Why do we use cylindrical shells instead of disks?

Shells are useful when you are rotating around the y-axis and it is easier to express the region as y = f(x) with x as the independent variable. Disks work better when you have a function of y, or when rotating around the x-axis. Choose whichever method requires less algebra.

What does the 2\pi x term represent?

It is the circumference of a thin cylindrical shell at radius x from the y-axis. When you "unwrap" a thin shell of height f(x) and thickness dx, its surface area is 2\pi x \cdot f(x), and multiplying by the thickness dx gives the volume of that shell.

Can I use the shell method to rotate around the x-axis?

Yes, but you must swap the roles of x and y: use V = 2\pi\int_c^d y \, g(y) \, dy, where g is expressed as a function of y and you integrate over the y-axis. If you apply the original formula to an x-axis rotation without this swap, you will get the wrong answer.

What if the function f(x) goes below the x-axis?

If f(x) is negative, it means the region extends below the x-axis. For rotation around the y-axis, only the distance from the y-axis matters, so you can use |f(x)| (the absolute value) or split the integral at each x-intercept and integrate each piece separately.

Volume by Shells

Volume by Shells calculates the volume of a solid of revolution around the y-axis by integrating thin cylindrical shells instead of disk slices.

V = 2\pi\int_a^b x\,f(x)\,dx

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What each symbol means

What Volume by Shells takes

Symbol	Meaning
$V$	The volume of the solid of revolution, measured in cubic units such as cubic centimeters or cubic meters.
$f$	A continuous function representing the height of the region being rotated; if you confuse it with the distance from the axis, you will integrate the wrong quantity.
$a$	The lower limit of integration, the x-value where the region begins; swapping $a$ and $b$ reverses the sign, and confusing this with a y-value gives an answer with incorrect units.
$b$	The upper limit of integration, the x-value where the region ends; it must be greater than $a$ for the standard orientation.
$x$	The variable of integration, representing the radius of each cylindrical shell (its distance from the y-axis); if you treat it as a constant, the integral collapses to zero.

When to use it

Use this method when you need to find the volume of a solid of revolution around the y-axis and the height of the region is easier to express as a function of $x$ than the radius would be.

Level

Usually taught in: Calculus II

Worked examples

1. Rotating a linear function with small integers

Problem

Find the volume of the solid obtained by rotating the region bounded by

y = x

x = 0

, and

x = 3

around the y-axis.

$V = 2\pi\int_0^3 x \cdot x \, dx$
Set up the shell method formula. The radius of each shell is $x$ (distance from the y-axis) and the height is $f(x) = x$ .
$V = 2\pi\int_0^3 x^2 \, dx$
Multiply the radius and height: $x \cdot x = x^2$ .
$V = 2\pi\left[\frac{x^3}{3}\right]_0^3$
Integrate using the power rule: the antiderivative of $x^2$ is $\frac{x^3}{3}$ .
$V = 2\pi\left(\frac{27}{3} - 0\right) = 2\pi \cdot 9 = 18\pi$
Evaluate at the bounds: $\frac{3^3}{3} = \frac{27}{3} = 9$ , then multiply by $2\pi$ .

Answer: $V = 18\pi \text{ cubic units}$

This is the straightforward case where the height function is simply $f(x) = x$ . The shell method works cleanly because we multiply the radius $x$ by the height $x$ to get $x^2$ , which integrates easily.

2. Rotating a linear function with a negative slope

Problem

Find the volume of the solid obtained by rotating the region bounded by

y = 2 - x

x = 0

x = 2

, and the x-axis around the y-axis.

$V = 2\pi\int_0^2 x(2 - x) \, dx$
Set up the shell method. The radius is $x$ and the height is $f(x) = 2 - x$ . Note that $2 - x$ is positive for $x < 2$ , so this is valid over our bounds.
$V = 2\pi\int_0^2 (2x - x^2) \, dx$
Distribute $x$ across the binomial: $x(2 - x) = 2x - x^2$ .
$V = 2\pi\left[x^2 - \frac{x^3}{3}\right]_0^2$
Integrate term by term: $\int 2x \, dx = x^2$ and $\int x^2 \, dx = \frac{x^3}{3}$ .
$V = 2\pi\left(4 - \frac{8}{3}\right) = 2\pi\left(\frac{12}{3} - \frac{8}{3}\right) = 2\pi \cdot \frac{4}{3} = \frac{8\pi}{3}$
Evaluate at the bounds: at $x = 2$ , we get $2^2 - \frac{2^3}{3} = 4 - \frac{8}{3} = \frac{4}{3}$ . At $x = 0$ , both terms are zero.

Answer: $V = \frac{8\pi}{3} \text{ cubic units}$

This example requires distributing inside the integral and handling fractions carefully. The negative slope means the height decreases as $x$ increases, which is fine as long as the height stays positive throughout the region.

3. Volume of a cone-shaped container

Problem

A decorative container is designed by rotating the curve

y = \frac{3}{2}x

from

x = 0

x = 4

feet around the vertical y-axis. Find the volume of the container.

$V = 2\pi\int_0^4 x \cdot \frac{3}{2}x \, dx$
Set up the shell method. The radius is $x$ feet and the height is $f(x) = \frac{3}{2}x$ feet.
$V = 2\pi\int_0^4 \frac{3}{2}x^2 \, dx$
Multiply: $x \cdot \frac{3}{2}x = \frac{3}{2}x^2$ .
$V = 3\pi\int_0^4 x^2 \, dx$
Factor out the constant: $2\pi \cdot \frac{3}{2} = 3\pi$ .
$V = 3\pi\left[\frac{x^3}{3}\right]_0^4$
Integrate using the power rule: $\int x^2 \, dx = \frac{x^3}{3}$ .
$V = 3\pi \cdot \frac{64}{3} = 64\pi$
Evaluate at the bounds: $\frac{4^3}{3} = \frac{64}{3}$ , then multiply by $3\pi$ .

Answer: $V = 64\pi \text{ cubic feet}$

Word problems often include fractional coefficients. This container problem demonstrates why the shell method is convenient: you can directly express the radius and height as functions of $x$ without needing to solve for $x$ in terms of $y$ .

Common mistakes

Where Volume by Shells usually goes wrong

The mistake	Why it is wrong	The fix
Writing $V = \pi\int_a^b x \, f(x) \, dx$ without the factor of 2.	The circumference of a cylindrical shell at radius $x$ is $2\pi x$ , not $\pi x$ ; the factor of 2 is essential.	Always include $V = 2\pi\int_a^b x \, f(x) \, dx$ . If you forget the 2, your answer will be half the correct volume.
Using the shell method formula when the region is being rotated around the x-axis.	The shell method as stated applies only to rotation around the y-axis. If you rotate around the x-axis, you must use the disk method or adapt the formula by treating $y$ as the variable instead of $x$ .	Read the problem carefully and sketch the region. If it says rotate around the x-axis, use the disk method; if around the y-axis, use shells.
Treating $f(x)$ as the radius instead of the height of each shell.	In $V = 2\pi\int_a^b x \, f(x) \, dx$ , the variable $x$ is always the radius (distance from the y-axis) and $f(x)$ is always the height. Swapping their roles means you integrate the wrong quantity and get a meaningless answer.	Remember: radius is $x$ (how far the shell is from the y-axis), height is $f(x)$ (how tall the region is at position $x$ ). Check this setup against your sketch.

Tips and when to use something else

If you are rotating around the x-axis instead of the y-axis, switch to the disk method or use $V = 2\pi\int_c^d y \, g(y) \, dy$ where you treat $y$ as the variable of integration.
When $f(x)$ is negative on part of the interval, only integrate over where it is positive, or split the integral at the point where $f(x) = 0$ .
The shell method avoids having to solve $y = f(x)$ for $x$ in terms of $y$ , which can be algebraically messy for complicated functions.
Always include the $2\pi$ factor—it comes from the circumference formula for a cylinder, and forgetting it is the most common error.

Related formulas

Volume by DisksVolume by Disks calculates the volume of a solid of revolution by integrating the areas of circular cross-sections perpendicular to the axis of rotation.Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.Power Rule for IntegrationPower Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.

Frequently asked questions

Why do we use cylindrical shells instead of disks?: Shells are useful when you are rotating around the y-axis and it is easier to express the region as $y = f(x)$ with $x$ as the independent variable. Disks work better when you have a function of $y$ , or when rotating around the x-axis. Choose whichever method requires less algebra.
What does the $2\pi x$ term represent?: It is the circumference of a thin cylindrical shell at radius $x$ from the y-axis. When you "unwrap" a thin shell of height $f(x)$ and thickness $dx$ , its surface area is $2\pi x \cdot f(x)$ , and multiplying by the thickness $dx$ gives the volume of that shell.
Can I use the shell method to rotate around the x-axis?: Yes, but you must swap the roles of $x$ and $y$ : use $V = 2\pi\int_c^d y \, g(y) \, dy$ , where $g$ is expressed as a function of $y$ and you integrate over the y-axis. If you apply the original formula to an x-axis rotation without this swap, you will get the wrong answer.
What if the function $f(x)$ goes below the x-axis?: If $f(x)$ is negative, it means the region extends below the x-axis. For rotation around the y-axis, only the distance from the y-axis matters, so you can use $|f(x)|$ (the absolute value) or split the integral at each x-intercept and integrate each piece separately.

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Reviewed 2026-09-18