Home / A-Level Maths 9709 / Integration / Volumes of revolution
Cambridge International AS & A Level · Topic 1.8
A-Level Maths 9709 / Topic 1.8 / AS & A Level

Spin a region, get a solid.

Turn a region a full circle about an axis and it sweeps out a solid. Slice that solid into thin discs and add them up: the integral of \(\pi y^{2}\) gives its volume.

Topic 1.8All integration lessons →
The key idea

About the \(x\)-axis: \(V = \pi\displaystyle\int_{a}^{b} y^{2}\,dx\). About the \(y\)-axis: \(V = \pi\displaystyle\int_{c}^{d} x^{2}\,dy\). For a region with a hole, subtract: \(V = \pi\displaystyle\int_{a}^{b} \left(y_{\text{outer}}^{2} - y_{\text{inner}}^{2}\right) dx\).

x y y = 5 y = 9 − x² slice at x = 1: outer radius 8 inner radius 5
Fig. 1: the shaded region between y = 9 − x² and y = 5 rotated about the x-axis; each slice is a washer, outer radius 9 − x² and inner radius 5 (8 and 5 at x = 1)
Section 01

Discs about the x-axis.

Rotate the region under \(y = f(x)\) through \(360^{\circ}\) about the \(x\)-axis. A thin slice of width \(\delta x\) becomes a disc of radius \(y\) and volume about \(\pi y^{2}\,\delta x\). Adding the discs and letting \(\delta x \to 0\) gives \(V = \pi\displaystyle\int_{a}^{b} y^{2}\,dx\).

Worked example A quadratic
The region between \(y = x^{2} + 1\), the \(x\)-axis, \(x = 0\) and \(x = 2\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume.
  1. Square first: \(y^{2} = (x^{2} + 1)^{2} = x^{4} + 2x^{2} + 1\).
  2. \(V = \pi\displaystyle\int_{0}^{2} (x^{4} + 2x^{2} + 1)\,dx = \pi\left[\tfrac{x^{5}}{5} + \tfrac{2x^{3}}{3} + x\right]_{0}^{2}\).
  3. \(= \pi\left(\tfrac{32}{5} + \tfrac{16}{3} + 2\right) = \pi \times \tfrac{96 + 80 + 30}{15}\).
Answer: \(\dfrac{206\pi}{15}\)
Worked example Checking with a cone
Rotate the region under \(y = \tfrac{1}{2}x\) from \(x = 0\) to \(x = 6\) about the \(x\)-axis. Show that the volume agrees with the cone formula \(\tfrac{1}{3}\pi r^{2}h\).
  1. \(V = \pi\displaystyle\int_{0}^{6} \tfrac{1}{4}x^{2}\,dx = \pi\left[\tfrac{x^{3}}{12}\right]_{0}^{6} = \pi \times \tfrac{216}{12} = 18\pi\).
  2. The solid is a cone with height 6 and base radius \(\tfrac{1}{2} \times 6 = 3\).
  3. \(\tfrac{1}{3}\pi \times 3^{2} \times 6 = 18\pi\).
Answer: Both give \(18\pi\).
Worked example A reciprocal of a bracket
The region under \(y = \dfrac{6}{2x + 1}\) from \(x = 0\) to \(x = 1\) is rotated about the \(x\)-axis. Find the exact volume.
  1. \(y^{2} = \dfrac{36}{(2x + 1)^{2}} = 36(2x + 1)^{-2}\).
  2. \(\displaystyle\int 36(2x + 1)^{-2}\,dx = \dfrac{36(2x + 1)^{-1}}{-1 \times 2} = -\dfrac{18}{2x + 1}\).
  3. \(V = \pi\left[-\dfrac{18}{2x + 1}\right]_{0}^{1} = \pi\left(-6 - (-18)\right) = 12\pi\).
Answer: \(12\pi\)
Examiner trap

Square \(y\) BEFORE integrating: \(\pi\displaystyle\int y^{2}\,dx\) is not \(\pi\left(\displaystyle\int y\,dx\right)^{2}\). And keep the \(\pi\): an exact answer is usually wanted, such as \(\tfrac{206\pi}{15}\).

Section 02

About the y-axis.

Swap the roles of \(x\) and \(y\). Rewrite the curve as \(x^{2}\) in terms of \(y\), use \(y\)-values for the limits, and integrate with respect to \(y\).

Worked example A parabola about the y-axis
The region bounded by \(y = x^{2}\) for \(x \ge 0\), the \(y\)-axis and the lines \(y = 1\) and \(y = 4\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Find the exact volume.
  1. \(x^{2} = y\), and the limits are \(y = 1\) to \(y = 4\).
  2. \(V = \pi\displaystyle\int_{1}^{4} y\,dy = \pi\left[\tfrac{y^{2}}{2}\right]_{1}^{4} = \pi\left(8 - \tfrac{1}{2}\right)\).
Answer: \(\dfrac{15\pi}{2}\)
Your turn
The region under \(y = \sqrt{x}\) from \(x = 0\) to \(x = 4\) is rotated about the \(x\)-axis. Find the exact volume.
Examiner trap

About the \(y\)-axis, the limits are \(y\)-values. Using \(x = 1\) to \(x = 2\) (where the curve meets \(y = 1\) and \(y = 4\)) with \(\displaystyle\int x^{2}\,dy\) mixes the two variables.

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Which formula gives the volume when the region under \(y = f(x)\) from \(x = a\) to \(x = b\) is rotated about the \(x\)-axis?

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The region under \(y = x\) from \(x = 0\) to \(x = 3\) is rotated about the \(x\)-axis. Find the volume.

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For a rotation about the \(y\)-axis, what do you integrate?

Quick check+10 XP

A region between an outer curve \(y_{1}\) and an inner curve \(y_{2}\) is rotated about the \(x\)-axis. Which integrand is correct?

Section 03

Regions with a hole.

If the region does not touch the axis of rotation, each slice is a washer: a disc with a smaller disc removed (Fig. 1). Find the volume swept by the outer boundary, and subtract the volume swept by the inner one.

Worked example Curve and line about the x-axis
The region between \(y = 9 - x^{2}\) and \(y = 5\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume.
  1. Meet where \(9 - x^{2} = 5\): \(x = \pm 2\). Outer radius \(9 - x^{2}\), inner radius \(5\).
  2. \((9 - x^{2})^{2} - 5^{2} = 81 - 18x^{2} + x^{4} - 25 = 56 - 18x^{2} + x^{4}\).
  3. By symmetry, \(V = 2\pi\displaystyle\int_{0}^{2} (56 - 18x^{2} + x^{4})\,dx = 2\pi\left[56x - 6x^{3} + \tfrac{x^{5}}{5}\right]_{0}^{2}\).
  4. \(= 2\pi\left(112 - 48 + \tfrac{32}{5}\right) = 2\pi \times \tfrac{352}{5}\).
Answer: \(\dfrac{704\pi}{5} \approx 442\) cubic units
Examiner trap

Washers: subtract the SQUARES, \(\pi\displaystyle\int (y_{1}^{2} - y_{2}^{2})\,dx\). Squaring the difference, \(\pi\displaystyle\int (y_{1} - y_{2})^{2}\,dx\), rotates the wrong shape and is the most common error.

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The region between \(y = x^{2} + 1\), the \(x\)-axis and the lines \(x = 0\) and \(x = 2\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the volume.

Exam style+20 XP

The region between \(y = 9 - x^{2}\) and the line \(y = 5\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the volume.

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The region under \(y = \dfrac{6}{2x + 1}\) between \(x = 0\) and \(x = 1\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the volume.

Exam style+20 XP

The region bounded by \(y = x^{2}\) for \(x \ge 0\), the \(y\)-axis and the lines \(y = 1\) and \(y = 4\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Find the volume.

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Six original Cambridge-style questions
Volumes about the x-axis and the y-axis, an infinite region with a finite volume, a washer, and a bowl filled with liquid.