Volumes: Disks, Washers & Cross Sections

You’ll be able to

Compute volumes of revolution with the disk method
Use the washer method when the solid has a hole
Find volumes of solids with known cross-sectional shapes

The disk method

Revolve a region under a curve around an axis and it sweeps out a solid. Slicing perpendicular to the axis gives thin disks (circles) of radius R = f(x) and thickness dx. Each disk has volume πR²·dx, so the total is V = π∫[a to b] [f(x)]² dx. The radius is the distance from the axis to the curve — when revolving around the x-axis, that is just f(x).

The washer method

When the region does not touch the axis, revolving it leaves a hole, and each slice is a washer — a disk with a smaller disk removed. Its area is π(R² − r²), where R is the outer radius (axis to the far curve) and r is the inner radius (axis to the near curve). So V = π∫[a to b] (R² − r²) dx. Crucially, you subtract the squares of the radii, not the radii themselves.

Known cross sections

Some solids are not solids of revolution: instead, cross sections perpendicular to an axis are a known shape (square, semicircle, equilateral triangle) built on the region as a base. Integrate the area of one cross section in terms of x: V = ∫[a to b] A(x) dx. For squares with side s = (top − bottom), A(x) = s²; for semicircles of diameter s, A(x) = (π/8)s².

Volume methods

Disk: V = π∫[a to b] R² dx · Washer: V = π∫[a to b] (R² − r²) dx · Cross sections: V = ∫[a to b] A(x) dx

Washers subtract the squares of the radii. For cross sections, integrate the area formula of the given shape.

Worked example

The region under y = √x from x = 0 to x = 4 is revolved about the x-axis. Find the volume.

1.Slices perpendicular to the x-axis are disks of radius R = √x, so use the disk method.
2.Square the radius: R² = (√x)² = x.
3.Set up the integral: V = π∫[0 to 4] x dx.
4.Integrate: π·[x²/2] from 0 to 4 = π·(16/2 − 0) = 8π.

Answer: The volume is 8π cubic units. Squaring √x conveniently gives x, making the integral a simple power-rule integration.

Watch out

In the washer method you subtract R² − r², the difference of the squared radii — never (R − r)². Squaring first and subtracting second is essential; (R − r)² gives a different, wrong volume.

Checkpoint

A region is revolved about the x-axis using the disk method with radius R = f(x). The volume integral is:

Checkpoint

A solid has a base bounded by curves, and each cross section perpendicular to the x-axis is a square with side length s = (top − bottom). The volume is:

On the exam

Identify the cross-section shape before writing the integral: disks and washers bring a π (they are circles), but squares, triangles, and rectangles do not. Slapping a π on a non-circular cross section is a classic AP volume error.

Answer the 2 checkpoints as you read.