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NotesMath AA HLTopic 5.17Volumes of revolution
Back to Math AA HL Topics
5.17.22 min read

Volumes of revolution

IB Mathematics: Analysis and Approaches • Unit 5

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Contents

  • Spin about the x-axis: V = π∫y² dx
  • Spin about the y-axis, and between two curves
A stack of thin discs: Take the region under a curve y = f(x) and spin it a full turn (360°, or 2π radians) around the x-axis. It sweeps out a solid.

Slice the solid into thin discs, each perpendicular to the x-axis. A disc at position x has radius y (the curve's height) and tiny thickness dx, so its volume is π·(radius)²·(thickness) = πy²·dx. Add up every disc and you get the formula below.
Volume when the region under y = f(x), from x = a to x = b, is rotated 360° about the x-axis.
Square the WHOLE of y: The radius is the function value y = f(x), so y² means [f(x)]² — square the entire expression before integrating.

For example, if y = x + 1 then y² = (x + 1)², which you expand to x² + 2x + 1 (or integrate directly as a chain-rule reverse).

IB-style question — rotate about the x-axis

The region under the curve y = √x, between x = 0 and x = 4, is rotated 360° about the x-axis.

Find the exact volume of the solid formed.

Step by step

  1. Use V = π∫y² dx. Square y first: (√x)² = x.
  2. Set up the integral with the x-limits 0 and 4.
  3. Integrate x.
  4. Substitute: 4²/2 = 16/2 = 8.

Final answer

V = 8π (≈ 25.1) cubic units.

About the y-axis: swap the roles: Spin a region around the y-axis instead, and the discs are now horizontal. Each disc has radius x and thickness dy, so its volume is πx²·dy.

It is the mirror of the x-axis case: write x² in terms of y and use y-limits.
Volume when a region is rotated 360° about the y-axis, between y = c and y = d.

IB-style question — rotate about the y-axis

The region bounded by y = x², the y-axis, and the line y = 4 (with x ≥ 0) is rotated 360° about the y-axis.

Find the exact volume.

Step by step

  1. Use V = π∫x² dy. We need x² in terms of y. From y = x², x² = y directly.
  2. Set up the integral with y-limits 0 and 4.
  3. Integrate y.
  4. Substitute: 4²/2 = 8.

Final answer

V = 8π (≈ 25.1) cubic units.

Between two curves: outer² − inner²: When the region sits between two curves and is rotated, each slice is a washer (a disc with a hole). Subtract the inner radius squared from the outer radius squared:

V = π∫(R² − r²) dx, where R is the outer (further) curve and r is the inner (nearer) curve. Subtract the squares, never (R − r)².

IB-style question — region between two curves

The region between y = √x (outer) and y = x (inner), from x = 0 to x = 1, is rotated 360° about the x-axis.

Find the exact volume.

Step by step

  1. Washers: V = π∫(R² − r²) dx with R = √x, r = x.
  2. Set up the integral from 0 to 1.
  3. Integrate term by term.
  4. Substitute: 1/2 − 1/3 = 1/6.

Final answer

V = π/6 (≈ 0.524) cubic units.

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The region under y = 3, from x = 1 to x = 5, is rotated 360° about the x-axis. Find the exact volume, and say what solid it is. [2 marks]

Related Math AA HL Topics

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5.1.1Derivative as gradient
5.10.1Reverse chain rule
5.10.2Substitution
5.11.1Definite integrals
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