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AP® Calculus AB · Unit 8 · Topic 8.9

Volume with the Disc Method: Revolving Around the x- or y-Axis: Study Guide (Calculus AB 8.9)

Learn why spinning a region about the x- or y-axis gives circular slices, how to write the disc integral π∫r² with the right variable and limits, and how to evaluate it with and without a calculator.

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so integrals are written in a compact form: π ∫ (a to b) [f(x)]² dx means π times the definite integral of [f(x)]² from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top. Keep the square brackets and the small ² outside them: the whole radius is squared. The dx or dy tells you which variable the limits belong to.

From known cross sections to solids of revolution

In Topics 8.7 and 8.8 you built a solid from thin slices. If every slice at position x has area A(x) and thickness Δx, the slice’s volume is about A(x) Δx. Adding the slices and letting Δx shrink gives

V = ∫ (a to b) A(x) dx

A solid of revolution is a special case. Take a flat region and spin it through a full turn about a line in its plane, called the axis of revolution. If the region sits right against the axis, every point on a thin vertical strip traces out a circle about the axis. So each slice, cut at right angles to the axis, is a solid circle: a disc.

The radius of that disc is the length of the strip, which is the distance from the axis out to the curve. For a region between y = f(x) and the x-axis, that distance is f(x) (or |f(x)| if the curve is below the axis). The slice’s area is

A(x) = π × (radius)² = π[f(x)]²

and the volume is the integral of that area. Nothing new is happening: the disc method is the cross-section method with the shape fixed as a circle.

The region under y = 3x − x² spun about the x-axis, with one disc slice The arch y = 3x − x² rises from (0, 0) to a peak of height 2.25 at x = 1.5 and returns to the x-axis at (3, 0). The region between the arch and the x-axis is shaded. A dashed mirror image of the arch below the x-axis shows the outline of the solid after a full turn about the x-axis. At x = 1 a vertical line segment of length 2 runs from the x-axis up to the curve; it is labelled radius r = f(x). A tall thin ellipse centred on the x-axis at x = 1, reaching from height 2 to height −2, shows the circular disc that this strip sweeps out. A curved arrow near the right end of the x-axis indicates rotation about the x-axis. 123x 12−1−2y radius r = f(x) y = 3x − x² (solid curve) dashed: outline of the solid after one full turn ellipse at x = 1: the disc swept out by one strip spin about the x-axis
Figure 1. The shaded region under y = 3x − x² (0 ≤ x ≤ 3) is spun once about the x-axis. The strip at x = 1 has length f(1) = 2, so it sweeps out a disc of radius 2 (drawn as an ellipse because the disc is seen at an angle). The dashed curve is the mirror image of the arch: together with the solid curve it outlines the football-shaped solid. Axes are unitless.

The disc method

About the x-axis. If the region lies between y = f(x) and the x-axis for a ≤ x ≤ b, the solid formed by spinning it about the x-axis has volume V = π ∫ (a to b) [f(x)]² dx.

About the y-axis. If the region lies between x = g(y) and the y-axis for c ≤ y ≤ d, the solid formed by spinning it about the y-axis has volume V = π ∫ (c to d) [g(y)]² dy.

A short way to remember both: V = π ∫ (radius)² d(variable along the axis).

Three points decide almost every mark in this topic.

  1. The radius is a distance from the axis. It runs at right angles to the axis, from the axis out to the curve. For the x-axis that is a vertical distance, f(x). For the y-axis it is a horizontal distance, g(y).
  2. Slice at right angles to the axis. Spinning about the x-axis gives vertical slices with thickness dx, so the limits are x-values. Spinning about the y-axis gives horizontal slices with thickness dy, so the limits are y-values and the radius must be written in terms of y.
  3. The region must touch the axis all the way along. If there is a gap between the region and the axis, each slice is a disc with a hole in it: a washer. That case is Topic 8.11. In this topic every slice is a full disc.

A curve below the axis. If f(x) is negative, the radius is |f(x)|. Because the radius is squared, π[f(x)]² gives the same area either way. A disc volume is never negative.

Revolving about the y-axis

When the axis is vertical, the slices are horizontal. Ask: “At height y, how far is the curve from the y-axis?” That distance is the x-coordinate of the curve at that height, so you need x as a function of y.

  • y = x², x ≥ 0, becomes x = √y.
  • y = x³ becomes x = y^(1/3).
  • y = eˣ becomes x = ln y.

The limits are the lowest and highest y-values of the region, not x-values. If the region is described by x-limits, convert them: an edge at x = 2 on y = x³ is the height y = 8.

The method, step by step

  1. Sketch the region and the axis. Check that the region touches the axis along its whole length.
  2. Draw one strip at right angles to the axis. Its length is the radius.
  3. Write the radius in terms of the variable along the axis (x for the x-axis, y for the y-axis).
  4. Find the limits along the axis: the first and last slice.
  5. Write the integral in full: V = π ∫ (limits) [radius]² d(variable). This line earns the setup marks.
  6. Evaluate, with antiderivatives or a calculator, and give cubic units.
  7. Check against a simple cylinder that contains the solid.

Worked example 1: about the x-axis (no calculator)

Question. The region R lies between the curve y = 3x − x² and the x-axis (Figure 1). Find the volume of the solid formed when R is spun about the x-axis.

  1. Limits. 3x − x² = x(3 − x) = 0 at x = 0 and x = 3. Between them the curve is above the axis (for example f(1) = 2), so R touches the x-axis from x = 0 to x = 3.
  2. Radius. A vertical strip at x reaches from the axis to the curve, so the radius is r = 3x − x².
  3. Integral. V = π ∫ (0 to 3) [3x − x²]² dx
  4. Expand before integrating. (3x − x²)² = 9x² − 6x³ + x⁴. You cannot integrate a square by squaring the antiderivative, so expand first.
  5. Antiderivative. 3x³ − (3/2)x⁴ + x⁵/5.
  6. Evaluate. At x = 3: 81 − 243/2 + 243/5 = 81 − 121.5 + 48.6 = 8.1. At x = 0 it is 0.

Answer. V = 8.1π = 81π/10 cubic units (about 25.447).

Check. The widest disc has radius 9/4 (at the peak, x = 1.5). A cylinder of radius 9/4 and length 3 has volume π(9/4)²(3) = 243π/16 ≈ 47.713, and the solid fits inside it. Our answer is a bit over half of that, which is reasonable for a shape that tapers to points at both ends.

A common slip. π ∫ (0 to 3) (3x − x²) dx = 9π/2 ≈ 14.137. That forgets to square the radius. It is π times the area of R, which is not a volume.

Worked example 2: about the y-axis (no calculator)

Question. The region S is bounded by the curve y = x³, the y-axis and the line y = 8. Find the volume of the solid formed when S is spun about the y-axis.

  1. Sketch. The curve y = x³ rises through (0, 0) and (2, 8). S is the region to the left of the curve, between the y-axis and the curve, from y = 0 up to y = 8. It touches the y-axis all the way, so discs work.
  2. Slices. The axis is vertical, so slice horizontally. Each slice has thickness dy.
  3. Radius. At height y, the curve is at x = y^(1/3). The radius is the horizontal distance from the y-axis to the curve: r = y^(1/3).
  4. Limits. y runs from 0 to 8. (The point (2, 8) gives the top limit as a y-value, 8, not the x-value 2.)
  5. Integral. V = π ∫ (0 to 8) [y^(1/3)]² dy = π ∫ (0 to 8) y^(2/3) dy
  6. Evaluate. The antiderivative of y^(2/3) is (3/5)y^(5/3). At y = 8: 8^(5/3) = 2⁵ = 32, so (3/5)(32) = 96/5.

Answer. V = 96π/5 cubic units (about 60.319).

Check. The solid sits inside a cylinder of radius 2 and height 8, of volume π(2²)(8) = 32π ≈ 100.531. Our solid is narrow at the bottom and full width at the top, so a value of 0.6 of the cylinder is sensible.

Why not use dx? Writing π ∫ (0 to 2) (x³)² dx spins a different region (the one under the curve, next to the x-axis) about a different axis (the x-axis). The variable must match the axis: y-axis, dy.

Worked example 3: a calculator-active profile

Question. A craft designer models a wooden spindle by spinning the region between r(x) = 1 + 0.3 sin x and the x-axis, for 0 ≤ x ≤ 6, about the x-axis. Lengths are in centimetres. (This context is invented.) Find the volume of wood in the spindle. Calculator allowed.

  1. Radius. The profile r(x) is always between 0.7 and 1.3, so it is positive and the region touches the x-axis from x = 0 to x = 6.
  2. Integral. Write it before pressing any keys: V = π ∫ (0 to 6) [1 + 0.3 sin x]² dx
  3. Evaluate in radian mode: V ≈ 19.811 cm³.

Notation on paper. A complete response shows “π ∫ (0 to 6) (1 + 0.3 sin x)² dx = 19.811”. The decimal alone is not enough; the integral shows the method.

Sense check. A cylinder of radius 1 cm and length 6 cm has volume 6π ≈ 18.850 cm³. The spindle is wider than 1 cm over slightly more than half of its length, so a volume a little above 18.850 makes sense. If you forget the square you get about 18.887, which happens to be close here. That is a warning: a sense check can miss a missing square when the radius is near 1, so always look at your integral too.

Common misconceptions

  • Forgetting π. Each slice is a circle of area πr². Without π you have integrated something else.
  • Forgetting to square. π ∫ f(x) dx is π times an area, not a volume.
  • Squaring after integrating. π[∫ f(x) dx]² is not the same as π ∫ [f(x)]² dx. Square the radius inside the integral.
  • Using 2π. 2πr is the circumference, not the area. The disc area is πr².
  • Mixing variables. About the y-axis, both the radius and the limits must be in y. A dx with y-limits (or the reverse) is a wrong integral.
  • Using the diameter. The radius runs from the axis to the curve, not across the whole solid.
  • Using discs when the region does not touch the axis. If there is a gap, the slices are washers (Topic 8.11).
  • Square units. A volume is in cubic units: cm³, m³ or “cubic units”.
  • Rounding too early on a calculator question. Store values and round only the final answer to three decimal places.

Where this leads

Topic 8.8 found volumes from slices of known shape; here the slice is always a circle. Topic 8.10 keeps the disc method but spins about other lines, such as y = 2 or x = −1, where the radius becomes a difference. Topic 8.11 then handles regions with a gap between them and the axis, where the slices are washers. Continue with Volume with the Disc Method: Revolving Around Other Axes, or return to Topic 8.8. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.

Quick questions

Is this page for Calculus AB or Calculus BC?

Both. Topic 8.9 is common content, so the same page serves AB and BC students.

Do I need the shell method?

No. Every volume of revolution in this course can be set up with discs (and, from Topic 8.11, washers). The shell method is not part of the course framework.

What if the curve is below the x-axis?

The radius is a distance, so it is |f(x)|. Because you square it, π[f(x)]² is still correct. The volume is always positive.

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