Short on time? This page is the recap. For the picture, the reasoning and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: π ∫ (a to b) [f(x)]² dx means π times the definite integral of [f(x)]² from x = a to x = b.
Recap
- Spinning a region about an axis it touches gives a solid whose slices, cut at right angles to the axis, are discs.
- Each disc has area π(radius)². The volume is the integral of that area along the axis. It is the cross-section method of Topics 8.7 and 8.8 with a circular slice.
- The radius is the distance from the axis to the curve, measured at right angles to the axis.
- About the y-axis, slices are horizontal: write the curve as x = g(y) and use y-limits.
- Volumes are in cubic units and are never negative.
Key relationships
| Axis | Slice | Radius | Volume |
|---|---|---|---|
| x-axis | vertical, thickness dx | f(x) | π ∫ (a to b) [f(x)]² dx |
| y-axis | horizontal, thickness dy | g(y), from x = g(y) | π ∫ (c to d) [g(y)]² dy |
| Any curve below the x-axis | vertical | |f(x)| | still π ∫ [f(x)]² dx (the square removes the sign) |
| Check | — | widest radius R, length L | V is less than the cylinder πR²L |
Useful rewrites for the y-axis: y = x² (x ≥ 0) → x = √y; y = x³ → x = y^(1/3); y = eˣ → x = ln y.
Assumptions behind the method
- The region touches the axis along the whole interval. If there is a gap, the slices are washers (Topic 8.11).
- The solid is formed by one full turn.
- The radius function is continuous on the interval.
Mistakes to avoid
- No π, or 2π instead of π.
- Not squaring the radius, or squaring the integral instead of the integrand.
- dx with a y-axis rotation (or y-limits with dx).
- Using the diameter as the radius.
- Integrating (3x − x²)² without expanding (or without a substitution).
- Square units on a volume.
- Rounding early on a calculator question; give three decimal places at the end.
Quick self-check
- Region under y = √x for 0 ≤ x ≤ 4, spun about the x-axis. Volume? (π ∫ (0 to 4) x dx = 8π)
- Region between y = x² (x ≥ 0), the y-axis and y = 4, spun about the y-axis. Volume? (x = √y, so π ∫ (0 to 4) y dy = 8π)
- True or false: π ∫ (0 to 2) x³ dx is the volume when the region under y = x³ on [0, 2] is spun about the x-axis. (False: the radius x³ must be squared, giving π ∫ (0 to 2) x⁶ dx.)
Next: practice questions.