Mathfolis

Volume: Disk & Washer

Unit 8 · Applications of Integration

Volumes of Revolution

When a region is revolved about an axis, the disk method (one curve) or washer method (two curves) gives the volume. For disks: V = π∫R² dx. For washers: V = π∫[R² − r²] dx, where R is the outer radius and r is the inner radius — both measured as distances from the axis of rotation.

Disk Method

About x-axis
V=π∫ab[f(x)]2 dxV = \pi\int_a^b [f(x)]^2\,dx
About y-axis
V=π∫cd[g(y)]2 dyV = \pi\int_c^d [g(y)]^2\,dy
About y = k: radius = distance from curve to line
V=π∫ab[f(x)−k]2 dxV = \pi\int_a^b [f(x)-k]^2\,dx

Washer Method (two curves)

V=π∫ab[R(x)2−r(x)2] dxV = \pi\int_a^b \bigl[R(x)^2 - r(x)^2\bigr]\,dx
R = outer radius, r = inner radius (both measured from axis)

Known Cross Sections

Square cross sections (side s)
A(x)=s2=[f(x)−g(x)]2A(x) = s^2 = [f(x)-g(x)]^2
Semicircle cross sections (diameter s)
A(x)=π8s2A(x) = \frac{\pi}{8}s^2
Caution: Radius is always a positive quantity — if revolving about y = k and f(x) < k, the radius is k − f(x), not f(x) − k.
AP Tip: Draw a picture and identify R and r at a generic x before setting up the integral.
Type 1

Disk Method

For a single boundary curve revolved about an axis (no hole), each cross section is a solid disk.

Example 1
Find the volume formed by revolving y = √x on [0, 4] about the x-axis.
Example 2
Find the volume formed by revolving y = 2 − x for 0 ≤ x ≤ 2 about y = −1.

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Type 2

Washer Method

When there are two boundary curves (creating a hole when revolved), use the washer formula with outer and inner radii.

Example 3
Find the volume formed by revolving the region between y = √x and y = x² on [0,1] about the x-axis.
Example 4
Revolve the region between y = x and y = x² on [0,1] about y = −1.

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Type 3

Volumes with Known Cross Sections

For non-revolution problems, find the cross-sectional area A(x), then integrate V = ∫A(x) dx.

Example 5
A solid has base bounded by y = 4 − x² and y = 0. Cross sections perpendicular to the x-axis are squares. Find the volume.

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