Volumes of revolution about either coordinate axis | Year 12

01 / LEARN

Understand the idea.

Square the radius, not the function difference

A slice perpendicular to the rotation axis becomes a disk or washer. Its area is $π R^{2}$ for a disk and $π (R^{2} - r^{2})$ for a washer. Measure both radii from the axis to the region's boundaries.

The washer is an outer disk with an inner disk removed, so subtract the squares of the radii. Squaring their difference would describe a different area. Check that $R \geq r \geq 0$ before setting up the integral.

Describe the bounded region completely

Describe the complete bounded region and its axis of rotation. About the $x$ axis, vertical slices commonly use $d x$ and radii expressed as $y$ values. About the $y$ axis, horizontal slices commonly use $d y$ and radii expressed as $x$ values.

Integrate the cross-sectional area over the slice thickness. The result has cubic units and cannot be negative. Compare it with a containing cylinder for scale, and retain exact multiples of $π$ unless the question asks for a decimal approximation.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Rotate the region 0<=x<=4, 0<=y<=sqrt(x) about the x axis.

Worked solution

1
Identify the disk radius.
$R = \sqrt{x}, R^{2} = x, 0 \leq x \leq 4$
The region meets the rotation axis, so there is no inner hole.
2
Integrate the cross-sectional area.
$V = π \int_{0}^{4} x d x = π {[\frac{x^{2}}{2}]}_{0}^{4}$
A vertical slice rotates into a disk perpendicular to the x axis.
3
Evaluate and check the scale.
$V = 8 π cubic units < π {(2)}^{2} (4) = 16 π cubic units$
The solid fits inside a cylinder of radius 2 and length 4, consistent with the calculated volume.

Answer

$8 π$ cubic units

EXAMPLE 02

Worked example 2

Rotate 0<=y<=2, 0<=x<=y squared about the y axis.

Worked solution

1
Use horizontal disks.
$R = y^{2}, R^{2} = y^{4}, 0 \leq y \leq 2$
The distance from the y axis to the outer boundary is the x coordinate, y squared.
2
Set up and integrate the volume.
$V = π \int_{0}^{2} y^{4} d y = π {[\frac{y^{5}}{5}]}_{0}^{2}$
Each horizontal cross-section is a filled disk, and its thickness is dy.
3
Evaluate with cubic units.
$V = \frac{32 π}{5} cubic units$
Squaring the radius before integration produces the fourth power and the correct volume.

Answer

$\frac{32 π}{5}$ cubic units

03 / YOUR TURN

Try it yourself.

Check your understanding

What units should a volume calculated from metre coordinates have?

Show answer & reasoning

Cubic metres.

Cross-sectional area is measured in $m^{2}$ , and slice thickness in $m$ . Their product has units $m^{3}$ , so the volume is measured in cubic metres.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Rotate $0 \leq x \leq 1$ , $0 \leq y \leq x^{2}$ about the $x$ axis.
2 marks
Rotate $0 \leq x \leq \sqrt{2}$ , $x^{2} \leq y \leq 2$ about the $x$ axis.
2 marks
Rotate $0 \leq y \leq 3$ , $0 \leq x \leq y$ about the $y$ axis.
2 marks
Rotate $1 \leq x \leq 2$ , $1 \leq y \leq x$ about the $x$ axis.
2 marks
Rotate the rectangle $1 \leq x \leq 2$ , $0 \leq y \leq 3$ about the $y$ axis.
2 marks
The region under $y = k x$ from $x = 0$ to 1 is rotated about the $x$ axis, with $k > 0$ . If volume is $12 π$ , find $k$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Rotate $0 \leq x \leq 1$ , $0 \leq y \leq x^{2}$ about the $x$ axis.
2 marks
The radius squared is $x^{4}$ .
Integrate $π$ times $x^{4}$ from 0 to 1.
$\frac{π}{5}$ cubic units
Rotate $0 \leq x \leq \sqrt{2}$ , $x^{2} \leq y \leq 2$ about the $x$ axis.
2 marks
Use washer radii 2 and $x^{2}$ .
Integrate $π (4 - x^{4})$ to $\sqrt{2}$ .
$\frac{16 \sqrt{2} π}{5}$ cubic units
Rotate $0 \leq y \leq 3$ , $0 \leq x \leq y$ about the $y$ axis.
2 marks
The radius is $y$ .
$π \int_{0}^{3} y^{2} d y = 9 π$ .
$9 π$ cubic units
Rotate $1 \leq x \leq 2$ , $1 \leq y \leq x$ about the $x$ axis.
2 marks
Use outer radius $x$ and inner radius 1.
Integrate $π (x^{2} - 1)$ from 1 to 2.
$\frac{4 π}{3}$ cubic units
Rotate the rectangle $1 \leq x \leq 2$ , $0 \leq y \leq 3$ about the $y$ axis.
2 marks
The annular cross-section area is $π (4 - 1) = 3 π$ .
Multiply by height 3.
$9 π$ cubic units
The region under $y = k x$ from $x = 0$ to 1 is rotated about the $x$ axis, with $k > 0$ . If volume is $12 π$ , find $k$ .
2 marks
The volume is $\frac{π k^{2}}{3}$ .
Solve $k^{2} = 36$ and use $k > 0$ .
$k = 6$ .

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

Official curriculum source ↗

In this lesson

Understand the idea.

Square the radius, not the function difference

Describe the bounded region completely

See the method in action.

Worked example 1

Identify the disk radius.

Integrate the cross-sectional area.

Evaluate and check the scale.

Worked example 2

Use horizontal disks.

Set up and integrate the volume.

Evaluate with cubic units.

Try it yourself.

Check your understanding

Keep practising