Integration, Applications of the Integral

Volumes of
Revolution

Spin a 2D region around an axis and it sweeps out a solid in 3D. Slice that solid into disks, washers, or cylindrical shells, and a single integral gives you the exact volume.

Disk method Washer method Shell method Interactive volume calculator
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§ 01What Is a Solid of Revolution?

Take a flat region bounded by a curve, rotate it a full 360° about a line, and you get a three-dimensional solid whose volume can be found exactly with a single integral.

Imagine the region under y = f(x), between x = a and x = b, spun around the x-axis. Every point on the curve traces out a circle as it rotates, and the whole region sweeps out a solid, think of a vase, a dome, or a horn shape, depending on f(x).

The slicing idea Instead of imagining the whole 3D solid at once, slice it into thin pieces perpendicular to the axis of rotation. Each thin slice is (approximately) a simple shape — a disk, a washer (a disk with a hole), or a cylindrical shell — whose volume you already know how to compute. Integrating adds up infinitely many infinitesimally thin slices.
Three methods, one idea
Disk  →  slice perpendicular to the axis, no gap
Washer  →  slice perpendicular to the axis, with a hole
Shell  →  slice parallel to the axis, as nested cylinders

Which method to use depends on the shape of the region and, especially, which axis you rotate around. Rotating about the x-axis usually favours disks/washers integrated in x; rotating about the y-axis usually favours shells integrated in x (or disks/washers integrated in y).

§ 02The Disk Method — Derivation

When the region touches the axis of rotation directly, each cross-section is a solid disk.

Rotate the region under y = f(x), from x = a to x = b, about the x-axis. A thin vertical strip of width dx at position x, when rotated, sweeps out a thin disk of radius r = f(x) and thickness dx.

Derivation Volume of one disk, then the whole solid
Vdisk = πr² · (thickness)
volume of a cylinder (disk) = area × thickness
dV = π[f(x)]² dx
radius r = f(x), thickness dx
V = ∫ab π[f(x)]² dx
sum (integrate) all disks from a to b
Disk Method Formula (rotation about the x-axis)
V = π∫ab [f(x)]² dx

If instead you rotate a region x = g(y) about the y-axis, the same idea gives V = π∫cd[g(y)]² dy, just swap the roles of x and y.

§ 03The Washer Method

When the region does not touch the axis of rotation, each slice is a disk with a hole punched out of it, a washer.

Suppose the region is bounded above by y = f(x) (outer curve) and below by y = g(x) (inner curve), both above the x-axis, rotated about the x-axis. Each slice is a disk of radius f(x) with a smaller disk of radius g(x) removed.

Washer Method Formula
V = π∫ab [ (f(x))² − (g(x))² ] dx
⚠ Square the whole radius, then subtract A very common error is computing [f(x) − g(x)]² instead of [f(x)]² − [g(x)]². These are not the same. The washer's cross-sectional area is (outer area) minus (inner area) — each radius must be squared separately before subtracting.

Disk vs. Washer — How to Tell Them Apart

  1. Sketch the region and the axis of rotation. If the axis passes through the region (or the region touches it), you likely need a disk. If there's a gap between the region and the axis, you need a washer.
  2. Identify the outer radius R(x) and, if needed, the inner radius r(x). These are the distances from the axis of rotation to the far and near boundary curves.
  3. Apply the formula. V = π∫[R(x)]² dx   (disk)    or    V = π∫([R(x)]²−[r(x)]²) dx   (washer)

§ 04The Shell Method

When the axis of rotation runs parallel to the strips you'd naturally draw, cylindrical shells are far easier than disks or washers.

Take a thin vertical strip at position x, of height f(x) and width dx, and rotate it about the y-axis (which is vertical, i.e. parallel to the strip). Instead of a disk, this sweeps out a thin cylindrical shell: like a tin can with no top or bottom.

Derivation Unrolling a cylindrical shell
Circumference = 2πr
r = distance from the axis to the strip = x
Unrolled shell = thin rectangle: 2πx × f(x) × dx
circumference × height × thickness
V = ∫ab 2πx·f(x) dx
integrate to sum all shells from a to b
Shell Method Formula (rotation about the y-axis)
V = 2π∫ab x·f(x) dx
When to reach for shells If solving for x in terms of y is difficult or produces multiple branches, but the region is naturally described with vertical strips in x, the shell method lets you integrate in x while still rotating about a vertical axis — avoiding an awkward change of variable.

§ 05Interactive Volume Calculator

Choose a function, bounds, and a method to numerically compute the volume of revolution about the x-axis (disk/washer style) or y-axis (shell style).

V for a Region Rotated About an Axis

Numerical Volume (Simpson's Rule, n = 200)

§ 06Worked Examples

Eight examples covering disk, washer, and shell methods about both axes.

Example 1 Disk method — a cone Pure

Find the volume when y = x, 0 ≤ x ≤ 3, is rotated about the x-axis.

Set up the disk integral
V = π∫03 x² dx
Integrate
V = π[x³/3]03 = π(27/3) = 9π
V = 9π

This matches the classic cone-volume formula V = ⅓πr²h with r = h = 3: ⅓π(9)(3) = 9π.

Example 2 Disk method — a paraboloid Pure

Find the volume when y = √x, 0 ≤ x ≤ 4, is rotated about the x-axis.

Set up: radius² = (√x)² = x
V = π∫04 x dx
Integrate
V = π[x²/2]04 = π(16/2) = 8π
V = 8π
Example 3 Washer method — region between two curves Pure

The region between y = x and y = x² (for 0 ≤ x ≤ 1) is rotated about the x-axis. Find the volume.

Identify outer and inner radii — on [0,1], x ≥ x²
R(x) = x   (outer),   r(x) = x²   (inner)
Set up the washer integral
V = π∫01 (x² − x⁴) dx
Integrate
V = π[x³/3 − x⁵/5]01 = π(1/3 − 1/5) = 2π/15
V = 2π/15
Example 4 Shell method — about the y-axis Pure

The region under y = x², 0 ≤ x ≤ 2, is rotated about the y-axis. Find the volume using shells.

Set up the shell integral: radius = x, height = x²
V = 2π∫02 x·x² dx = 2π∫02 x³ dx
Integrate
V = 2π[x⁴/4]02 = 2π(16/4) = 8π
V = 8π
Example 5 Same region, disk method with respect to y (cross-check) Pure

Repeat Example 4 using the disk method integrated in y, and confirm the same answer.

Solve for x: x = √y, with y ranging 0 to 4
V = π∫04 (√y)² dy = π∫04 y · dy
Integrate
V = π[y²/2]04 = π(16/2) = 8π
V = 8π (matches Example 4)

Both methods must agree, this is a useful way to check your work when a region can be sliced either way.

Example 6 Rotation about a line other than an axis Applied

The region under y = x², 0 ≤ x ≤ 1, is rotated about the line y = −1. Find the volume.

Shift the radius: distance from the curve to y=−1 is x²+1
R(x) = x² − (−1) = x² + 1
Set up the disk integral (no inner radius — solid disk)
V = π∫01 (x²+1)² dx = π∫01 (x⁴ + 2x² + 1) dx
Integrate
V = π[x⁵/5 + 2x³/3 + x]01 = π(1/5 + 2/3 + 1) = 28π/15
V = 28π/15
Example 7 Washer about the y-axis Pure

The region between x = y and x = y² (for 0 ≤ y ≤ 1) is rotated about the y-axis. Find the volume.

On [0,1], y ≥ y², so outer radius R = y, inner radius r = y²
V = π∫01 (y² − y⁴) dy
Integrate
V = π[y³/3 − y⁵/5]01 = π(1/3−1/5) = 2π/15
V = 2π/15
Example 8 Applied — designing a wine glass bowl Applied

A designer models the inside of a glass bowl by rotating y = √x, 0 ≤ x ≤ 9 (cm), about the x-axis. What volume of liquid (in cm³) does the bowl hold?

Disk method: radius² = x
V = π∫09 x dx = π[x²/2]09 = π(81/2)
V = 40.5π ≈ 127.2 cm³

§ 07Common Mistakes

⚠ Squaring the difference instead of subtracting squares For the washer method, [R(x)]² − [r(x)]² is correct. Writing [R(x) − r(x)]² is a different (and wrong) expression — expand a numeric example to see they disagree.
⚠ Forgetting the factor of π (disk/washer) or 2π (shell) The constant out front comes directly from the geometry (area of a circle, or circumference of a shell) — it is easy to drop when writing the integral quickly.
⚠ Using the wrong variable of integration Rotating about the x-axis with disks/washers integrates in x; rotating about the y-axis with disks/washers requires solving for x(y) and integrating in y. Mixing up dx and dy — or the bounds that go with them — is the single most common setup error.
⚠ Not shifting the radius for an off-axis line When rotating about a line like y = k (not the x-axis itself), the radius is the vertical distance to that line — f(x) − k, not f(x) alone. Forgetting to shift produces the wrong solid entirely.
⚠ Mixing up disk/washer with shell setups In the shell method, the "radius" is the horizontal distance from the axis to the strip, and the "height" is the function value — these play a different geometric role than the disk method's radius. Sketch the region before committing to a formula.

§ 08Practice Quiz — 10 Questions

Identify the correct method and set up (or evaluate) the integral.

Score: 0 / 0
Question 1 — Choosing the method

A region touches the x-axis directly and is rotated about the x-axis. Which method applies?

Question 2 — Disk volume

y = 2, 0 ≤ x ≤ 5, rotated about the x-axis. What is V/π?

V = π∫₀⁵ 2² dx = π(4)(5).
Question 3 — Washer setup

For the washer method with outer radius R(x) and inner radius r(x), the correct integrand is:

Question 4 — Shell method formula

Rotating the region under y = f(x), 0 ≤ x ≤ b, about the y-axis using shells gives:

Question 5 — Cone volume by disks

y = x, 0 ≤ x ≤ 2, rotated about the x-axis. Find V/π (should match ⅓πr²h ÷ π with r=h=2).

  1. V = π∫₀² x² dx = π[x³/3]₀² = π(8/3)
  2. V/π = 8/3, matching ⅓π(4)(2)/π = 8/3.
Question 6 — Off-axis rotation

A region under y = f(x) is rotated about the line y = 3 (with f(x) < 3 throughout). The radius of each disk is:

Question 7 — Consistency check

A region gives V = 8π by the shell method integrated in x. If the disk method (integrated in y) is set up correctly for the same solid, the result should be:

Question 8 — Washer numeric

Region between y = 3 and y = x² (for −√3 ≤ x ≤ √3... but restrict to 0 ≤ x ≤ 1 here), rotated about the x-axis: what is R(x) and r(x) on [0,1]?

  1. On [0,1], the line y=3 lies above the curve y=x², so it is the outer boundary: R(x) = 3.
  2. The curve y = x² is the inner boundary: r(x) = x².
Question 9 — Shell radius

In the shell method for rotation about the y-axis, the "radius" of a shell at position x is:

Question 10 — Full computation

y = x², 0 ≤ x ≤ 1, rotated about the y-axis using shells. What is V/π?

V = 2π∫₀¹ x·x² dx = 2π∫₀¹ x³ dx = 2π[x⁴/4]₀¹.

§ 09What to Study Next

Volumes of revolution are one of several classic applications of integration. Continue exploring how the definite integral measures geometric quantities:

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