The shape of the lesson
By the end of the lesson, learners can
- Discs about the x-axis: Rotate the region under \(y = f(x)\) through \(360^{\circ}\) about the \(x\)-axis.
- About the y-axis: Swap the roles of \(x\) and \(y\).
- 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).
- About the \(x\)-axis: \(V = \pi\displaystyle\int_{a}^{b} y^{2}\,dx\), \(x\)-limits.
- About the \(y\)-axis: \(V = \pi\displaystyle\int_{c}^{d} x^{2}\,dy\), \(y\)-limits.
- Square first: expand \(y^{2}\) or write it as a power of a linear bracket.
- Washers: \(\pi\displaystyle\int (\text{outer}^{2} - \text{inner}^{2})\).
- Exact: leave \(\pi\) in unless a decimal is asked for.
One picture every idea returns to
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: \(V = \pi\displaystyle\int_{a}^{b} \left(y_{\text{outer}}^{2} - y_{\text{inner}}^{2}\right) dx\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Spin a region, get a solid.. Turn a region a full circle about an axis and it sweeps out a solid. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Discs about the x-axis | Rotate the region under \(y = f(x)\) through \(360^{\circ}\) about the \(x\)-axis. Model the worked example A quadratic on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: About the y-axis | Swap the roles of \(x\) and \(y\). Model the worked example A parabola about the y-axis on the board, one line at a time. Your turn: The region under \(y = \sqrt{x}\) from \(x = 0\) to \(x = 4\) is rotated about the \(x\)-axis. Find the exact volume. | Pairs |
| 21 to 30 min | Teach and check: 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). Model the worked example Curve and line about the x-axis on the board, one line at a time. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 27 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Regions with a hole, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| 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 \(\tfrac{206\pi}{15}\). |
| 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. |
| 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. |
Volumes of revolution: worksheet
6 questions, 27 marks. Written in the style of the papers; no past paper question is reproduced.
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.
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.
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.
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.
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.
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.
Answer key and marking notes
- \(V = \pi\displaystyle\int_{0}^{3} 4x^{2}\,dx = \pi\left[\tfrac{4x^{3}}{3}\right]_{0}^{3} = 36\pi\).
- Cone: radius \(6\), height \(3\): \(\tfrac{1}{3}\pi \times 36 \times 3 = 36\pi\). They agree.
- Limits: \(x(2 - x) = 0\) at \(x = 0, 2\).
- \(y^{2} = 4x^{2} - 4x^{3} + x^{4}\).
- \(V = \pi\left[\tfrac{4x^{3}}{3} - x^{4} + \tfrac{x^{5}}{5}\right]_{0}^{2} = \pi\left(\tfrac{32}{3} - 16 + \tfrac{32}{5}\right)\).
- \(= \dfrac{16\pi}{15}\)
- (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.
- \(x^{2} = y - 1\); the region runs from \(y = 1\) (where \(x = 0\)) to \(y = 5\).
- \(V = \pi\displaystyle\int_{1}^{5} (y - 1)\,dy = \pi\left[\tfrac{y^{2}}{2} - y\right]_{1}^{5}\).
- \(= \pi\left(\tfrac{15}{2} - \left(-\tfrac{1}{2}\right)\right) = 8\pi\)
- Meet where \(x^{2} = 2x\): \(x = 0, 2\). The line is the outer boundary.
- \(V = \pi\displaystyle\int_{0}^{2} \left((2x)^{2} - (x^{2})^{2}\right) dx = \pi\displaystyle\int_{0}^{2} (4x^{2} - x^{4})\,dx\).
- \(= \pi\left[\tfrac{4x^{3}}{3} - \tfrac{x^{5}}{5}\right]_{0}^{2} = \pi\left(\tfrac{32}{3} - \tfrac{32}{5}\right)\).
- \(= \dfrac{64\pi}{15}\)
- (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.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The region under \(y = x^{2}\) from \(x = 0\) to \(x = 1\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume generated.
- A\(\dfrac{\pi}{5}\)
- B\(\dfrac{\pi}{3}\)
- C\(\dfrac{1}{5}\)
- D\(\dfrac{\pi}{9}\)
The region under \(y = 2\sqrt{x}\) from \(x = 0\) to \(x = 4\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume generated.
- A\(32\pi\)
- B\(\dfrac{32\pi}{3}\)
- C\(16\pi\)
- D\(32\)
The region bounded by \(y = x^{3}\) for \(x \ge 0\), the \(y\)-axis and the lines \(y = 1\) and \(y = 8\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Which integral gives the volume?
- A\(\pi\displaystyle\int_{1}^{8} y^{\frac{2}{3}}\,dy\)
- B\(\pi\displaystyle\int_{1}^{2} x^{6}\,dx\)
- C\(\pi\displaystyle\int_{1}^{8} y^{\frac{1}{3}}\,dy\)
- D\(\pi\displaystyle\int_{1}^{2} y^{\frac{2}{3}}\,dy\)
The region bounded by \(y = 4 - x\) and the coordinate axes is rotated through \(360^{\circ}\) about the \(x\)-axis to form a cone. Find the exact volume of the cone.
- A\(\dfrac{64\pi}{3}\)
- B\(\dfrac{128\pi}{3}\)
- C\(8\pi\)
- D\(64\pi\)
The region under \(y = \dfrac{3}{x + 2}\) from \(x = 0\) to \(x = 1\) is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume generated.
- A\(\dfrac{3\pi}{2}\)
- B\(-\dfrac{3\pi}{2}\)
- C\(\dfrac{\pi}{2}\)
- D\(\dfrac{15\pi}{2}\)
The region bounded by \(y = x^{2} + 1\) for \(x \ge 0\), the \(y\)-axis and the lines \(y = 2\) and \(y = 5\) is rotated through \(360^{\circ}\) about the \(y\)-axis. Find the exact volume generated.
- A\(\dfrac{15\pi}{2}\)
- B\(\dfrac{14\pi}{3}\)
- C\(\dfrac{27\pi}{2}\)
- D\(\dfrac{178\pi}{15}\)
The region enclosed by the line \(y = x + 2\) and the curve \(y = x^{2}\), which meet where \(x = -1\) and \(x = 2\), is rotated through \(360^{\circ}\) about the \(x\)-axis. Find the exact volume generated.
- A\(\dfrac{72\pi}{5}\)
- B\(\dfrac{81\pi}{10}\)
- C\(21\pi\)
- D\(\dfrac{72}{5}\)
The region under \(y = kx\), where \(k > 0\), from \(x = 0\) to \(x = 3\) is rotated through \(360^{\circ}\) about the \(x\)-axis. The volume generated is \(36\pi\). Find \(k\).
- A\(k = 2\)
- B\(k = 4\)
- C\(k = 8\)
- D\(k = \dfrac{2}{\sqrt{3}}\)
Quiz key
\(V = \pi\displaystyle\int_{0}^{1} (x^{2})^{2}\,dx = \pi\left[\tfrac{1}{5}x^{5}\right]_{0}^{1} = \dfrac{\pi}{5}\). \(\dfrac{\pi}{3}\) integrates \(y\) instead of \(y^{2}\); \(\dfrac{1}{5}\) forgets the \(\pi\); \(\dfrac{\pi}{9}\) squares the area \(\tfrac{1}{3}\) instead of squaring \(y\) before integrating.
\(y^{2} = 4x\), so \(V = \pi\displaystyle\int_{0}^{4} 4x\,dx = \pi\left[2x^{2}\right]_{0}^{4} = 32\pi\). \(\dfrac{32\pi}{3}\) integrates \(y = 2\sqrt{x}\) without squaring; \(16\pi\) squares only the root, giving \(2x\); \(32\) drops the \(\pi\).
\(x = y^{\frac{1}{3}}\), so \(x^{2} = y^{\frac{2}{3}}\), and the limits are \(y = 1\) to \(y = 8\). \(\pi\displaystyle\int_{1}^{2} x^{6}\,dx\) is a rotation about the \(x\)-axis; \(y^{\frac{1}{3}}\) is \(x\), not \(x^{2}\); limits \(1\) to \(2\) are \(x\)-values, wrong for an integral in \(y\).
\(\pi\displaystyle\int_{0}^{4} (4 - x)^{2}\,dx = \pi\left[-\tfrac{1}{3}(4 - x)^{3}\right]_{0}^{4} = \pi\left(0 + \tfrac{64}{3}\right) = \dfrac{64\pi}{3}\), matching \(\tfrac{1}{3}\pi \times 4^{2} \times 4\). \(\dfrac{128\pi}{3}\) squares \(4 - x\) as \(16 - x^{2}\); \(8\pi\) does not square at all; \(64\pi\) is the cylinder \(\pi r^{2}h\), three times too big.
\(V = \pi\displaystyle\int_{0}^{1} 9(x + 2)^{-2}\,dx = \pi\left[-\dfrac{9}{x + 2}\right]_{0}^{1} = \pi\left(-3 + \dfrac{9}{2}\right) = \dfrac{3\pi}{2}\). \(-\dfrac{3\pi}{2}\) subtracts the limits the wrong way round; \(\dfrac{\pi}{2}\) squares the bracket but not the 3; \(\dfrac{15\pi}{2}\) drops the minus sign at the upper limit, working out \(\pi\left(3 + \tfrac{9}{2}\right)\).
\(x^{2} = y - 1\), so \(V = \pi\displaystyle\int_{2}^{5} (y - 1)\,dy = \pi\left[\tfrac{1}{2}y^{2} - y\right]_{2}^{5} = \pi\left(\tfrac{15}{2} - 0\right) = \dfrac{15\pi}{2}\). \(\dfrac{14\pi}{3}\) integrates \(x = \sqrt{y - 1}\) instead of \(x^{2}\); \(\dfrac{27\pi}{2}\) rearranges to \(x^{2} = y + 1\); \(\dfrac{178\pi}{15}\) rotates about the \(x\)-axis instead, \(\pi\displaystyle\int_{1}^{2} (x^{2} + 1)^{2}\,dx\).
The line is outer, the curve inner: \(V = \pi\displaystyle\int_{-1}^{2} \left((x + 2)^{2} - x^{4}\right) dx = \pi\left(\left[\tfrac{1}{3}(x + 2)^{3}\right]_{-1}^{2} - \left[\tfrac{1}{5}x^{5}\right]_{-1}^{2}\right) = \pi\left(21 - \tfrac{33}{5}\right) = \dfrac{72\pi}{5}\). \(\dfrac{81\pi}{10}\) squares the difference, \(\pi\displaystyle\int (x + 2 - x^{2})^{2}\,dx\), which is not the same as the difference of the squares; \(21\pi\) is the outer solid only, with no hole; \(\dfrac{72}{5}\) forgets the \(\pi\).
\(V = \pi k^{2}\left[\tfrac{1}{3}x^{3}\right]_{0}^{3} = 9\pi k^{2} = 36\pi\), so \(k^{2} = 4\) and \(k = 2\) (as \(k > 0\)). \(k = 4\) stops at \(k^{2} = 4\); \(k = 8\) does not square \(y\) (\(\pi\displaystyle\int kx\,dx = \tfrac{9}{2}\pi k\)); \(\dfrac{2}{\sqrt{3}}\) integrates \(x^{2}\) as \(x^{3}\) without dividing by 3, giving \(27\pi k^{2} = 36\pi\).
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.