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AS & A Level · Practice questions · Volumes of revolution

Discs, washers and bowls.

Six original Cambridge-style questions on volumes of revolution, from a cone check to a bowl problem in context.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Square the radius, then integrate along the axis.

01
Recall
[3 marks]

The region under \(y = 2x\) from \(x = 0\) to \(x = 3\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the volume, and check it with the formula for the volume of a cone.

  • \(V = \pi\displaystyle\int_{0}^{3} 4x^{2}\,dx = \pi\left[\tfrac{4x^{3}}{3}\right]_{0}^{3} = 36\pi\). ✓
  • Cone: radius \(6\), height \(3\): \(\tfrac{1}{3}\pi \times 36 \times 3 = 36\pi\). They agree. ✓
02
Calculation
[4 marks]

The region enclosed by \(y = 2x - x^{2}\) and the \(x\)-axis is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume.

  • Limits: \(x(2 - x) = 0\) at \(x = 0, 2\). ✓
  • \(y^{2} = 4x^{2} - 4x^{3} + x^{4}\). ✓
  • \(V = \pi\left[\tfrac{4x^{3}}{3} - x^{4} + \tfrac{x^{5}}{5}\right]_{0}^{2} = \pi\left(\tfrac{32}{3} - 16 + \tfrac{32}{5}\right)\). ✓
  • \(= \dfrac{16\pi}{15}\) ✓
03
Calculation
[5 marks]

The region under \(y = \dfrac{2}{x}\) from \(x = 1\) to \(x = N\) is rotated through \(360^{\circ}\) about the \(x\)-axis.

(a) Find the volume when \(N = 4\).

(b) Show that the volume stays finite as \(N \to \infty\), and state its limiting value.

  • (a) \(V = \pi\displaystyle\int_{1}^{4} 4x^{-2}\,dx = \pi\left[-\dfrac{4}{x}\right]_{1}^{4} = \pi(-1 + 4) = 3\pi\). ✓
  • (b) \(V = \pi\left[-\dfrac{4}{x}\right]_{1}^{N} = \pi\left(4 - \dfrac{4}{N}\right)\). ✓
  • As \(N \to \infty\), \(\dfrac{4}{N} \to 0\), so \(V \to 4\pi\), a finite value. ✓
04
Calculation
[4 marks]

The region bounded by \(y = x^{2} + 1\) for \(x \ge 0\), the \(y\)-axis and the line \(y = 5\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Find the exact volume.

  • \(x^{2} = y - 1\); the region runs from \(y = 1\) (where \(x = 0\)) to \(y = 5\). ✓
  • \(V = \pi\displaystyle\int_{1}^{5} (y - 1)\,dy = \pi\left[\tfrac{y^{2}}{2} - y\right]_{1}^{5}\). ✓
  • \(= \pi\left(\tfrac{15}{2} - \left(-\tfrac{1}{2}\right)\right) = 8\pi\) ✓
05
Analysis
[5 marks]

The region enclosed by the curve \(y = x^{2}\) and the line \(y = 2x\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume of the solid formed.

  • Meet where \(x^{2} = 2x\): \(x = 0, 2\). The line is the outer boundary. ✓
  • \(V = \pi\displaystyle\int_{0}^{2} \left((2x)^{2} - (x^{2})^{2}\right) dx = \pi\displaystyle\int_{0}^{2} (4x^{2} - x^{4})\,dx\). ✓
  • \(= \pi\left[\tfrac{4x^{3}}{3} - \tfrac{x^{5}}{5}\right]_{0}^{2} = \pi\left(\tfrac{32}{3} - \tfrac{32}{5}\right)\). ✓
  • \(= \dfrac{64\pi}{15}\) ✓
06
Analysis
[6 marks]

A bowl is modelled by rotating the curve \(y = x^{2}\), for \(0 \le y \le 9\), through \(360^{\circ}\) about the \(y\)-axis. Units are centimetres.

(a) Show that the volume of liquid in the bowl when the depth is \(h\) cm is \(\tfrac{1}{2}\pi h^{2}\ \text{cm}^{3}\).

(b) Find the volume of the full bowl.

(c) Find the depth of liquid when the bowl is half full.

  • (a) \(V = \pi\displaystyle\int_{0}^{h} x^{2}\,dy = \pi\displaystyle\int_{0}^{h} y\,dy = \pi\left[\tfrac{y^{2}}{2}\right]_{0}^{h} = \tfrac{1}{2}\pi h^{2}\). ✓
  • (b) \(h = 9\): \(V = \tfrac{81\pi}{2} = 127\ \text{cm}^{3}\) (3 s.f.). ✓
  • (c) \(\tfrac{1}{2}\pi h^{2} = \tfrac{81\pi}{4}\) gives \(h^{2} = 40.5\). ✓
  • \(h = 6.36\) cm (3 s.f.). Not 4.5 cm: the bowl is wider at the top. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.