IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~12 min read

Modelling with Volumes of Revolution

Real-world objects with axial symmetry — vases, wine glasses, bowls, perfume bottles, lampshades, water tanks — can be modelled as solids of revolution. The volume integrals are the same as the previous note, but now you’ll handle additional layers: unit conversions (cm³ ↔ mL ↔ L ↔ m³), hollow objects (subtract inner from outer), multi-part shapes (sum the integrals piece by piece), and “working backwards” problems (given a target volume, find a parameter). The mathematics doesn’t change; the modelling skill is reading the context and assembling the right integral.

📘 What you need to know

Same formula as before: V = π ∫ y² dx (x-axis) or V = π ∫ x² dy (y-axis). The new layer is using it in a real-world context with units and modelling decisions.
Unit cheat-sheet (CRITICAL): 1 cm³ = 1 mL; 1 L = 1000 mL = 1000 cm³; 1 m³ = 1000 L = 10⁶ cm³. Most volume-of-revolution problems mix these — keep them straight.
Hollow objects use SUBTRACTION: V_material = V_outer − V_inner, both computed as separate volumes of revolution with the same axis and limits. This is the “annulus method”.
Multi-part shapes use ADDITION: split into pieces, integrate each separately, sum them. Common for objects like perfume bottles (curved body + cylindrical neck).
“Region between two curves” rotated uses V = π ∫ (outer² − inner²) dx. Outer and inner refer to distance from the axis of revolution, not “upper” vs “lower” in the usual sense.
Working backwards problems: given a target volume, find a parameter (height, radius, scaling factor). Set up the integral with the parameter symbolic, solve the equation V(parameter) = target.
Standard modelling assumption: axial symmetry. The real object is treated as if generated exactly by rotating a 2D curve about an axis. Real vases have wobbles, finger marks, etc. — the model ignores these.
Sketch the cross-section first — identify what’s being rotated, what’s hollow, where pieces join. Skipping the sketch is the #1 source of setup errors.

Multi-part objects: addition

Multi-part shape — sum of partial volumes V_total = V₁ + V₂ + … = π ∫_{section 1} y₁² dx + π ∫_{section 2} y₂² dx + …

Many real objects have distinct sections — a perfume bottle has a curved body and a cylindrical neck; a wine glass has a paraboloid bowl and a thin cylindrical stem. Treat each section separately, then sum.

Cross-section of a perfume bottle (in cm), shown in 2D before rotation. The body (green) follows y = √x from x = 0 to x = 9, joining smoothly to the cylindrical neck (orange) with constant y = 3 from x = 9 to x = 12. Both pieces are rotated about the x-axis to form the 3D bottle. Adding the two partial volumes gives 81π/2 + 27π = 135π/2 cm³ ≈ 212 mL of perfume capacity.

The “smooth join” matters: at x = 9, the body curve gives y = √9 = 3, which matches the constant neck value y = 3. If the curves DIDN’T match at the boundary, the model would be physically broken — a sudden change in radius would mean a ridge or gap. When constructing your own model, always check continuity at section boundaries.

Two modelling templates — add vs subtract

Multi-part: add partial volumes

V_total = V_{part 1} + V_{part 2} + …

For shapes with distinct sections joining at boundaries (e.g., bottle body + neck, wine glass bowl + stem). Each section uses its own function over its own x-interval.

Hollow: subtract inner from outer

V_material = V_outer − V_inner

For hollow objects with wall thickness (e.g., the glass of a vase, the wall of a trough). Outer curve generates the OUTSIDE surface; inner curve generates the cavity. Same axis, same limits.

🧭 Recipe — modelling problems

Read the context and sketch the cross-section. Note units in the problem (cm, m, mm). Identify the axis of revolution and which features generate the solid (outer surface, inner cavity, multiple sections).
Classify the problem: single piece, multi-part (addition), hollow (subtraction), or region between curves (annulus formula π∫(outer² − inner²) dx).
For y-axis revolution, rearrange each function: y = f(x) → x = g(y). For x-axis revolution, keep as given.
Set up and evaluate the integral: square the function(s), apply limits, compute. For multi-part, sum; for hollow, subtract.
Convert units if requested: 1 cm³ = 1 mL; 1 L = 1000 cm³ = 1000 mL; 1 m³ = 1000 L. State the answer with correct units and appropriate significant figures (usually 3 s.f. for decimals).

Worked examples

WE 1

Drinking glass — basic with mL conversion

The interior of a drinking glass is modelled by rotating the curve y = √x (where x, y are in cm) about the x-axis from x = 0 to x = 9. Find the glass’s capacity in millilitres.

Step 1 — identify axis and limits x-axis revolution, x from 0 to 9 cm Step 2 — single piece, use V = π ∫ y² dx y² = (√x)² = x ← square eliminates the root Step 3 — set up and evaluate V = π ∫ from 0 to 9 of x dx = π [x²/2] from 0 to 9 = π (81/2 – 0) = 81π/2 cm³ Step 4 — convert cm³ → mL (1 cm³ = 1 mL) V = 81π/2 mL ≈ 127 mL (3 s.f.) Capacity = 81π2 mL ≈ 127 mL capacity ≈ 127 mL is realistic for a small drinking glass. The conversion cm³ → mL is 1-to-1 — this is the most useful unit identity in modelling problems. For LITRES, divide by 1000: 127 mL = 0.127 L.

WE 2

Hollow trough — V_outer minus V_inner

A water trough has a cross-section with outer radius y = √(x + 1) and inner radius y = √x, both rotated about the x-axis from x = 0 to x = 8 (measurements in metres). Find the volume of material needed to make the trough.

Step 1 — hollow object → annulus method V_material = V_outer – V_inner Step 2 — compute outer volume y_outer² = x + 1 V_outer = π ∫ from 0 to 8 of (x + 1) dx = π [x²/2 + x] from 0 to 8 = π (32 + 8) = 40π m³ Step 3 — compute inner volume y_inner² = x V_inner = π ∫ from 0 to 8 of x dx = π [x²/2] from 0 to 8 = 32π m³ Step 4 — subtract V_material = 40π – 32π = 8π m³ V_material = 8π m³ ≈ 25.1 m³ the annulus method: square BOTH radii, subtract, then integrate (or compute each volume separately and subtract — same answer). Common trap: subtracting before squaring, i.e., π∫(y_outer – y_inner)² dx, which gives a different (incorrect) answer.

WE 3

Multi-part perfume bottle — addition

A perfume bottle is modelled with two sections (all measurements in cm):
Body: y = √x, for x ∈ [0, 9]
Neck: y = 3, for x ∈ [9, 12]
The bottle is rotated about the x-axis. Find its capacity in mL.

Step 1 — multi-part shape → addition V_total = V_body + V_neck Step 2 — body volume (curved section) y_body² = (√x)² = x V_body = π ∫ from 0 to 9 of x dx = π [x²/2] from 0 to 9 = 81π/2 cm³ Step 3 — neck volume (cylinder, constant y = 3) y_neck² = 9 V_neck = π ∫ from 9 to 12 of 9 dx = 9π [x] from 9 to 12 = 9π (12 – 9) = 27π cm³ Step 4 — total V_total = 81π/2 + 27π = 81π/2 + 54π/2 = 135π/2 cm³ Convert to mL: 135π/2 mL ≈ 212 mL Capacity = 135π2 mL ≈ 212 mL the neck section is constant y = 3, giving y² = 9. The integral ∫9 dx is just 9 times the interval width — the cylinder volume formula πr²h = π(3)²(3) = 27π emerges naturally. Check continuity: at x = 9, body gives y = √9 = 3 and neck gives y = 3 — matches.

WE 4

Working backwards — find r given target volume

A hemispherical lampshade is modelled by rotating the upper semicircle y = √(r² − x²) about the x-axis from x = −r to x = r. The lampshade has volume 36π cm³. Find r.

Step 1 — set up V as a function of the unknown r y² = (√(r² – x²))² = r² – x² V(r) = π ∫ from -r to r of (r² – x²) dx Step 2 — integrate (r is constant w.r.t. x) V(r) = π [r²x – x³/3] from -r to r Step 3 — evaluate at limits At x = r: r²·r – r³/3 = r³ – r³/3 = 2r³/3 At x = -r: -r³ + r³/3 = -2r³/3 Difference: 2r³/3 – (-2r³/3) = 4r³/3 V(r) = π · 4r³/3 = (4πr³)/3 Step 4 — set V = 36π and solve (4πr³)/3 = 36π r³ = 36π · 3/(4π) = 27 r = ∛27 = 3 r = 3 cm notice that V(r) = (4πr³)/3 is exactly the formula for the volume of a SPHERE. This confirms the geometric fact: rotating a semicircle of radius r about its diameter generates a sphere of radius r. The “working backwards” technique — solve V(parameter) = target — is a common IB exam pattern.

WE 5

Region between two curves — annulus formula

An ornament is modelled by rotating the region between the curves y = √x and y = x² (where x, y are in cm) about the x-axis. The curves intersect at x = 0 and x = 1. Find the volume of the ornament.

Step 1 — identify outer and inner curves For x ∈ (0, 1): √x > x² (e.g., at x=0.5: √0.5 ≈ 0.71 vs 0.25) ⟹ outer is y = √x, inner is y = x² Step 2 — apply annulus formula V = π ∫ (outer² – inner²) dx outer² = (√x)² = x inner² = (x²)² = x⁴ V = π ∫ from 0 to 1 of (x – x⁴) dx Step 3 — integrate V = π [x²/2 – x⁵/5] from 0 to 1 = π (1/2 – 1/5 – 0) = π (5/10 – 2/10) = 3π/10 V = 3π10 cm³ ≈ 0.94 cm³ small ornament — final volume under 1 cm³. The annulus formula π∫(outer² – inner²) dx is the same as V_outer – V_inner; the formula just combines them into one integral for efficiency. Choose outer/inner based on DISTANCE FROM AXIS, not on “which curve is higher” in some other sense.

WE 6

Applied — water tank capacity in litres

A water tank is modelled by rotating y = x² + 1 (with x, y in metres) about the y-axis. The tank’s lowest point is at y = 1 m (the narrow base) and it is filled to y = 5 m. Find the tank’s water capacity in litres. (1 m³ = 1000 L)

Step 1 — y-axis revolution → rearrange to x = g(y) y = x² + 1 ⟹ x² = y – 1 Step 2 — set up V = π ∫ x² dy V = π ∫ from 1 to 5 of (y – 1) dy Step 3 — integrate V = π [y²/2 – y] from 1 to 5 At y = 5: 25/2 – 5 = 15/2 At y = 1: 1/2 – 1 = -1/2 V = π [15/2 – (-1/2)] = π · 16/2 = 8π m³ Step 4 — convert m³ → litres (×1000) V = 8π m³ × 1000 L/m³ = 8000π L ≈ 25 133 L (5 s.f.) Capacity = 8000π L ≈ 25 100 L (3 s.f.) the y-axis revolution requires the rearrangement x² = y – 1 — analogous to the inverse-function step from area-between-curve-and-y-axis. Unit conversion m³ → L is multiplication by 1000 (NOT by 1 000 000 which would be m³ → mL). At 25 000+ litres, this is realistic for a residential water storage tank.

💡 Top tips

Sketch the cross-section before integrating. Identify outer/inner curves, multi-part boundaries, axis of revolution, and limits. Skipping the sketch causes most setup errors.
Unit cheat-sheet: 1 cm³ = 1 mL exactly; 1 L = 1000 mL = 1000 cm³; 1 m³ = 1000 L = 10⁶ cm³ = 10⁶ mL. Always check which conversion the problem needs.
Annulus method: V = π ∫ (outer² − inner²) dx — SQUARE FIRST, then subtract. Subtracting before squaring is a common error.
Multi-part shapes: integrate each section over its own x-range and sum. Check continuity at section boundaries (functions should match in value).
Working backwards: set up V(parameter) symbolically, then solve V = target. The parameter usually appears as a simple cube (gives clean ∛) or square (gives ±√) in the equation.

⚠ Common mistakes

Confusing mL and L: 127 mL is NOT 127 L (off by a factor of 1000). Always check whether the question asks for “millilitres” or “litres” specifically — answer in the requested unit.
Subtracting BEFORE squaring: π∫(y_outer − y_inner)² dx ≠ π∫(y_outer² − y_inner²) dx. The annulus formula squares each, THEN subtracts.
Boundary mismatch in multi-part shapes: if section 1 ends at y = 2.8 and section 2 starts at y = 3, the model has a physical step. Check continuity always.
Wrong axis formula: using V = π∫y² dx when the rotation is about the y-axis (should be V = π∫x² dy). Identify the axis FIRST.
Forgetting to convert units: an answer of 127 cm³ when the question asks “litres” should be 0.127 L, not 127 L. Read what units the question wants in the FINAL answer.

You’ve finished Further Integration — congratulations! Up next: Differential Equations. A differential equation links a function to its derivative(s), and “solving” means finding the function. AA HL covers first-order separable equations (rearrange to separate x and y sides, integrate both), Euler’s method for numerical solutions (a step-by-step approximation when the analytic solution is hard), and the integrating factor method for first-order linear DEs. Same integration tools you’ve just mastered (substitution, parts, partial fractions) — now applied to differential equations.

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