. The 3D solid here is a paraboloid opening UP, vertex at origin, cap circle at y = 4 with radius 2. Same volume formula but different orientation than WE 1 (which rotated about x-axis instead).

WE 3

Exponential — x-axis revolution with linear-arg exp

The region bounded by y = e^x, the x-axis, and the lines x = 0 and x = 1 is rotated 2π about the x-axis. Find the volume.

Step 1 — limits and axis x-axis revolution, x from 0 to 1 Step 2 — y = e^x already in form Step 3 — square the function (linear-arg exp appears) y² = (e^x)² = e^(2x) Step 4 — apply formula V = π ∫ from 0 to 1 of e^(2x) dx = π · [e^(2x)/2] from 0 to 1 (linear-arg integration) = (π/2)(e² – e⁰) = (π/2)(e² – 1) V = π(e² − 1)2 ≈ 10.04 cubic units squaring e^x gives e^(2x), which is linear-arg exp integration ∫e^(ax) dx = e^(ax)/a + c. The 1/2 outside is from a = 2 in the antiderivative. Always quote the exact form first; the decimal is only for verification.

WE 4

Trigonometric — squaring needs an identity

The region bounded by y = sin x, the x-axis, and the lines x = 0 and x = π is rotated 2π about the x-axis. Find the volume in exact form.

Step 1 — limits and axis x-axis revolution, x from 0 to π Step 2 — y = sin x as given Step 3 — square: need the double-angle identity to integrate sin² y² = sin²(x) = (1 – cos(2x))/2 (Pythagorean/double-angle identity) Step 4 — apply formula and integrate term-by-term V = π ∫ from 0 to π of (1 – cos(2x))/2 dx = (π/2) ∫ from 0 to π of (1 – cos(2x)) dx = (π/2) [x – sin(2x)/2] from 0 to π Step 5 — evaluate At x = π: π – sin(2π)/2 = π – 0 = π At x = 0: 0 – 0 = 0 V = (π/2)(π – 0) = π²/2 V = π²2 cubic units sin²x and cos²x always need the half-angle identity to integrate: sin²x = (1-cos 2x)/2, cos²x = (1+cos 2x)/2. The answer π²/2 has TWO factors of π — one from the V = π∫ outside, one from the bounded interval of length π. The solid is a “lemon” or “American football” shape.

WE 5

Square-root function — squaring eliminates the radical

The region bounded by y = √(x² + 4), the x-axis, and the lines x = 0 and x = 2 is rotated 2π about the x-axis. Find the volume.

Step 1 — limits and axis x-axis revolution, x from 0 to 2 Step 2 — y as given Step 3 — square the function y² = (√(x² + 4))² = x² + 4 ← square root vanishes Step 4 — apply formula (now a simple polynomial integration) V = π ∫ from 0 to 2 of (x² + 4) dx = π [x³/3 + 4x] from 0 to 2 = π (8/3 + 8 – 0) = π (8/3 + 24/3) = 32π/3 V = 32π3 cubic units if the problem had asked for the original integral ∫√(x²+4) dx, you’d need trig substitution (outside AA HL syllabus). But the SQUARE makes it a 2-term polynomial — fully integrable. This is why √ functions are the textbook favourite for volumes of revolution.

WE 6

Inverse trig — y-axis revolution with rearrangement

The region bounded by y = arctan x, the y-axis, and the line y = π/4 is rotated 2π about the y-axis. Find the volume in exact form.

Step 1 — limits and axis y-axis revolution, y from 0 to π/4 Step 2 — rearrange (inverse of arctan) y = arctan x ⟹ x = tan y Step 3 — square the function x² = tan²(y) Need identity to integrate: tan²(y) = sec²(y) – 1 Step 4 — apply V = π ∫ x² dy V = π ∫ from 0 to π/4 of tan²(y) dy = π ∫ from 0 to π/4 of (sec²(y) – 1) dy = π [tan(y) – y] from 0 to π/4 Step 5 — evaluate At y = π/4: tan(π/4) – π/4 = 1 – π/4 At y = 0: tan(0) – 0 = 0 V = π · (1 – π/4) = π – π²/4 V = π − π²4 ≈ 0.674 cubic units three techniques combined: y-axis rearrangement (x = tan y), trig identity (tan² = sec² – 1), and reciprocal trig integration (∫sec² = tan). When you see arctan rotated about y-axis, expect this exact chain. The π² in the answer is the “double-π” signature of volumes involving inverse trig functions.

💡 Top tips

The axis of revolution determines the formula: x-axis → V = π∫y²dx with x-limits; y-axis → V = π∫x²dy with y-limits. Match the differential to the limits.
Square roots in the integrand vanish when squared. ∫(√x)² dx = ∫x dx — trivial. This is why textbook problems love √-functions.
For y-axis revolution, REARRANGE FIRST: solve y = f(x) for x = g(y). Same inverse-function step as area-between-curve-and-y-axis.
Trig functions need identities when squared: sin² = (1 − cos 2θ)/2, cos² = (1 + cos 2θ)/2, tan² = sec² − 1, cot² = cosec² − 1.
Keep π in exact form: 8π, 32π/3, π²/2 — these ARE the answers. Convert to decimal (10.04, 4.93, etc.) only when explicitly asked for “3 s.f.” or similar.

⚠ Common mistakes

Forgetting the π: writing V = ∫y²dx (no π) gives a number that’s the integral, not the volume. The π factor is essential — each disk has area π·(radius)².
Squaring incorrectly: (sin x)² ≠ sin(x²); (e^x)² = e^2x not e^x². Apply the SQUARE to the whole function value.
Using x-axis formula for y-axis revolution (or vice versa) — the integration differential MUST match the axis. Always state which axis BEFORE writing the formula.
Forgetting to rearrange for y-axis revolution: integrating y² dy is wrong — you need x² dy with x expressed as g(y).
Skipping the trig identity on ∫sin²x dx: you can’t integrate sin²x directly — convert to (1 − cos 2x)/2 first. Same for cos²x, tan²x, cot²x.

Up next: Modelling with Volumes of Revolution. Real-world objects — vases, bowls, lampshades, sport trophies — can be modelled as solids of revolution generated by curves. You’ll set up the integral from a contextual description, often involving adding or subtracting volumes (e.g., a hollow bowl = outer solid − inner solid). Unit conversions (cm³ ↔ litres ↔ ml) appear frequently. Same formulas as this note; the new skill is translating “a vase is shaped like…” into y = f(x) with appropriate limits.

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