Volumes of revolution (WACE Year 12 Mathematics Specialist Unit 4)
“Find the volume of a solid generated by rotating a region about the x-axis or y-axis using the disc method”
Rotating about the -axis from to sweeps out discs of radius , so the volume is . Rotating about the -axis instead integrates with written as a function of .
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What this dot point is asking
SCSA wants you to set up and evaluate volumes of revolution by the disc method, choosing the axis, the variable of integration and the limits correctly.
What a solid of revolution is
A solid of revolution is generated by rotating a plane region a full turn about a fixed straight line, the axis of revolution. Because of the rotation, every cross-section perpendicular to the axis is a circle (or, between two curves, an annulus), so the solid is rotationally symmetric. Familiar shapes arise this way: rotating a rectangle gives a cylinder, rotating a right triangle gives a cone, and rotating a semicircle gives a sphere. The disc method computes the volume of any such solid by summing the areas of these circular cross-sections along the axis, which is why the technique reduces to a single definite integral of .
The disc method
When a region under is rotated about the -axis, each thin slice becomes a disc of radius and thickness , with area . Summing gives
The radius is the distance from the axis to the curve, which here is just . The squared radius is essential: forgetting to square is the classic error.
The reasoning behind the formula is a slicing argument. Cut the solid into thin slabs perpendicular to the axis of rotation. Each slab is approximately a cylinder (a disc) of radius and thickness , so its volume is approximately . Adding the slabs and letting turns the sum into the definite integral . Seeing the volume as an accumulation of disc areas explains why the integrand is the cross-sectional area and why the variable of integration must run along the axis of rotation.
Rotating about the y-axis
Rotating about the -axis makes the radius the horizontal distance , so you express as a function of and integrate over :
The limits and are now -values. Match the variable of integration to the axis of rotation.
Rotating the region between two curves
If the region between an outer curve and an inner curve is rotated about the -axis, each slice is an annulus, so . Subtract the squares, not the radii.
Setting limits
The limits are where the region begins and ends along the axis of rotation, often the intersection points of the bounding curves with each other or with the axis. A sketch confirms them.
Matching the integral to the axis
The single most important set-up decision is matching the variable of integration to the axis of rotation. For rotation about the -axis, the discs are stacked along , so integrate with -limits. For rotation about the -axis, the discs are stacked along , so express the radius as a function of and integrate with -limits. Mixing the two, for instance integrating in for a -axis rotation, gives a meaningless answer. A clear sketch with a single representative disc drawn perpendicular to the axis makes the correct choice obvious.
Rotation about a line other than an axis
A harder variant rotates the region about a horizontal or vertical line that is not a coordinate axis, such as or . The method is the same, but the radius becomes the distance from that line to the curve rather than from the axis. Rotating about the line uses a radius of , so the volume is . Reading the radius as a distance, rather than just as the function value, is the key adjustment, and it also covers the washer case where an inner and outer distance are each measured from the same line.
Find the volume when for is rotated about the -axis.
Step 1: Identify the disc radius and set up the integral
When a curve is rotated about the -axis, each thin vertical slice sweeps out a disc whose radius equals the -value of the curve at that position. The volume formula sums infinitely many disc areas across the interval.
The radius of each disc is , so . The volume is
Step 2: Evaluate the integral
We integrate with respect to using the power rule, then apply the limits of integration.
Final answer: The volume of the solid of revolution is cubic units.
- Forgetting to square the radius
- The integrand is , not .
- Subtracting radii instead of squares
- For a region between two curves, subtract , never .
- Mismatched variable and axis
- Rotation about the -axis means integrate in with as a function of .
Exam-style questions
Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.
WACE 2022-style7 marksCalculator-assumed. The region between and is rotated about the -axis. Find the volume of the solid generated.Show worked answer →
A washer (annulus) volume between two curves.
Intersections: gives and . On the line is the outer radius and the parabola is the inner radius .
Volume .
At : . So the volume is cubic units.
Markers reward identifying the outer and inner radii, subtracting the squares (not the radii), the integration, and the volume .
WACE 2024-style5 marksCalculator-free. The region bounded by , the -axis and is rotated about the -axis. Find the volume of the solid generated.Show worked answer →
A rotation about the -axis.
Rotation about the -axis means the radius is , expressed in terms of . From , . The region runs from to .
Volume cubic units.
Markers reward expressing , integrating in with the correct limits, and the volume .