Both methods compute volumes of revolution. They give the same answer, take different amounts of work depending on the problem, and confuse students who try to memorise when to use which. The choice is not arbitrary — it comes down to which direction you slice the solid, and that choice determines whether the algebra simplifies or gets messier.

The Two Formulas

Disk/Washer Method — slice perpendicular to axis: V = π ∫ₐᵇ [f(x)]² dx (rotation around x-axis) V = π ∫ₐᵇ [g(y)]² dy (rotation around y-axis) Shell Method — slice parallel to axis: V = 2π ∫ₐᵇ x·f(x) dx (rotation around y-axis) V = 2π ∫ₐᵇ y·g(y) dy (rotation around x-axis)

Decision Guide

📋 How to Choose
DiskUse when you can express the boundary curve easily as a function of the integration variable, slicing perpendicular to axis.
ShellUse when disk would require solving for the inverse function, or when the region extends both above and below an axis.
Rule of thumbRotating around x-axis: try disk with dx. Rotating around y-axis: try shell with dx (avoids solving for x in terms of y).

Example 1 — Disk Method (rotation around x-axis)

📋 Rotate y = x² on [0,1] around x-axis
SetupV = π∫₀¹ (x²)² dx = π∫₀¹ x⁴ dx
Integrate= π[x⁵/5]₀¹ = π/5

Example 2 — Shell Method (same region, rotation around y-axis)

📋 Rotate y = x² on [0,1] around y-axis
SetupV = 2π∫₀¹ x·x² dx = 2π∫₀¹ x³ dx
Integrate= 2π[x⁴/4]₀¹ = π/2

Example 3 — Washer Method (region between two curves)

📋 Rotate region between y=√x and y=x² around x-axis (they intersect at x=0,1)
SetupV = π∫₀¹ [(√x)² − (x²)²] dx = π∫₀¹ [x − x⁴] dx
Integrate= π[x²/2 − x⁵/5]₀¹ = π(1/2 − 1/5) = 3π/10

Example 4 — When Shell is Easier

📋 Rotate y = e^(x²) on [0,1] around y-axis
Disk attemptWould need x = √(ln y) — messy. Disk not suitable here.
Shell setupV = 2π∫₀¹ x·e^(x²) dx
u-subu = x², du = 2x dx → π∫₀¹ eᵘ du = π[eᵘ]₀¹ = π(e−1)

Example 5 — Rotation Around y = −1

📋 Rotate y = x on [0,2] around y = −1
RadiusDistance from curve to axis = x − (−1) = x + 1
SetupV = π∫₀² (x+1)² dx = π∫₀² (x²+2x+1) dx
Answer= π[x³/3 + x² + x]₀² = π(8/3 + 4 + 2) = 26π/3

Example 6 — Shell Around x = 2

📋 Rotate y = x² on [0,1] around x = 2
Shell radiusDistance from x to the axis x=2 is (2−x)
SetupV = 2π∫₀¹ (2−x)·x² dx = 2π∫₀¹ (2x²−x³) dx
Answer= 2π[2x³/3 − x⁴/4]₀¹ = 2π(2/3 − 1/4) = 5π/6

Frequently Asked Questions

Do both methods always give the same answer?▾
Yes — they compute the same physical volume, just with different slicing strategies. If you get different answers, you have made an error in one of the setups. Both are mathematically equivalent.
When is the washer method needed?▾
When rotating a region bounded by two curves, not just one curve and an axis. The washer method subtracts the inner radius from the outer radius: V = π∫(R² − r²)dx, where R is the distance to the outer boundary and r to the inner boundary.

Side-by-Side Algebra Comparison

The best way to understand when each method is simpler is to set up both and compare. Here is the region between y = x and y = x² rotated around the y-axis:

📋 Disk vs shell for y-axis rotation of y=x vs y=x² on [0,1]
Disk setupNeed to integrate over y: x = √y (outer) and x = y (inner). V = π∫₀¹ (y − y²) dy = π[y²/2 − y³/3]₀¹ = π/6
Shell setupIntegrate over x: V = 2π∫₀¹ x(x−x²)dx = 2π∫₀¹ (x²−x³)dx = 2π[x³/3 − x⁴/4]₀¹ = 2π·(1/12) = π/6
VerdictBoth give π/6 ✓. Shell is simpler here — no need to solve for x in terms of y.

Decision Guide

📋 How to choose every time
Step 1Identify the axis of rotation
Step 2Is the region described naturally in terms of x or y?
Axis = x-axis, region in x→ Disk method, integrate over x
Axis = y-axis, region in x→ Shell method, integrate over x (avoids solving for x in terms of y)
If solving for inverse is easy→ Either method works; pick simpler integrand

Common Setup Mistakes

⚠ Mistakes that cost full marks

Washer method: forgetting to subtract the inner radius. The volume of the hole must be removed: π∫(R²−r²)dx, NOT π∫(R−r)²dx.

Shell method: the radius of the shell is the distance from the axis, not from the origin. For rotation around x=2, the shell radius is |2−x|, not x.

Limits: find the intersections first. The limits of integration are where the boundary curves cross, not arbitrary points.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026