Volumes of revolution

AS · P1 · 12 min

Spin a region of the plane through a full turn about an axis and it sweeps out a solid: a curve becomes the outline of a vase, a bowl or a nose cone. The volume of such a solid of revolution can be found by integration, using the same idea of thin slices that gives areas. The syllabus asks for the volume of revolution about either the $x$ -axis or the $y$ -axis, and explicitly includes regions not bounded by the axis of rotation, such as the region between $y = 9 - x^2$ and $y = 5$ rotated about the $x$ -axis. A volume question appears on most Paper 1s, usually as the last part of a longer integration question.

Slicing into discs

Take the region under $y = f(x)$ from $x = a$ to $x = b$ and rotate it through $360^\circ$ about the $x$ -axis. Cut the solid into thin slices perpendicular to the axis. A slice at position $x$ , of thickness $\delta x$ , is almost a flat cylinder (a disc) of radius $y$ , the height of the curve there. Its volume is

\delta V \approx \pi y^2\, \delta x

Adding all the discs and letting their thickness tend to zero turns the sum into an integral, just as thin strips gave the area under a curve.

Rotating the curve about the x-axis. One thin slice is a disc of radius y and thickness δx, with volume about πy²δx.

Key result

Rotating the region between $y = f(x)$ , the $x$ -axis and the lines $x = a$ , $x = b$ through $360^\circ$ about the $x$ -axis:

V = \pi\int_a^b y^2\, dx

Rotating the region between $x = g(y)$ , the $y$ -axis and the lines $y = c$ , $y = d$ through $360^\circ$ about the $y$ -axis:

V = \pi\int_c^d x^2\, dy

The $\pi$ comes from the area of the circular face, $\pi r^2$ . It is a constant, so take it outside the integral and multiply at the end.

As a check, rotate $y = x$ from $0$ to $3$ about the $x$ -axis. The solid is a cone of radius $3$ and height $3$ , so its volume should be $\tfrac{1}{3}\pi r^2 h = 9\pi$ . The formula gives $\pi\displaystyle\int_0^3 x^2\, dx = \pi\left[\tfrac{x^3}{3}\right]_0^3 = 9\pi$ .

About the y-axis

For rotation about the $y$ -axis the slices are horizontal, so the radius of each disc is $x$ and the thickness is $\delta y$ . Everything must be in terms of $y$ :

rearrange the equation of the curve to get $x^2$ in terms of $y$ (you often do not need $x$ itself);
the limits are $y$ -values, so convert any $x$ -limits using the curve.

For $y = x^2 + 2$ , for example, $x^2 = y - 2$ directly, which is all the formula needs.

Regions not touching the axis: washers

If the region does not reach the axis of rotation, the solid has a hole through the middle. Each slice is a washer: a disc of outer radius $y_1$ (the curve further from the axis) with a disc of inner radius $y_2$ removed. Its volume is $\pi y_1^2\,\delta x - \pi y_2^2\,\delta x$ .

Region between two graphs

If $y_1 \ge y_2 \ge 0$ for $a \le x \le b$ , the region between them rotated about the $x$ -axis has volume

V = \pi\int_a^b \left(y_1^2 - y_2^2\right) dx

This is not $\pi\displaystyle\int_a^b (y_1 - y_2)^2\, dx$ .

Equivalently, find the volume generated by the outer boundary and subtract the volume generated by the inner boundary. When the inner boundary is a horizontal line, the inner solid is a cylinder, and when it is a line through the origin it is a cone, so you may be able to use $\pi r^2 h$ or $\tfrac{1}{3}\pi r^2 h$ instead of integrating.

y = 9 - x^2 y = 5 (-2, 5) (2, 5)

The region between $y = 9 - x^2$ and $y = 5$ (above the line, below the curve) does not touch the $x$ -axis. Rotated about the $x$ -axis, it makes a solid ring with a cylindrical hole of radius $5$ . This is the syllabus example, worked in Example 4.

Volume of revolution

Sketch the region and identify the axis of rotation.
For the $x$ -axis: write $y^2$ in terms of $x$ , with $x$ -limits. For the $y$ -axis: write $x^2$ in terms of $y$ , with $y$ -limits.
If the region does not touch the axis, use (outer) $^2$ $-$ (inner) $^2$ , or subtract the volume of a simple solid.
Expand $y^2$ fully before integrating (unless it is a single $(ax + b)^n$ bracket).
Integrate, substitute the limits, and multiply by $\pi$ . Leave the answer as an exact multiple of $\pi$ unless told otherwise.

Worked examples

A polynomial about the x-axis

The region bounded by the curve $y = x^2 + 1$ , the axes and the line $x = 2$ is rotated through $360^\circ$ about the $x$ -axis. Find the exact volume of the solid formed.

Solution

$y^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1$ .

V = \pi\int_0^2 \left(x^4 + 2x^2 + 1\right) dx = \pi\left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^2 = \pi\left(\frac{32}{5} + \frac{16}{3} + 2\right) = \frac{206\pi}{15}

A common error is to square $x^2 + 1$ as $x^4 + 1$ . The middle term $2x^2$ is essential.

y = x^2 + 1 fill 0 2 y = x^2 + 1 x = 2

A bracket to a power

The region bounded by the curve $y = \dfrac{6}{3x + 2}$ , the axes and the line $x = 2$ is rotated through $360^\circ$ about the $x$ -axis. Find the volume generated.

Solution

$y^2 = \dfrac{36}{(3x + 2)^2} = 36(3x + 2)^{-2}$ . Using the $(ax + b)^n$ rule:

\int 36(3x + 2)^{-2}\, dx = \frac{36(3x + 2)^{-1}}{3 \times (-1)} = -\frac{12}{3x + 2}

V = \pi\left[-\frac{12}{3x + 2}\right]_0^2 = \pi\left(-\frac{12}{8} + \frac{12}{2}\right) = \pi\left(-\frac{3}{2} + 6\right) = \frac{9\pi}{2}

About the y-axis

The region bounded by the curve $y = x^2 + 2$ , the $y$ -axis and the lines $y = 2$ and $y = 6$ is rotated through $360^\circ$ about the $y$ -axis. Find the volume generated.

Solution

Rearrange for $x^2$ : $x^2 = y - 2$ . The limits are $y = 2$ and $y = 6$ .

V = \pi\int_2^6 (y - 2)\, dy = \pi\left[\frac{y^2}{2} - 2y\right]_2^6 = \pi\big[(18 - 12) - (2 - 4)\big] = 8\pi

A region above a line

The region enclosed between the curve $y = 9 - x^2$ and the line $y = 5$ is rotated through $360^\circ$ about the $x$ -axis. Find the exact volume of the solid formed.

Solution

Intersections: $9 - x^2 = 5$ , so $x = \pm 2$ . The curve is the outer boundary and the line the inner one.

V = \pi\int_{-2}^{2} \left[(9 - x^2)^2 - 5^2\right] dx = \pi\int_{-2}^{2} \left(81 - 18x^2 + x^4 - 25\right) dx = \pi\int_{-2}^{2} \left(56 - 18x^2 + x^4\right) dx

= \pi\left[56x - 6x^3 + \frac{x^5}{5}\right]_{-2}^{2} = 2\pi\left(112 - 48 + \frac{32}{5}\right) = 2\pi \times \frac{352}{5} = \frac{704\pi}{5}

(The integrand is even, so the integral from $-2$ to $2$ is twice the integral from $0$ to $2$ .)

Alternatively: the curve alone gives $\pi\displaystyle\int_{-2}^{2}(9 - x^2)^2\, dx = \dfrac{1204\pi}{5}$ , and the inner solid is a cylinder of radius $5$ and length $4$ , volume $100\pi$ . Then $\dfrac{1204\pi}{5} - 100\pi = \dfrac{704\pi}{5}$ .

A reciprocal curve

The region bounded by the curve $y = \dfrac{2}{x}$ , the $x$ -axis and the lines $x = 1$ and $x = 4$ is rotated through $360^\circ$ about the $x$ -axis. Find the volume generated.

Solution

$y^2 = \dfrac{4}{x^2} = 4x^{-2}$ .

V = \pi\int_1^4 4x^{-2}\, dx = \pi\Big[-4x^{-1}\Big]_1^4 = \pi(-1 + 4) = 3\pi

Between a curve and a line, about both axes

The curve $y = 2\sqrt{x}$ and the line $y = x$ meet at the origin and at the point $A$ .

(a) Find the coordinates of $A$ .

(b) The region $R$ enclosed between the curve and the line is rotated through $360^\circ$ about the $x$ -axis. Find the volume generated.

Solution

(a) $2\sqrt{x} = x$ gives $4x = x^2$ , so $x = 0$ or $x = 4$ . $A$ is $(4, 4)$ .

(b) For $0 < x < 4$ , $2\sqrt{x} > x$ (at $x = 1$ : $2 > 1$ ), so the curve is the outer boundary. $y_1^2 = 4x$ , $y_2^2 = x^2$ :

V = \pi\int_0^4 \left(4x - x^2\right) dx = \pi\left[2x^2 - \frac{x^3}{3}\right]_0^4 = \pi\left(32 - \frac{64}{3}\right) = \frac{32\pi}{3}

(c) In terms of $y$ : the curve is $x = \dfrac{y^2}{4}$ , the line is $x = y$ . For $0 < y < 4$ , $y > \dfrac{y^2}{4}$ (at $y = 2$ : $2 > 1$ ), so now the line is further from the $y$ -axis and is the outer boundary.

V = \pi\int_0^4 \left(y^2 - \frac{y^4}{16}\right) dy = \pi\left[\frac{y^3}{3} - \frac{y^5}{80}\right]_0^4 = \pi\left(\frac{64}{3} - \frac{64}{5}\right) = \frac{128\pi}{15}

The roles of "outer" and "inner" swap between the two axes, so always check with a value.

y = 2 sqrt(x) y = x (4, 4)

Watch out

Forgetting $\pi$ . Every volume of revolution has a factor $\pi$ . Write it in the first line.

Not squaring. The formula uses $y^2$ , not $y$ . An integral of $y$ is an area.

Squaring a sum wrongly. $(x^2 + 1)^2 = x^4 + 2x^2 + 1$ , not $x^4 + 1$ ; $(2\sqrt{x})^2 = 4x$ , not $2x$ .

Using $(y_1 - y_2)^2$ . For a region between two graphs, the integrand is $y_1^2 - y_2^2$ . Squaring the difference gives the wrong solid.

Mixing variables about the $y$ -axis. The integrand must be $x^2$ written in $y$ , with $y$ -limits. Using $x$ -limits is a very common error.

Rotating only half a symmetrical region. If the region runs from $x = -2$ to $x = 2$ , integrate over the whole interval, or double the integral from $0$ to $2$ , not both.

Exam tip

Write the formula with your function in it first, for example $V = \pi\displaystyle\int_0^2 (x^2 + 1)^2\, dx$ . This usually earns a method mark for "use of $\pi\int y^2\, dx$ ".
Expand before integrating and show the expanded form. Mark schemes award a mark for the correct expansion.
Exact answers like $\dfrac{206\pi}{15}$ are expected when the question says "exact". Otherwise give $3$ significant figures, but an exact multiple of $\pi$ is always safe.
Volume questions often follow an area question on the same diagram. The limits found earlier (intersection points) are usually the limits for the volume.
Subtracting cones and cylinders is a legitimate shortcut. State the shape and the formula you use, such as "cylinder, $\pi r^2 h = 100\pi$ ".
Infinite regions: if the region extends to infinity, the integral is improper; use a letter for the limit.

Summary

About the $x$ -axis: $V = \pi\displaystyle\int_a^b y^2\, dx$ . About the $y$ -axis: $V = \pi\displaystyle\int_c^d x^2\, dy$ .
Each thin slice is a disc of volume $\pi(\text{radius})^2 \times \text{thickness}$ .
About the $y$ -axis, write $x^2$ in terms of $y$ and use $y$ -limits.
Region between two graphs: $V = \pi\displaystyle\int (y_1^2 - y_2^2)\, dx$ , outer squared minus inner squared.
Which graph is outer can change when you switch axes; check with a value.
Expand $y^2$ carefully; use the $(ax + b)^n$ rule for brackets.
Leave answers as exact multiples of $\pi$ .

Practice questions

Question

The region under $y = x$ from $x = 0$ to $x = 3$ is rotated through $360^\circ$ about the $x$ -axis. Find the volume by integration and check it with the formula for a cone.
Find the volume generated when the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated through $360^\circ$ about the $x$ -axis.
Find the volume generated when the region bounded by $y = \dfrac{1}{x}$ , the $x$ -axis, $x = 1$ and $x = 3$ is rotated through $360^\circ$ about the $x$ -axis.
The region bounded by $y = x^2 + 1$ , the $y$ -axis and the lines $y = 1$ and $y = 5$ is rotated through $360^\circ$ about the $y$ -axis. Find the volume.
Find the volume generated when the region bounded by $y = (x + 1)^2$ , the axes and the line $x = 1$ is rotated through $360^\circ$ about the $x$ -axis.
The region enclosed between the curve $y = 4 - x^2$ and the line $y = 3$ is rotated through $360^\circ$ about the $x$ -axis. Find the exact volume.
By rotating the region between $y = \sqrt{r^2 - x^2}$ and the $x$ -axis about the $x$ -axis, show that the volume of a sphere of radius $r$ is $\tfrac{4}{3}\pi r^3$ .
The region $R$ is enclosed between the curve $y = x^2$ and the line $y = 2x$ . Find the volume generated when $R$ is rotated through $360^\circ$ (a) about the $x$ -axis, (b) about the $y$ -axis.
The region bounded by $y = \dfrac{k}{x}$ , where $k > 0$ , the $x$ -axis and the lines $x = 1$ and $x = 4$ is rotated through $360^\circ$ about the $x$ -axis. The volume generated is $27\pi$ . Find $k$ .

Answers

$V = \pi\displaystyle\int_0^3 x^2\, dx = \pi\left[\tfrac{x^3}{3}\right]_0^3 = 9\pi$ . A cone of radius $3$ and height $3$ : $\tfrac{1}{3}\pi(9)(3) = 9\pi$ .
$y^2 = x$ : $V = \pi\displaystyle\int_0^4 x\, dx = \pi\left[\tfrac{x^2}{2}\right]_0^4 = 8\pi$ .
$y^2 = x^{-2}$ : $V = \pi\Big[-x^{-1}\Big]_1^3 = \pi\left(-\tfrac{1}{3} + 1\right) = \tfrac{2\pi}{3}$ .
$x^2 = y - 1$ : $V = \pi\displaystyle\int_1^5 (y - 1)\, dy = \pi\left[\tfrac{(y - 1)^2}{2}\right]_1^5 = 8\pi$ .
$y^2 = (x + 1)^4$ : $V = \pi\left[\dfrac{(x + 1)^5}{5}\right]_0^1 = \pi\left(\dfrac{32}{5} - \dfrac{1}{5}\right) = \dfrac{31\pi}{5}$ .
Intersections: $4 - x^2 = 3$ , $x = \pm 1$ . $V = \pi\displaystyle\int_{-1}^{1}\left[(4 - x^2)^2 - 9\right] dx = \pi\int_{-1}^{1}\left(7 - 8x^2 + x^4\right) dx = 2\pi\left[7x - \tfrac{8x^3}{3} + \tfrac{x^5}{5}\right]_0^1 = 2\pi\left(7 - \tfrac{8}{3} + \tfrac{1}{5}\right) = 2\pi \times \tfrac{68}{15} = \tfrac{136\pi}{15}$ .
$y^2 = r^2 - x^2$ . $V = \pi\displaystyle\int_{-r}^{r} (r^2 - x^2)\, dx = \pi\left[r^2x - \tfrac{x^3}{3}\right]_{-r}^{r} = \pi\left[\left(r^3 - \tfrac{r^3}{3}\right) - \left(-r^3 + \tfrac{r^3}{3}\right)\right] = \pi \times \tfrac{4r^3}{3} = \tfrac{4}{3}\pi r^3$ .
Intersections: $x^2 = 2x$ , $x = 0$ or $2$ , so the points are $(0, 0)$ and $(2, 4)$ . (a) For $0 < x < 2$ the line is above the curve. $V = \pi\displaystyle\int_0^2 \left(4x^2 - x^4\right) dx = \pi\left(\tfrac{32}{3} - \tfrac{32}{5}\right) = \tfrac{64\pi}{15}$ . (b) In terms of $y$ : the curve is $x = \sqrt{y}$ and the line is $x = \tfrac{y}{2}$ . For $0 < y < 4$ , $\sqrt{y} > \tfrac{y}{2}$ (at $y = 1$ : $1 > \tfrac{1}{2}$ ), so the curve is outer. $V = \pi\displaystyle\int_0^4 \left(y - \tfrac{y^2}{4}\right) dy = \pi\left[\tfrac{y^2}{2} - \tfrac{y^3}{12}\right]_0^4 = \pi\left(8 - \tfrac{16}{3}\right) = \tfrac{8\pi}{3}$ .
$V = \pi\displaystyle\int_1^4 k^2x^{-2}\, dx = \pi k^2\Big[-x^{-1}\Big]_1^4 = \pi k^2\left(1 - \tfrac{1}{4}\right) = \tfrac{3\pi k^2}{4}$ . Setting this equal to $27\pi$ : $k^2 = 36$ , $k = 6$ .

Slicing into discs

About the y-axis

Regions not touching the axis: washers

Worked examples

Practice questions

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