The Shell Method

A region turned about the y-axis can be sliced across the axis into washers or along it into thin cylinders, shells, each of radius x and height h(x). The volume is the same number either way, and the page computes both, independently. What changes is how hard the integral is, and which way you slice decides it.

Shells, about the y-axis	$V = 2 π \int_{a}^{b} x h (x) d x$
Pappus	$V = A \cdot 2 π \overset{r}{ˉ}$

Worked example — The Shell Method

Lateral area of the shell at x	0.7854
Volume by shells	1.5708
Volume by washers, sliced across y	1.5708
Shell at x	2πx·h(x)

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Shell Method

Turning the region under y = h(x), a ≤ x ≤ b, about the y-axis gives a solid of volume V = 2π∫ₐᵇ x h(x) dx by cylindrical shells: each strip at x sweeps a thin cylinder of radius x, height h(x) and thickness dx.

What each symbol means

h(x): the height of the region at x, top edge minus bottom edge
x: the shell's radius: its distance from the axis of rotation

Where the formula comes from

One shell. Unrolled, a thin cylinder of radius x, height h and thickness Δx is almost a flat slab, 2πx long and h high, so its volume is close to 2πx·h·Δx.
$Δ V \approx (2 π x) \cdot h (x) \cdot Δ x$
Add the shells. Sum over the strips and let Δx shrink.
$V = lim i \sum 2 π x_{i} h (x_{i}) Δ x = 2 π \int_{a}^{b} x h (x) d x$

Problems solved in full

The same solid, both ways 2 steps
The region under y = x² on [0, 1] is turned about the y-axis. Find the volume by shells and by washers.

Open this problem in the tool
1. Shells: radius x, height x².
  
  $V = 2 π \int_{0}^{1} x \cdot x^{2} d x = 2 π \cdot \frac{1}{4} = \frac{π}{2}$
2. Washers: across y the region runs from x = √y out to x = 1.
  
  $V = π \int_{0}^{1} (1^{2} - (y)^{2}) d y = π \cdot \frac{1}{2} = \frac{π}{2}$
Answer

π/2 ≈ 1.5708, both ways.

$V = \frac{π}{2} \approx 1.5708$
Where shells are the easy way 2 steps
Turn the region under y = sin x on [0, π] about the y-axis. Find the volume.

Open this problem in the tool
Show solution
1. Shells: radius x, height sin x.
  
  $V = 2 π \int_{0}^{π} x sin x d x$
2. ∫x sin x dx by parts, u = x, dv = sin x dx:
  
  $\int_{0}^{π} x sin x d x = [- x cos x]_{0}^{π} + \int_{0}^{π} cos x d x = π + 0$
Answer

2π² ≈ 19.7392. By washers this needs arcsin y, which is why nobody does it that way.

$V = 2 π^{2} \approx 19.7392$
A region between two curves 1 steps
The region between y = x and y = x² on [0, 1] is turned about the y-axis. Find the volume.

Open this problem in the tool
Show solution
1. Shells: radius x, height x − x².
  
  $V = 2 π \int_{0}^{1} x (x - x^{2}) d x = 2 π (\frac{1}{3} - \frac{1}{4}) = \frac{π}{6}$
Answer

π/6 ≈ 0.5236.

$V = \frac{π}{6} \approx 0.5236$
A torus 2 steps
The disc (x − 2)² + y² ≤ 1 is turned about the y-axis. Find the volume.

Open this problem in the tool
Show solution
1. By Pappus: the area times the distance the centre travels.
  
  $V = A \cdot d = π \cdot 1^{2} \cdot 2 π \cdot 2 = 4 π^{2}$
2. By shells, the height at x is 2√(1 − (x − 2)²):
  
  $V = 2 π \int_{1}^{3} x \cdot 2 1 - (x - 2)^{2} d x = u = x - 2 4 π \int_{- 1}^{1} (u + 2) 1 - u^{2} d u = 4 π \cdot 2 \cdot \frac{π}{2}$
Answer

4π² ≈ 39.4784. The shell integral needs a substitution and the area of a half disc; Pappus needs one line.

$V = 4 π^{2} \approx 39.4784$
A root, by shells 1 steps
The region under y = √x on [0, 4] is turned about the y-axis. Find the volume.
Show solution
1. Shells: radius x, height √x, so the integrand is a power of x.
  
  $V = 2 π \int_{0}^{4} x x d x = 2 π [\frac{2}{5} x^{5/2}]_{0}^{4} = 2 π \cdot \frac{64}{5}$
Answer

128π/5 ≈ 80.4248.

$V = \frac{128 π}{5} \approx 80.4248$
Shells about the x-axis 2 steps
Use shells to find the volume when the region under y = x² on [0, 1] is turned about the x-axis.
Show solution
1. About the x-axis the shells run along x: radius y, and length from x = √y to x = 1.
  
  $V = 2 π \int_{0}^{1} y (1 - y) d y = 2 π (\frac{1}{2} - \frac{2}{5}) = \frac{π}{5}$
2. Check by discs:
  
  $V = π \int_{0}^{1} (x^{2})^{2} d x = \frac{π}{5}$
Answer

π/5 ≈ 0.6283, both ways. Shells always run parallel to the axis of rotation; their radius is the distance to it.

$V = \frac{π}{5} \approx 0.6283$

Two slicings, one volume 🖖

Press y = x², both ways. Shells cut the solid into nested cylinders, radius x and height x², and add 2πx·x² over [0, 1]. Washers cut it across the axis, discs of radius 1 with holes of radius √y, and add π(1 − y) over [0, 1]. Neither calculation uses the other, and both give 1.5708 = π/2. They have to: they are two ways of chopping the same solid into thin pieces, and the volume does not depend on the knife.

Slice along the direction the region is easy to describe 🖖

Press sin x: shells win. The region is easy to describe in x, 0 ≤ y ≤ sin x for 0 ≤ x ≤ π, so shells give 2π∫x sin x dx, one integration by parts, and 2π². Across y the region runs from arcsin y to π − arcsin y, so washers need π∫((π − arcsin y)² − (arcsin y)²) dy. The page computes it, and gets the same 19.7392, but nobody would choose to by hand. Before picking a method, ask which variable the region's edges are simple functions of; the slices should run along the other one.

A torus is a disc times the distance its centre travels 🖖

Press A torus. The disc of radius 1 centred at x = 2 sweeps out a ring, and shells give its volume as 39.4784 = 4π². That is the disc's area, π, times the distance its centre travels round the axis, 2π·2 = 4π. Pappus of Alexandria stated this around 340, in Book VII of his Mathematical Collection, for any region turned about a line that does not cut it; Paul Guldin published it again in 1640, and it is often called by his name.

Learning path

Applications of integration

References (3)

The method of cylindrical shells as the page uses it (equation 6.6, V = ∫ₐᵇ 2πx f(x) dx about the y-axis), and the section's own comparison of shells with discs and washers, the choice depending on which integral is easier: G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 6.3, “Volumes of Revolution: Cylindrical Shells”.
Insight block 3: Theorem 6.14, Theorem of Pappus for Volume (a region turned about a line that does not meet it has volume equal to its area times the distance its centroid travels): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 6.6, “Moments and Centers of Mass”.
Insight block 3: the attribution, “what is now known as Guldin's theorem was proved by Pappus in Book VII of the Mathematical Collection”, the Collection “thought to have been written in around 340”, and Guldin's publication in 1640: J. J. O'Connor and E. F. Robertson, “Pappus of Alexandria”, MacTutor History of Mathematics Archive, University of St Andrews (last updated April 1999).

By the 42Math team · Reviewed against published sources. Corrections welcome.