Volumes by Cylindrical Shells

Quick answer

The shell method slices a solid of revolution parallel to its axis instead of across it. A thin vertical strip at x, revolved around a vertical axis, sweeps out a cylindrical shell of radius r(x), height h(x) and thickness dx. Unrolled, the shell is a thin slab with volume 2π r h dx, so V = 2π∫ₐᵇ r(x) h(x) dx. Shells suit a region given by y = f(x) and revolved around a vertical axis, where washers would need x written in terms of y.

What you'll learn

Derive the volume of a thin cylindrical shell
Find volumes of revolution with the shell method
Measure the radius of a shell from an axis other than the y-axis
Choose between shells and washers for a given region and axis

Slicing parallel to the axis

Disks and washers slice a solid of revolution across its axis. Shells slice it along the axis. Take a thin vertical strip of the region, at $x$ , with height $h(x)$ and width $\Delta x$ , and revolve it around a vertical axis. It sweeps out a thin-walled tube, a cylindrical shell, whose radius $r(x)$ is the strip’s distance from the axis.

y = x²

The strip that becomes a shell

The shells nest inside one another like the layers of an onion, and together they fill the solid. Adding their volumes gives the shell method:

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

Why a shell holds 2πrh times its thickness

Cut a shell straight down its side and unroll it. It flattens into a thin slab: its length is the circumference $2\pi r$ , its height is $h$ , and its thickness is $\Delta x$ . So its volume is about $2\pi r h\,\Delta x$ .

With $r$ measured to the middle of the strip, the formula is exact. The shell is a cylinder of radius $r + \tfrac{\Delta x}{2}$ with one of radius $r - \tfrac{\Delta x}{2}$ removed, and

\pi\left(r + \tfrac{\Delta x}{2}\right)^2 h - \pi\left(r - \tfrac{\Delta x}{2}\right)^2 h = 2\pi r h\,\Delta x

A shell is a rectangle rolled into a tube: circumference times height times thickness. Summing the shells gives a Riemann sum, and its limit is the integral.

Other axes, and choosing a method

Around a vertical line $x = k$ , the radius is the distance to that line: $x - k$ when the region lies to its right, and $k - x$ when it lies to the left. The height is still top minus bottom.

Shells and washers slice the same solid in two directions, so they always agree. Choose the one whose integral is easier to set up:

Region given as $y = f(x)$ , vertical axis: shells integrate in $x$ directly. Washers would need $x$ in terms of $y$ .
Region given as $y = f(x)$ , horizontal axis: disks and washers integrate in $x$ directly.

Shells work around a horizontal axis too. The strips are then horizontal, with radius $y$ measured from the axis and length in terms of $y$ , and the integral is in $y$ .

Worked examples

Common mistakes

Practice problems

Revolve the region under $y = x$ , $0 \le x \le 2$ , around the $y$ -axis. Use shells.

Answer

$\tfrac{16\pi}{3}$

Full solution

$V = 2\pi\int_0^2 x \cdot x\,dx = 2\pi \cdot \tfrac{8}{3} = \tfrac{16\pi}{3}$ .
Revolve the region under $y = \tfrac{1}{x}$ , $1 \le x \le 3$ , around the $y$ -axis.

Answer

$4\pi$

Full solution

$V = 2\pi\int_1^3 x \cdot \tfrac{1}{x}\,dx = 2\pi\int_1^3 1\,dx = 4\pi$ .
Revolve the region under $y = 4 - x^2$ , $0 \le x \le 2$ , around the $y$ -axis.

Answer

$8\pi$

Full solution

$V = 2\pi\int_0^2 x(4 - x^2)\,dx = 2\pi\big[2x^2 - \tfrac{x^4}{4}\big]_0^2 = 2\pi(8 - 4) = 8\pi$ .
Revolve the region between $y = 2x$ and $y = x^2$ around the $y$ -axis. Use shells.

Answer

$\tfrac{8\pi}{3}$

Full solution

The height is $2x - x^2$ for $0 \le x \le 2$ . $V = 2\pi\int_0^2 (2x^2 - x^3)\,dx = 2\pi\left(\tfrac{16}{3} - 4\right) = \tfrac{8\pi}{3}$ , the same as with washers.
Revolve the region under $y = x^2$ , $0 \le x \le 1$ , around the line $x = 2$ .

Answer

$\tfrac{5\pi}{6}$

Full solution

The region lies left of the axis, so the radius is $2 - x$ . $V = 2\pi\int_0^1 (2 - x)x^2\,dx = 2\pi\left(\tfrac{2}{3} - \tfrac{1}{4}\right) = \tfrac{5\pi}{6}$ .
Revolve the region under $y = \sqrt{x}$ , $0 \le x \le 4$ , around the $y$ -axis.

Answer

$\tfrac{128\pi}{5}$

Full solution

$V = 2\pi\int_0^4 x^{3/2}\,dx = 2\pi \cdot \tfrac{2}{5} \cdot 4^{5/2} = 2\pi \cdot \tfrac{64}{5}$ .
Revolve the region under $y = x^3$ , $0 \le x \le 1$ , around the $y$ -axis.

Answer

$\tfrac{2\pi}{5}$

Full solution

$V = 2\pi\int_0^1 x^4\,dx = \tfrac{2\pi}{5}$ .
Revolve the region under $y = \sqrt{x}$ , $0 \le x \le 4$ , around the $x$ -axis, using shells. Check against the disk method.

Answer

$8\pi$

Full solution

Use horizontal strips. At height $y$ , for $0 \le y \le 2$ , the strip runs from $x = y^2$ to $x = 4$ , so its length is $4 - y^2$ , and its radius is $y$ . $V = 2\pi\int_0^2 y(4 - y^2)\,dy = 2\pi(8 - 4) = 8\pi$ . The disk method gives $\pi\int_0^4 x\,dx = 8\pi$ as well.
Revolve the region under $y = x^2$ , $1 \le x \le 3$ , around the $y$ -axis.

Answer

$40\pi$

Full solution

$V = 2\pi\int_1^3 x^3\,dx = 2\pi \cdot \tfrac{81 - 1}{4} = 40\pi$ .
A student revolves the region under $y = x^2$ , $0 \le x \le 1$ , around the line $x = -1$ and writes $V = 2\pi\int_0^1 x \cdot x^2\,dx = \tfrac{\pi}{2}$ . What went wrong?

Hint

How far is the strip at $x$ from the line $x = -1$ ?

Answer

The radius is $x + 1$ , not $x$ . The volume is $\tfrac{7\pi}{6}$ .

Full solution

The student measured the radius from the $y$ -axis. Measured from $x = -1$ it is $x + 1$ :

$V = 2\pi\int_0^1 (x + 1)x^2\,dx = 2\pi\left(\tfrac{1}{4} + \tfrac{1}{3}\right) = \tfrac{7\pi}{6}$ .

Frequently asked questions

What is the shell method formula?

V = 2π ∫ from a to b of r(x) h(x) dx, where r(x) is the distance from the axis to the strip at x and h(x) is the strip's height.

Why is a shell's volume 2πrh times its thickness?

Cut the shell along its height and flatten it: it becomes a thin slab 2πr long, h tall and dx thick.

When should I use shells instead of washers?

When the region is given as y = f(x) and the axis is vertical, or as x = g(y) and the axis is horizontal. Shells then avoid solving for the other variable.

Do shells and washers give the same volume?

Yes. They slice the same solid in two directions, so both give its volume. Use whichever integral is easier.

What is the radius of a shell around the line x = −1?

The distance from the strip at x to that line: x + 1 for x to the right of it.

What to learn next

Arc LengthFind the length of a curve with an integral, L = ∫√(1 + f′(x)²) dx, built from the Pythagorean theorem on tiny pieces of the curve.

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