Shell Method: Formula & Volume Examples

The shell method finds the volume of a solid of revolution by slicing the region into thin strips parallel to the axis. Each strip spins into a cylindrical shell with volume about 2π · radius · height · thickness. Adding the shells gives $V = \int_a^b 2\pi r(x)h(x)\,dx$ .

Before this lesson: Disk Method, Washer Method and Area Between Curves

Why shells?

With disks and washers, you slice across the axis. For a region spun around the y-axis, that means writing everything in terms of y, which can mean solving a hard equation for x.

Shells avoid that. Cut the region into thin vertical strips and spin each one around the y-axis. A strip becomes a thin tube, like the label on a soup can. Unroll the tube and it is almost a flat box: 2πr long, h tall and dx thick.

The region under y = 4x − x² from x = 0 to x = 4. A thin vertical strip at x is shaded darker. Its distance from the y-axis is labeled radius r = x, and its height is labeled h = 4x − x². The y-axis is marked as the axis of rotation. — A strip at distance x from the y-axis sweeps out a shell with radius x and height 4x − x².

Shell method formula

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

Here r(x) is the distance from the strip to the axis, and h(x) is the strip’s height (top curve minus bottom curve). The strips must be parallel to the axis.

Axis of rotation	Strips	Radius r	Height h
y-axis	Vertical, in x	x	top − bottom
x = c, with the region to the left (c bigger than x)	Vertical, in x	c − x	top − bottom
x = c, with the region to the right (c smaller than x)	Vertical, in x	x − c	top − bottom
x-axis	Horizontal, in y	y	right − left

Sketch the region, the axis and one strip parallel to the axis.
Write the radius r and the height h in terms of the strip’s variable.
Find the limits of integration.
Integrate 2πrh.

Shell method examples

Example 1: Around the y-axis

Find the volume when the region under $y = 4x - x^2$ , from x = 0 to x = 4, is rotated around the y-axis.

r = x, \; h = 4x - x^2

Vertical strips; the radius is the distance to the y-axis.

V = \int_0^4 2\pi x(4x - x^2)\,dx

Set up the shell integral.

= 2\pi\int_0^4 (4x^2 - x^3)\,dx

Distribute x (see distributive property).

= 2\pi\Big[\dfrac{4x^3}{3} - \dfrac{x^4}{4}\Big]_0^4

Integrate with the power rule (see antiderivatives).

= 2\pi\left(\dfrac{256}{3} - 64\right)

Plug in 4; the 0 terms are 0.

= 2\pi\cdot\dfrac{64}{3} = \dfrac{128\pi}{3}

256/3 − 192/3 = 64/3.

Answer: $V = \dfrac{128\pi}{3} \approx 134.04$ . With washers, you would have to solve y = 4x − x² for x.

Example 2: Between two curves

The region between y = x and y = x² is rotated around the y-axis. Find the volume.

x = x^2 \Rightarrow x = 0 \text{ or } x = 1

The curves meet at x = 0 and x = 1.

r = x, \; h = x - x^2

On (0, 1), x is above x².

V = 2\pi\int_0^1 x(x - x^2)\,dx = 2\pi\int_0^1 (x^2 - x^3)\,dx

Set up and distribute.

= 2\pi\left(\dfrac{1}{3} - \dfrac{1}{4}\right)

Integrate and plug in.

= 2\pi\cdot\dfrac{1}{12} = \dfrac{\pi}{6}

Subtract the fractions.

Answer: $V = \dfrac{\pi}{6}$ .

Example 3: Around a line that isn’t an axis

The region under $y = \sqrt{x}$ from x = 0 to x = 4 is rotated around the line x = −1. Find the volume.

r = x - (-1) = x + 1, \; h = \sqrt{x}

The radius is the distance from x to the line x = −1.

V = 2\pi\int_0^4 (x + 1)x^{1/2}\,dx

Set up.

= 2\pi\int_0^4 \left(x^{3/2} + x^{1/2}\right)dx

Distribute; x · x^(1/2) = x^(3/2) (see rational exponents).

= 2\pi\Big[\dfrac{2}{5}x^{5/2} + \dfrac{2}{3}x^{3/2}\Big]_0^4

Integrate.

= 2\pi\left(\dfrac{2}{5}\cdot 32 + \dfrac{2}{3}\cdot 8\right)

4^(5/2) = 32 and 4^(3/2) = 8.

= 2\pi\left(\dfrac{192}{15} + \dfrac{80}{15}\right) = \dfrac{544\pi}{15}

Use the common denominator 15.

Answer: $V = \dfrac{544\pi}{15} \approx 113.94$ .

Example 4: Around the x-axis

The region in the first quadrant bounded by $x = 4 - y^2$ and the axes is rotated around the x-axis. Find the volume with shells.

r = y, \; h = 4 - y^2

Rotating around the x-axis, so use horizontal strips in y; h is right minus left, 4 − y² − 0.

0 \le y \le 2

The curve meets the y-axis where 4 − y² = 0.

V = 2\pi\int_0^2 y(4 - y^2)\,dy = 2\pi\int_0^2 (4y - y^3)\,dy

Set up and distribute.

= 2\pi\Big[2y^2 - \dfrac{y^4}{4}\Big]_0^2 = 2\pi(8 - 4)

Integrate and plug in.

= 8\pi

Multiply.

Answer: $V = 8\pi \approx 25.13$ .

Example 5: A napkin ring

A hole of radius 1 cm is drilled straight through the center of a wooden ball of radius 2 cm. What volume of wood is left?

x^2 + y^2 = 4 \Rightarrow y = \pm\sqrt{4 - x^2}

Model the ball by spinning a circle of radius 2 around the y-axis (see equation of a circle).

r = x, \; h = 2\sqrt{4 - x^2}, \; 1 \le x \le 2

The wood left is between the hole (x = 1) and the outside (x = 2).

V = \int_1^2 2\pi x\cdot 2\sqrt{4 - x^2}\,dx = 4\pi\int_1^2 x\sqrt{4 - x^2}\,dx

Set up.

u = 4 - x^2, \; du = -2x\,dx; \; x = 1 \to u = 3, \; x = 2 \to u = 0

Substitute and change the limits (see u-substitution with definite integrals).

= 4\pi\int_3^0 \sqrt{u}\left(-\dfrac{1}{2}\right)du = 2\pi\int_0^3 u^{1/2}\,du

Flip the limits to remove the minus sign.

= 2\pi\cdot\dfrac{2}{3}u^{3/2}\Big|_0^3 = \dfrac{4\pi}{3}\cdot 3\sqrt{3}

Integrate; 3^(3/2) = 3√3.

= 4\sqrt{3}\,\pi \approx 21.77

Simplify.

Answer: about 21.8 cubic centimeters of wood are left.

Common mistakes

Mistake	Example of the error	Fix
Using x as the radius for every axis	r = x when rotating around x = 3	The radius is the distance to the axis: r = 3 − x.
Forgetting 2π	$\int_0^4 x(4x - x^2)\,dx$	The shell’s length is its circumference, 2πr: $\int_0^4 2\pi x(4x - x^2)\,dx$ .
Strips in the wrong direction	dx shells for rotation around the x-axis	Shells are parallel to the axis. For the x-axis, use horizontal strips and dy.
Using only the top curve	h = x for the region between y = x and y = x²	Height is top minus bottom: h = x − x².
Mixing in the disk formula	$\int 2\pi r^2 h\,dx$	Shells use 2πrh, with no square. Squares belong to disks: πr².

Related methods

Shells or washers?

Slices are…	Method	Formula
Perpendicular to the axis	Disks or washers	$\int \pi(R^2 - r^2)$
Parallel to the axis	Shells	$\int 2\pi rh$

Pick the method whose strips are easier to describe. If the curves are given as y = f(x) and the axis is vertical, shells usually win. Both methods give the same volume.

Checking with a cone

Spin the region under y = 3 − x, from 0 to 3, around the y-axis. Shells give 9π, which matches the cone formula (1/3)π(3)²(3) = 9π (see volume of pyramids and cones).

Practice problems

Use the shell method to find each volume.

The region under y = x, from 0 to 2, around the y-axis.
The region under y = x³, from 0 to 1, around the y-axis.
The region under y = 1/x, from 1 to 3, around the y-axis.
The region under y = √x, from 0 to 1, around the y-axis.
The region between y = 2x and y = x², around the y-axis.
The region under y = x², from 0 to 2, around the line x = 2.
The region under $y = e^{-x^2}$ , from 0 to 1, around the y-axis.
The region under y = sin x, from 0 to π, around the y-axis.
The region bounded by y = x, y = 0 and x = 1, around the x-axis (use shells in y).
A pile of sand is made by spinning the region under y = 3 − x, from 0 to 3, around the y-axis (in meters). Find its volume.

Show answers

V = 2\pi\int_0^2 x\cdot x\,dx = 2\pi\Big[\dfrac{x^3}{3}\Big]_0^2

r = x and h = x.

= 2\pi\cdot\dfrac{8}{3} = \dfrac{16\pi}{3}

Plug in and multiply.

Answer: $\dfrac{16\pi}{3}$

V = 2\pi\int_0^1 x\cdot x^3\,dx = 2\pi\Big[\dfrac{x^5}{5}\Big]_0^1

r = x and h = x³.

= \dfrac{2\pi}{5}

Plug in.

Answer: $\dfrac{2\pi}{5}$

V = 2\pi\int_1^3 x\cdot\dfrac{1}{x}\,dx = 2\pi\int_1^3 1\,dx

r = x and h = 1/x; they cancel.

= 2\pi(3 - 1) = 4\pi

Integrate and plug in.

Answer: $4\pi$

V = 2\pi\int_0^1 x\cdot x^{1/2}\,dx = 2\pi\int_0^1 x^{3/2}\,dx

r = x and h = √x.

= 2\pi\cdot\dfrac{2}{5} = \dfrac{4\pi}{5}

Integrate to (2/5)x^(5/2) and plug in.

Answer: $\dfrac{4\pi}{5}$

x^2 = 2x \Rightarrow x = 0 \text{ or } x = 2

Find where the curves meet.

V = 2\pi\int_0^2 x(2x - x^2)\,dx = 2\pi\Big[\dfrac{2x^3}{3} - \dfrac{x^4}{4}\Big]_0^2

h = 2x − x², since 2x is on top.

= 2\pi\left(\dfrac{16}{3} - 4\right) = \dfrac{8\pi}{3}

Plug in and simplify.

Answer: $\dfrac{8\pi}{3}$

r = 2 - x, \; h = x^2

The region is to the left of x = 2.

V = 2\pi\int_0^2 (2 - x)x^2\,dx = 2\pi\Big[\dfrac{2x^3}{3} - \dfrac{x^4}{4}\Big]_0^2

Distribute and integrate.

= 2\pi\left(\dfrac{16}{3} - 4\right) = \dfrac{8\pi}{3}

Plug in and simplify.

Answer: $\dfrac{8\pi}{3}$

V = 2\pi\int_0^1 xe^{-x^2}\,dx

r = x and h = e^(−x²).

= 2\pi\Big[-\dfrac{1}{2}e^{-x^2}\Big]_0^1

Use u = −x², du = −2x dx.

= \pi\left(1 - e^{-1}\right) \approx 1.986

Plug in 1 and 0.

Answer: $\pi\left(1 - e^{-1}\right) \approx 1.986$

V = 2\pi\int_0^{\pi} x\sin x\,dx

r = x and h = sin x.

= 2\pi\Big[-x\cos x + \sin x\Big]_0^{\pi}

Integration by parts with u = x (see integration by parts).

= 2\pi\big[(-\pi)(-1) + 0 - 0\big] = 2\pi^2

cos π = −1 and sin π = 0.

Answer: $2\pi^2 \approx 19.74$

r = y, \; h = 1 - y, \; 0 \le y \le 1

Horizontal strips run from x = y to x = 1.

V = 2\pi\int_0^1 (y - y^2)\,dy = 2\pi\left(\dfrac{1}{2} - \dfrac{1}{3}\right) = \dfrac{\pi}{3}

Distribute, integrate and plug in.

Answer: $\dfrac{\pi}{3}$ , which matches the disk method.

10.

V = 2\pi\int_0^3 x(3 - x)\,dx = 2\pi\Big[\dfrac{3x^2}{2} - \dfrac{x^3}{3}\Big]_0^3

r = x and h = 3 − x.

= 2\pi\left(\dfrac{27}{2} - 9\right) = 9\pi

Plug in and simplify.

Answer: 9π ≈ 28.3 cubic meters, the volume of a cone with radius 3 and height 3.

Shell method FAQ

What is the shell method formula?

V = ∫ 2π r h dx, where r is the distance from a strip to the axis and h is the strip’s height. For rotation around the x-axis, use V = ∫ 2π y h dy.

When should I use the shell method instead of disks?

Use shells when strips parallel to the axis are easier to describe. A common case is a region under y = f(x) spun around the y-axis, where disks would need x in terms of y.

How do you find the radius in the shell method?

The radius is the distance from the strip to the axis of rotation. For the y-axis it is x; for the line x = c it is |x − c|.

Can you use the shell method around the x-axis?

Yes. Use horizontal strips, so the radius is y, the height is the right curve minus the left curve, and you integrate with respect to y.

Summary

Shells come from strips parallel to the axis of rotation.
V = ∫ 2π r h, with r the distance to the axis and h the strip’s height.
Around a vertical axis, use dx; around a horizontal axis, use dy.
For an axis like x = c, the radius is |x − c|.
Shells and washers give the same volume; pick the easier setup.

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