Year 12 Maths Extension 1 (2027) Further applications of calculus

Volumes of solids of revolution

Q: How do you find a volume of revolution?

About the x-axis, V equals pi times the integral of y squared dx; about the y-axis, V equals pi times the integral of x squared dy.

Q: How do you rotate about the y-axis?

Rewrite x in terms of y, square it, and integrate pi times x squared dy over the y-limits.

Q: What is the formula between two curves?

V equals pi times the integral of f squared minus g squared dx, outer squared minus inner squared.

Q: Why square before integrating?

Each disc has area pi times radius squared, so you integrate pi y squared, not the square of an integral.

Q: Do you always include pi?

Yes, the disc area pi r squared carries the factor pi into the volume.

20 practice questions 2 video lessons Theory + worked examples

Explore volumes of solids of revolution in NSW Year 12 Mathematics Extension 1 (NESA outcome ME1-12-05). Rotating a region about an axis sweeps out a solid whose volume can be found by integration.

You will learn to set up and evaluate the volume integral for rotation about either the $x$- or $y$-axis, including regions between two curves — a powerful application of integration used in practical and purely mathematical problems in Extension 1.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)

Volumes of solids of revolution introduction Watch
Volume of solids of revolution: between 2 curves Watch

Theory

A solid of revolution is built from thin discs: about the $x$ -axis $V = π \int_{a}^{b} y^{2} d x$ , about the $y$ -axis $V = π \int_{c}^{d} x^{2} d y$ , and between two curves $V = π \int_{a}^{b} (f^{2} - g^{2}) d x$ .

A solid of revolution is formed by rotating a region about an axis. Each thin slice perpendicular to the axis is a disc, and summing their volumes gives an integral.

About the $x$ -axis (region under $y = f (x)$ ): $V = π \int_{a}^{b} y^{2} d x$ . About the $y$ -axis (write $x$ in terms of $y$ ): $V = π \int_{c}^{d} x^{2} d y$ .

Between two curves about the $x$ -axis (outer $f$ , inner $g$ ): $V = π \int_{a}^{b} (f (x)^{2} - g (x)^{2}) d x$ .

Students rotate an arc about the $x$ - or $y$ -axis and find the volume, including the region between two curves.

Rotating the region under

y = f (x)

about the

x

-axis sweeps out a solid.

A representative disc has radius

y

and thickness

d x

;

V = π \int y^{2} d x

V = π \int_{a}^{b} y^{2} d x (about the x -axis), V = π \int_{c}^{d} x^{2} d y (about the y -axis) .

V = pi integral y^2 dx (x-axis); V = pi integral x^2 dy (y-axis)

Between two curves about the $x$ -axis (outer $f$ , inner $g$ ):

V = π \int_{a}^{b} (f (x)^{2} - g (x)^{2}) d x .

V = pi integral (f^2 - g^2) dx

Square first. Use $π \int y^{2} d x$ , not ${(π \int y d x)}^{2}$ . Between curves subtract $f^{2} - g^{2}$ (outer squared minus inner squared), never $(f - g)^{2}$ .

How to find a volume of revolution

Identify the axis and write the radius: $y = f (x)$ about the $x$ -axis, or $x$ in terms of $y$ about the $y$ -axis.
Square the radius and set up $π \int (radius)^{2}$ .
For two curves, subtract $f^{2} - g^{2}$ (outer minus inner, both squared).
Integrate over the correct limits for an exact volume.

Example 1 — About the x-axis

The region under

y = x^{2}

from

x = 0

x = 2

is rotated about the

x

-axis. Find the volume.

Solution

$V$	$=$	$π \int_{0}^{2} (x^{2})^{2} d x = π [\frac{x^{5}}{5}]_{0}^{2}$
	$=$	$\frac{32 π}{5}$

V = 32 pi / 5

$V = \frac{32 π}{5}$ units $^{3}$ .

Example 2 — An exponential

The region under

y = e^{x}

from

x = 0

x = 1

is rotated about the

x

-axis. Find the volume.

Solution

$V$	$=$	$π \int_{0}^{1} e^{2 x} d x = π [\frac{1}{2} e^{2 x}]_{0}^{1}$
	$=$	$\frac{π}{2} (e^{2} - 1)$

V = (pi/2)(e^2 - 1)

$V = \frac{π}{2} (e^{2} - 1)$ units $^{3}$ .

Example 3 — Between two curves

The region between

y = x

and

y = x^{2}

(first quadrant) is rotated about the

x

-axis. Find the volume.

Solution

They meet at $x = 0, 1$ ; outer $y = x$ .

$V$	$=$	$π \int_{0}^{1} (x^{2} - x^{4}) d x$
	$=$	$π [\frac{x^{3}}{3} - \frac{x^{5}}{5}]_{0}^{1} = \frac{2 π}{15}$

V = 2 pi / 15

$V = \frac{2 π}{15}$ units $^{3}$ .

Example 4 — About the y-axis

The region under

y = x^{2}

from

y = 0

y = 4

is rotated about the

y

-axis. Find the volume.

Solution

$x^{2} = y$ .

$V$	$=$	$π \int_{0}^{4} x^{2} d y = π \int_{0}^{4} y d y$
	$=$	$π [\frac{y^{2}}{2}]_{0}^{4} = 8 π$

V = 8 pi

$V = 8 π$ units $^{3}$ .

Common pitfalls

Squaring order. Square the function before integrating:

π \int y^{2} d x

, not

{(π \int y d x)}^{2}

Wrong axis setup. About the

y

-axis, rewrite

x

in terms of

y

and integrate in

y

$(f - g)^{2}$ error. Between two curves subtract

f^{2} - g^{2}

, not

(f - g)^{2}

Dropping $π$ . Every disc volume carries the factor

π

Frequently asked questions

How do you find a volume of revolution?

About the $x$ -axis, $V = π \int_{a}^{b} y^{2} d x$ ; about the $y$ -axis, $V = π \int_{c}^{d} x^{2} d y$ .

How do you rotate about the y-axis?

Rewrite $x$ in terms of $y$ , square it, and integrate $π \int x^{2} d y$ over the $y$ -limits.

What is the formula between two curves?

$V = π \int_{a}^{b} (f^{2} - g^{2}) d x$ — outer squared minus inner squared.

Why square before integrating?

Each disc has area $π (radius)^{2}$ , so you integrate $π y^{2}$ , not the square of an integral.

Do you always include pi?

Yes — the disc area $π r^{2}$ carries the factor $π$ into the volume.

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More in Further applications of calculus

Volumes of solids of revolution

📖 Theory

How to find a volume of revolution

Common pitfalls

Frequently asked questions

Theory