Mathematics guide, Unit 8, step 27 of 33Previous step Next step

Step 27: Volumes of Revolution and Arc Length

By the EconTutorials team. 11-minute read with 6 practice questions.

SyllabusAP 8.7–8.13 · 9709 P1 1.8 · IB HL 5.17

In 30 seconds

A volume is a stack of thin slices: $ V = \displaystyle\int_a^b A(x)\,dx $, where $ A(x) $ is the area of the cross-section at $ x $. On AP, the slices are often squares, triangles or semicircles standing on a base region.
Rotating a region about the $ x $-axis gives circular slices: the disc method $ V = \pi\displaystyle\int_a^b y^2\,dx $. About the $ y $-axis, use $ V = \pi\displaystyle\int_c^d x^2\,dy $.
If the region does not touch the axis, each slice is a washer: $ V = \pi\displaystyle\int (R^2-r^2) $, with each radius measured from the axis. Other axes, such as $ y = -1 $, are AP only; 9709 Paper 1 and IB HL use the $ x $- and $ y $-axes.
Exam point: arc length $ \displaystyle\int_a^b \sqrt{1 + [f'(x)]^2}\,dx $ is AP BC only. And the Gini coefficient, $ G = 1-2\displaystyle\int_0^1 L(x)\,dx $, turns a country’s inequality into an area.

Video lesson coming soonA short animated lesson for this step will appear here.

Volumes by slicing: known cross-sections

Slice a loaf of bread. If you know the area of every slice, you know the volume: add up area times thickness. A solid whose cross-section at position $ x $ has area $ A(x) $ has volume

\[ V = \int_a^b A(x)\,dx. \]

On AP, the solid usually stands on a base region in the $ xy $-plane, and each cross-section perpendicular to the $ x $-axis is a shape whose side $ s $ lies across the base. The side is the width of the base at $ x $, top minus bottom. Then the shape gives the area:

Cross-section, side $ s $ in the base	Area $ A $
square	$ s^2 $
rectangle of height $ h $	$ sh $
equilateral triangle	$ \tfrac{\sqrt{3}}{4}s^2 $
isosceles right triangle with a leg $ s $	$ \tfrac{1}{2}s^2 $
semicircle with diameter $ s $	$ \tfrac{\pi}{8}s^2 $

Worked example

The base of a solid is the region bounded by $ y = \sqrt{x} $, the $ x $-axis and the line $ x = 4 $. Find its volume if the cross-sections perpendicular to the $ x $-axis are (a) squares, (b) semicircles, (c) equilateral triangles.

Step 1The side at $ x $ runs from the axis up to the curve: $ s = \sqrt{x} $, so $ s^2 = x $, for $ 0 \le x \le 4 $.

Step 2The key integral: $ \displaystyle\int_0^4 s^2\,dx = \int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = 8 $.

Step 3(a) Squares: $ V = 8 $. (b) Semicircles: $ V = \tfrac{\pi}{8} \times 8 = \pi $. (c) Equilateral triangles: $ V = \tfrac{\sqrt{3}}{4} \times 8 = 2\sqrt{3} $.

One integral serves all three parts; only the shape factor changes.

The base region: under y = √x from x = 0 to x = 4

y = √x

Figure 1: looking down on the base. The strip at x has width s = √x. A square cross-section standing on it has area s² = x, and a slab of thickness dx has volume x dx.

Common mistakeUsing the radius formula for a semicircle whose diameter is $ s $. The radius is $ \tfrac{s}{2} $, so the area is $ \tfrac{1}{2}\pi\left(\tfrac{s}{2}\right)^2 = \tfrac{\pi s^2}{8} $, not $ \tfrac{\pi s^2}{2} $. Read the question for which length lies in the base. Known cross-sections are an AP topic (8.7 and 8.8); Cambridge 9709 and IB do not set them.

Discs: rotating about the $ x $- and $ y $-axes

Turn the region under $ y = f(x) $, for $ a \le x \le b $, through $ 360^\circ $ about the $ x $-axis. Each thin vertical strip sweeps out a disc of radius $ f(x) $ and thickness $ dx $, with volume $ \pi[f(x)]^2\,dx $. Add the discs:

\[ \text{about the } x\text{-axis: } V = \pi\int_a^b y^2\,dx \]

\[ \text{about the } y\text{-axis: } V = \pi\int_c^d x^2\,dy \]

For the $ y $-axis the slices are horizontal discs of radius $ x $ at height $ y $, so write $ x $ in terms of $ y $ and use $ y $-limits.

Key termSolid of revolution: the solid swept out when a plane region turns through $ 360^\circ $ about a line in its plane that does not pass through the interior of the region.

Worked example

The region under $ y = x^2 + 1 $ for $ 0 \le x \le 1 $ is rotated through $ 360^\circ $ about the $ x $-axis. Find the exact volume.

Step 1The radius of the disc at $ x $ is $ y = x^2 + 1 $, so $ y^2 = x^4 + 2x^2 + 1 $.

Step 2$ V = \pi\displaystyle\int_0^1 (x^4 + 2x^2 + 1)\,dx = \pi\left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^1 $.

Step 3$ = \pi\left(\tfrac{1}{5} + \tfrac{2}{3} + 1\right) = \tfrac{28\pi}{15} $.

Square the whole of $ y $ before integrating: $ (x^2 + 1)^2 $ is not $ x^4 + 1 $.

Rotating y = x² + 1, 0 ≤ x ≤ 1, about the x-axis

y = x² + 1 (the edge that turns)one disc: volume π y² dx

Figure 2: every vertical strip under the curve sweeps out a thin disc of radius y = x² + 1. Adding the discs from x = 0 to x = 1 gives V = 28π/15.

Worked example

The region in the first quadrant bounded by $ y = 4-x^2 $ and the axes is rotated through $ 360^\circ $ about the $ y $-axis. Find the exact volume.

Step 1Horizontal slices: at height $ y $ the radius is $ x $, where $ x^2 = 4-y $. The region runs from $ y = 0 $ to $ y = 4 $.

Step 2$ V = \pi\displaystyle\int_0^4 (4-y)\,dy = \pi\left[4y-\frac{y^2}{2}\right]_0^4 = \pi(16-8) = 8\pi $.

Sense check: the solid is a paraboloid that fits inside a cylinder of radius 2 and height 4, of volume $ 16\pi $. A paraboloid fills exactly half its cylinder, and $ 8\pi $ is half of $ 16\pi $.

Washers and other axes of rotation

If the region does not touch the axis of rotation, each slice is a disc with a hole in it: a washer. Its outer radius $ R $ reaches the far edge of the region and its inner radius $ r $ the near edge. The face has area $ \pi R^2-\pi r^2 $, so

\[ V = \pi\int_a^b \big(R^2-r^2\big)\,dx \quad \text{(or } dy \text{ for a vertical axis).} \]

Why $ R^2-r^2 $ and not $ (R-r)^2 $? A washer is a big disc with a small disc removed. The ring of width $ R-r $ is not a disc of radius $ R-r $: it is wrapped round a much larger circle. For $ R = 3 $ and $ r = 2 $ the ring has area $ 5\pi $, five times the $ \pi $ that $ (R-r)^2 $ gives. Every radius is a distance from the axis: about the line $ y = k $ it is $ |y-k| $, and about $ x = h $ it is $ |x-h| $.

Worked example

$ \mathcal{R} $ is the region enclosed by $ y = 2x $ and $ y = x^2 $. Find the volume when $ \mathcal{R} $ is rotated through $ 360^\circ $ about (a) the $ x $-axis, (b) the line $ y = -1 $, (c) the $ y $-axis.

Step 1The curves meet where $ x^2 = 2x $: at $ x = 0 $ and $ x = 2 $. On $ [0, 2] $, $ 2x \ge x^2 $, so the line is the upper edge.

Step 2(a) $ R = 2x $, $ r = x^2 $: $ V = \pi\displaystyle\int_0^2 (4x^2-x^4)\,dx = \pi\left(\tfrac{32}{3}-\tfrac{32}{5}\right) = \tfrac{64\pi}{15} $.

Step 3(b) The axis is 1 unit below the $ x $-axis, so add 1 to each height: $ R = 2x + 1 $, $ r = x^2 + 1 $. Then $ R^2-r^2 = 2x^2 + 4x-x^4 $ and $ V = \pi\left(\tfrac{16}{3} + 8-\tfrac{32}{5}\right) = \tfrac{104\pi}{15} $.

Step 4(c) A vertical axis needs horizontal slices. Write the edges as $ x = \tfrac{y}{2} $ (from the line) and $ x = \sqrt{y} $ (from the parabola), for $ 0 \le y \le 4 $. The parabola is farther from the $ y $-axis: $ R = \sqrt{y} $, $ r = \tfrac{y}{2} $. $ V = \pi\displaystyle\int_0^4 \left(y-\frac{y^2}{4}\right)dy = \pi\left(8-\tfrac{16}{3}\right) = \tfrac{8\pi}{3} $.

Same region, three different solids. Parts (a) and (c) are within Cambridge 9709 Paper 1 and IB HL; part (b), an axis other than the coordinate axes, is AP only.

Region between y = 2x and y = x², rotated about y = −1

y = 2x (outer edge)y = x² (inner edge)axis of rotation y = −1

Figure 3: the strip at x turns into a washer. Its outer radius reaches the far edge, y = 2x, and its inner radius the near edge, y = x², both measured from the axis y = −1.

Common mistakeGetting the radii backwards when the axis is above the region. Rotate the same region about $ y = 5 $: now the lower curve $ y = x^2 $ is farther from the axis, so it gives the outer radius, $ R = 5-x^2 $, and the line gives $ r = 5-2x $. The volume is $ \pi\displaystyle\int_0^2 \big[(5-x^2)^2-(5-2x)^2\big]\,dx = \tfrac{136\pi}{15} $. Always draw the axis on your sketch and measure each radius from it.

Exam tipCambridge 9709 Paper 1 (topic 1.8) and IB HL (AHL 5.17) ask only for rotation about the $ x $-axis or the $ y $-axis, usually through $ 360^\circ $ and usually with an exact answer in terms of $ \pi $. AP (topics 8.9 to 8.12) also rotates about lines such as $ y = -1 $ or $ x = 3 $, and on the calculator section expects three decimal places. AP free-response questions often ask you to “write, but do not evaluate, an integral expression”: then the set-up, with correct limits and radii, is the whole answer.

Arc length (AP BC only)

How long is the curve $ y = f(x) $ from $ x = a $ to $ x = b $? Zoom in on a tiny piece. It is almost straight, with run $ dx $ and rise $ dy = f'(x)\,dx $, so by Pythagoras its length is $ \sqrt{dx^2 + dy^2} = \sqrt{1 + [f'(x)]^2}\,dx $. Add the pieces:

\[ L = \int_a^b \sqrt{1 + \big[f'(x)\big]^2}\,dx \]

The formula needs $ f’ $ to be continuous on $ [a, b] $.

Worked example

Find the length of the curve $ y = \tfrac{1}{3}\left(x^2 + 2\right)^{3/2} $ from $ x = 0 $ to $ x = 3 $.

Step 1Chain rule: $ f'(x) = \tfrac{1}{3} \cdot \tfrac{3}{2}\left(x^2 + 2\right)^{1/2} \cdot 2x = x\sqrt{x^2 + 2} $.

Step 2$ 1 + [f'(x)]^2 = 1 + x^2(x^2 + 2) = x^4 + 2x^2 + 1 = (x^2 + 1)^2 $, a perfect square.

Step 3$ L = \displaystyle\int_0^3 (x^2 + 1)\,dx = \left[\frac{x^3}{3} + x\right]_0^3 = 9 + 3 = 12 $.

Sense check: the straight chord between the end points has length about 11.61, and a curve can never be shorter than the chord joining its ends.

Most arc-length integrals have no simple antiderivative. The length of $ y = \sin x $ from 0 to $ \pi $ is $ \displaystyle\int_0^\pi \sqrt{1 + \cos^2 x}\,dx \approx 3.820 $, found on a calculator. Exam questions that want an exact answer are built, as above, so that $ 1 + [f’]^2 $ is a perfect square or a simple power. Arc length is not in Cambridge 9709 or IB Analysis and Approaches.

Common mistakeWriting $ \sqrt{1 + f'(x)} $ or $ \sqrt{1 + [f(x)]^2} $. It is the derivative that is squared, because the rise of a tiny piece is $ f'(x)\,dx $.

Economics: the Lorenz curve and the Gini coefficient

How can one number measure the inequality of a whole country? Line the population up from poorest to richest. The Lorenz curve $ L(x) $ is the share of total income received by the poorest fraction $ x $ of people. So $ L(0) = 0 $, $ L(1) = 1 $, and $ L(0.5) = 0.25 $ would mean the poorer half earns a quarter of all income. If everyone earned the same, the poorest $ x $ would earn exactly $ x $: the line of equality $ y = x $. Real Lorenz curves sag below it, and the more they sag, the more unequal the country.

Key termGini coefficient: the area between the line of equality and the Lorenz curve, divided by the whole area under the line of equality. Since that whole area is $ \tfrac{1}{2} $, $ G = \dfrac{\tfrac{1}{2}-\int_0^1 L(x)\,dx}{\tfrac{1}{2}} = 1-2\displaystyle\int_0^1 L(x)\,dx $. $ G = 0 $ is perfect equality; $ G $ close to 1 means almost all income goes to very few people.

Worked example

A country’s Lorenz curve is $ L(x) = 0.3x + 0.7x^3 $. (a) What share of income does the poorest half receive? (b) Find the Gini coefficient. (c) After a new tax-and-transfer scheme the curve becomes $ L(x) = 0.6x + 0.4x^3 $. Find the new Gini coefficient.

Step 1(a) $ L(0.5) = 0.15 + 0.7 \times 0.125 = 0.2375 $: the poorest half receives 23.75% of income.

Step 2(b) $ \displaystyle\int_0^1 (0.3x + 0.7x^3)\,dx = 0.15 + 0.175 = 0.325 $, so $ G = 1-2(0.325) = 0.35 $.

Step 3(c) $ \displaystyle\int_0^1 (0.6x + 0.4x^3)\,dx = 0.3 + 0.1 = 0.4 $, so $ G = 1-0.8 = 0.2 $.

The scheme lifts the Lorenz curve towards the line of equality and cuts the Gini coefficient from 0.35 to 0.2. Check that each curve is a genuine Lorenz curve: $ L(0) = 0 $, $ L(1) = 1 $, and $ L $ is increasing and concave up.

Lorenz curves before and after a tax-and-transfer scheme

line of equality y = xafter: L = 0.6x + 0.4x³, G = 0.2before: L = 0.3x + 0.7x³, G = 0.35

Figure 4: the shaded area between the line of equality and the original Lorenz curve is 0.175, which is 35% of the triangle under y = x, so G = 0.35. The new curve sits closer to the line, so its Gini coefficient is lower.

In practice Lorenz curves come from household surveys, so the area is found numerically, often with the trapezium rule. The same “area between curves” idea gives consumer and producer surplus (Step 26).

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. The region under $ y = 2x $ for $ 0 \le x \le 3 $ is rotated through $ 360^\circ $ about the $ x $-axis. Find the exact volume. Show worked answer

$ V = \pi\displaystyle\int_0^3 (2x)^2\,dx = \pi\int_0^3 4x^2\,dx = \pi\left[\frac{4x^3}{3}\right]_0^3 = 36\pi $.
Check: the solid is a cone of radius 6 and height 3, and $ \tfrac{1}{3}\pi(6^2)(3) = 36\pi $.

EasyQ2. (AP) The base of a solid is the triangle bounded by $ y = x $, the $ x $-axis and $ x = 3 $. Cross-sections perpendicular to the $ x $-axis are squares. Find the volume. Show worked answer

Side $ s = x $, so $ A(x) = x^2 $.
$ V = \displaystyle\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = 9 $.
(This is a pyramid with a $ 3 \times 3 $ square face and length 3: $ \tfrac{1}{3} \times 9 \times 3 = 9 $.)

MediumQ3. The region bounded by $ y = \sqrt{x} $, the $ y $-axis and the line $ y = 2 $ is rotated through $ 360^\circ $ about the $ y $-axis. Find the exact volume. Show worked answer

Horizontal discs: at height $ y $ the radius is $ x = y^2 $, for $ 0 \le y \le 2 $.
$ V = \pi\displaystyle\int_0^2 (y^2)^2\,dy = \pi\int_0^2 y^4\,dy = \pi\left[\frac{y^5}{5}\right]_0^2 = \tfrac{32\pi}{5} $.

MediumQ4. The region enclosed by $ y = x^2 $ and $ y = 4 $ is rotated through $ 360^\circ $ about the $ x $-axis. Find the exact volume. Show worked answer

The curves meet at $ x = \pm 2 $. Washers: $ R = 4 $ (the line), $ r = x^2 $ (the parabola).
$ V = \pi\displaystyle\int_{-2}^{2} \left(16-x^4\right)dx = \pi\left[16x-\frac{x^5}{5}\right]_{-2}^{2} $.
$ = \pi\left(64-\tfrac{64}{5}\right) = \tfrac{256\pi}{5} $.

MediumQ5. (AP BC) Find the exact length of the curve $ y = x^{3/2} $ from $ x = 0 $ to $ x = 4 $. Show worked answer

$ f'(x) = \tfrac{3}{2}x^{1/2} $, so $ 1 + [f'(x)]^2 = 1 + \tfrac{9}{4}x $.
$ L = \displaystyle\int_0^4 \left(1 + \tfrac{9}{4}x\right)^{1/2}dx = \left[\tfrac{4}{9} \cdot \tfrac{2}{3}\left(1 + \tfrac{9}{4}x\right)^{3/2}\right]_0^4 $.
$ = \tfrac{8}{27}\left(10^{3/2}-1\right) = \tfrac{8}{27}\left(10\sqrt{10}-1\right) \approx 9.073 $.

ExamQ6. The region $ \mathcal{R} $ is enclosed by $ y = 4x-x^2 $ and $ y = x $. (a) Find the area of $ \mathcal{R} $. [3] (b) Find the exact volume of the solid formed when $ \mathcal{R} $ is rotated through $ 360^\circ $ about the $ x $-axis. [4] (c) (AP only) Write, but do not evaluate, an integral expression for the volume when $ \mathcal{R} $ is rotated about the line $ y = 4 $. [2] Show worked answer

(a) $ 4x-x^2 = x $ gives $ x(3-x) = 0 $, so $ x = 0 $ and $ x = 3 $. [B1]
$ A = \displaystyle\int_0^3 (3x-x^2)\,dx = \left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3 $ [M1] $ = \tfrac{27}{2}-9 = \tfrac{9}{2} $. [A1]
(b) $ R = 4x-x^2 $, $ r = x $: $ V = \pi\displaystyle\int_0^3 \big[(4x-x^2)^2-x^2\big]\,dx $. [M1]
$ (4x-x^2)^2-x^2 = 15x^2-8x^3 + x^4 $. [A1]
$ V = \pi\left[5x^3-2x^4 + \frac{x^5}{5}\right]_0^3 = \pi\left(135-162 + \tfrac{243}{5}\right) $ [M1] $ = \tfrac{108\pi}{5} $. [A1]
(c) The axis $ y = 4 $ lies on or above $ \mathcal{R} $, so the lower edge $ y = x $ is farther away: $ R = 4-x $, $ r = 4-(4x-x^2) = (x-2)^2 $. [M1]
$ V = \pi\displaystyle\int_0^3 \big[(4-x)^2-(x-2)^4\big]\,dx $. [A1] (Its value is $ \tfrac{72\pi}{5} $.)
Total: 9 marks.

Check yourself

1. Which integral gives the volume when the region under $ y = f(x) $, $ a \le x \le b $, is rotated about the $ x $-axis?

Each slice is a disc of radius $ y $ and thickness $ dx $, with volume $ \pi y^2\,dx $.

2. A washer has outer radius 3 and inner radius 2. What is the area of its face?

$ \pi R^2-\pi r^2 = 9\pi-4\pi = 5\pi $. Using $ (R-r)^2 $ gives the wrong answer $ \pi $.

3. What is the Gini coefficient for the Lorenz curve $ L(x) = x^3 $?

$ \displaystyle\int_0^1 x^3\,dx = \tfrac{1}{4} $, so $ G = 1-2 \times \tfrac{1}{4} = \tfrac{1}{2} $.

Your score: 0 out of 3

The short version

Known cross-sections (AP): $ V = \displaystyle\int_a^b A(x)\,dx $, with $ A $ from the side $ s $: $ s^2 $ for squares, $ \tfrac{\pi}{8}s^2 $ for semicircles on a diameter, $ \tfrac{\sqrt{3}}{4}s^2 $ for equilateral triangles.
Discs: $ \pi\displaystyle\int y^2\,dx $ about the $ x $-axis; $ \pi\displaystyle\int x^2\,dy $ about the $ y $-axis, with $ y $-limits.
Washers: $ \pi\displaystyle\int (R^2-r^2) $, with each radius measured from the axis. Axes other than the coordinate axes are AP only.
Arc length (AP BC only): $ \displaystyle\int_a^b \sqrt{1 + [f'(x)]^2}\,dx $, usually on a calculator.
Gini coefficient: $ G = 1-2\displaystyle\int_0^1 L(x)\,dx $, between 0 (equality) and 1.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

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Next: step 28 of 33Parametric, Polar and Vector-Valued Functions

Cross-section, side \( s \) in the base	Area \( A \)
square	\( s^2 \)
rectangle of height \( h \)	\( sh \)
equilateral triangle	\( \tfrac{\sqrt{3}}{4}s^2 \)
isosceles right triangle with a leg \( s \)	\( \tfrac{1}{2}s^2 \)
semicircle with diameter \( s \)	\( \tfrac{\pi}{8}s^2 \)

Step 27: Volumes of Revolution and Arc Length

Volumes by slicing: known cross-sections

Discs: rotating about the \( x \)- and \( y \)-axes

Washers and other axes of rotation

Arc length (AP BC only)

Economics: the Lorenz curve and the Gini coefficient

Practise: 6 questions

Check yourself

The short version

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