IB Maths AI HL Further Integration Paper 1 & 2 ~6 min read

Volumes of Revolution

Spin a region around an axis and it sweeps out a 3D solid. Slice that solid into thin discs: each is a circle of radius y (or x) and thickness dx (or dy). Add up the disc volumes with an integral and you get the volume of revolution — a definite integral with a Ļ€ out front.

šŸ“˜ What you need to know

The disc idea

Volume of revolution V = Ļ€ ∫ab y2 dx   (about x-axis)  ā€¢  V = Ļ€ ∫cd x2 dy   (about y-axis) āœ“ both forms are in the formula booklet
A region spun into a solid
x y disc y = f(x)
Each slice is a disc of radius y, area πy2, thickness dx. Summing them along the axis gives the volume.

šŸ¤” Why y2 and where does Ļ€ come from?

Each thin slice is a cylinder (disc) of radius r = y and thickness dx, so its volume is area Ɨ thickness = Ļ€y2 dx. The Ļ€r2 for a circle’s area is exactly where both the Ļ€ and the square come from. Integrating sums infinitely many such discs.

🧠 “Pi, integral, radius squared”

Ļ€ out front, integral over the limits, the radius squared inside. Rotating about x the radius is y; about y the radius is x. Don’t forget to square it.

Carrying it out

🧭 Recipe — volume of revolution

  1. Identify the axis of rotation — sets the variable and which radius to square.
  2. Get the radius: y = f(x) for the x-axis; rearrange to x = g(y) for the y-axis.
  3. Square it and multiply by π.
  4. Integrate between the correct limits.
  5. Leave π for an exact answer, or evaluate to a decimal.
Squaring tidies things up: a square root like y = √x becomes y2 = x, and an exponential y = ex becomes y2 = e2x — both far easier to integrate.

Worked examples

WE 1

The region under y = x from x = 0 to 2 is rotated about the x-axis. Find the volume.

Radius y = x, so y2 = x2. (This is a cone.)

V = Ļ€ ∫02 x² dx = Ļ€[x³3]02 = Ļ€ Ā· 83 V = 8Ļ€3 (ā‰ˆ 8.38)
WE 2

The region under y = √x from x = 0 to 4 is rotated about the x-axis. Find the volume.

Squaring the root simplifies it: y2 = x.

V = Ļ€ ∫04 x dx = Ļ€[x²2]04 = Ļ€ Ā· 8 V = 8Ļ€ (ā‰ˆ 25.13)
WE 3

The region under y = x2 from x = 0 to 1 is rotated about the x-axis. Find the volume.

Radius y = x2, so y2 = x4.

V = Ļ€ ∫01 x⁓ dx = Ļ€[x⁵5]01 = Ļ€ Ā· 15 V = Ļ€5 (ā‰ˆ 0.63)
WE 4

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

About the y-axis — rearrange to x2 = y and use y-limits.

x² = y; V = Ļ€ ∫04 y dy = Ļ€[y²2]04 = Ļ€ Ā· 8 V = 8Ļ€ (ā‰ˆ 25.13)
WE 5

The region under y = ex from x = 0 to 1 is rotated about the x-axis. Find the exact volume.

Square first: y2 = e2x, then integrate (divide by 2).

V = Ļ€ ∫01 e2x dx = Ļ€[12 e2x]01 = Ļ€2(e² āˆ’ e⁰) = Ļ€2(e² āˆ’ 1) V = Ļ€(e² āˆ’ 1)2 (ā‰ˆ 10.04)

šŸ’” Top tips

⚠ Common mistakes

That wraps up Further Integration. The unit started with the standard integrals and reverse-chain techniques, then turned the definite integral into geometry — areas under curves, below the axis, against the y-axis, and between graphs — before lifting it into three dimensions here with volumes of revolution. The disc method is the same definite-integral move once more: sum infinitely many thin slices, this time circles of area Ļ€r2, between two limits.

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