Volumes of revolution about either coordinate axis | Year 12

01 / LEARN

The idea, step by step

Square the radius, not the function difference

Rotating a filled region around an axis creates circular cross-sections. For disks, $V = π \int R^{2}$ . For washers, $V = π \int (R^{2} - r^{2})$ , with outer radius R and inner radius r measured from the rotation axis.

The washer integrand is not ${(R - r)}^{2}$ . Draw a slice and identify both radii. For the x axis, vertical slices commonly use dx; for the y axis, horizontal slices commonly use dy.

Describe the bounded region completely

State all boundaries and the interval before integrating. A curve alone does not describe a solid of revolution. When working in y, express the radius as x in terms of y. Check that the outer radius is at least the inner radius.

Volume uses cubic units and cannot be negative. A sensible rough cylinder bound is a useful check. Exact multiples of pi are often more informative than rounded decimals; round only when requested.

A common trap

For a washer, subtract the radii and then square.

Instead: Subtract their squared areas: outer radius squared minus inner radius squared.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

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

The disk radius is sqrt(x), so its square is x.
$V = π \int_{0}^{4} x d x = 8 π$ .

8 π

cubic units

Example 2

Worked example 2

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

Horizontal disk radius is x=y squared.
$V = π \int_{0}^{2} y^{4} d y = 32 π / 5$ .

32 π / 5

cubic units

Check your understanding

What units should a volume calculated from metre coordinates have?

Show answer

Cubic metres.

Cross-section area contributes square metres and the slice thickness contributes metres.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Rotate 0<=x<=1, 0<=y<=x squared about the x axis.
2 marks
Rotate 0<=x<=sqrt(2), x squared<=y<=2 about the x axis.
2 marks
Rotate 0<=y<=3, 0<=x<=y about the y axis.
2 marks
Rotate 1<=x<=2, 1<=y<=x about the x axis.
2 marks
Rotate the rectangle 1<=x<=2, 0<=y<=3 about the y axis.
2 marks
The region under y=kx from x=0 to 1 is rotated about the x axis, with k>0. If volume is 12pi, find k.
2 marks

Read the worked solutions

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

Rotate 0<=x<=1, 0<=y<=x squared about the x axis.
2 marks
The radius squared is x to the fourth.
Integrate pi times x to the fourth from 0 to 1.
$π / 5$ cubic units
Rotate 0<=x<=sqrt(2), x squared<=y<=2 about the x axis.
2 marks
Use washer radii 2 and x squared.
Integrate $π (4 - x^{4})$ to sqrt(2).
$16 \sqrt{2} π / 5$ cubic units
Rotate 0<=y<=3, 0<=x<=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<=x<=2, 1<=y<=x about the x axis.
2 marks
Use outer radius x and inner radius 1.
Integrate $π (x^{2} - 1)$ from 1 to 2.
$4 π / 3$ cubic units
Rotate the rectangle 1<=x<=2, 0<=y<=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=kx from x=0 to 1 is rotated about the x axis, with k>0. If volume is 12pi, find k.
2 marks
The volume is $π k^{2} / 3$ .
Solve k squared=36 and use k>0.
$k = 6$ .

Checking saved progress…

Continue learning

Interpreting a result in context

Next lesson

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 ↗

Back to Integration techniques and applications