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Cylindrical Shell Method

Cylindrical Shell Method
Shell Method

A technique for finding the volume of a solid of revolution.

 

Graph showing y=f(x) curve with shaded region between x=a and x=b, rotated around y-axis. Formula: Volume=∫a to b 2πyf(x)dx

 

 

See also

Disk method, washer method, axis of rotation

Key Formula

V=2π∫abr(x) h(x) dxV = 2\pi \int_a^b r(x)\, h(x)\, dx
Where:
  • VV = Volume of the solid of revolution
  • a,ba, b = Limits of integration along the axis perpendicular to the axis of rotation
  • r(x)r(x) = Shell radius — the distance from the axis of rotation to the shell
  • h(x)h(x) = Shell height — the length of the shell, determined by the function(s) defining the region
  • dxdx = Shell thickness — an infinitesimally thin strip

Worked Example

Problem: Find the volume of the solid formed by revolving the region bounded by y = x², y = 0, and x = 2 about the y-axis.
Step 1: Identify the shell components. When revolving about the y-axis and integrating with respect to x, a vertical strip at position x has radius r(x) = x (distance to the y-axis) and height h(x) = x² (the function value).
r(x)=x,h(x)=x2r(x) = x, \quad h(x) = x^2
Step 2: Determine the limits of integration. The region extends from x = 0 to x = 2.
a=0,b=2a = 0, \quad b = 2
Step 3: Set up the shell method integral.
V=2π∫02x⋅x2 dx=2π∫02x3 dxV = 2\pi \int_0^2 x \cdot x^2\, dx = 2\pi \int_0^2 x^3\, dx
Step 4: Evaluate the integral.
V=2π[x44]02=2π⋅164=2π⋅4=8πV = 2\pi \left[\frac{x^4}{4}\right]_0^2 = 2\pi \cdot \frac{16}{4} = 2\pi \cdot 4 = 8\pi
Answer: The volume is 8π8\pi cubic units.

Another Example

This example differs because the axis of rotation is not the y-axis but the vertical line x = 4, requiring a modified shell radius r(x) = 4 − x.

Problem: Find the volume of the solid formed by revolving the region bounded by y = x, y = 0, and x = 3 about the line x = 4.
Step 1: Identify the shell components. The axis of rotation is x = 4, so the radius of each shell is the distance from x to x = 4. Since the region lies to the left of x = 4, the radius is r(x) = 4 − x. The height of each shell is h(x) = x.
r(x)=4−x,h(x)=xr(x) = 4 - x, \quad h(x) = x
Step 2: Set the limits of integration from x = 0 to x = 3.
a=0,b=3a = 0, \quad b = 3
Step 3: Write and expand the integral.
V=2π∫03(4−x)(x) dx=2π∫03(4x−x2) dxV = 2\pi \int_0^3 (4 - x)(x)\, dx = 2\pi \int_0^3 (4x - x^2)\, dx
Step 4: Evaluate the integral.
V=2π[2x2−x33]03=2π(18−9)=2π⋅9=18πV = 2\pi \left[2x^2 - \frac{x^3}{3}\right]_0^3 = 2\pi \left(18 - 9\right) = 2\pi \cdot 9 = 18\pi
Answer: The volume is 18π18\pi cubic units.

Frequently Asked Questions

When should you use the shell method instead of the disk/washer method?
Use the shell method when the axis of rotation is parallel to the strips you naturally draw. For example, if you revolve around the y-axis but the function is easier to express as y = f(x), shells let you integrate with respect to x, avoiding the need to solve for x = g(y). As a rule of thumb, if setting up the disk method forces you to invert a complicated function or split into multiple integrals, the shell method is likely simpler.
What is the difference between the shell method and the disk method?
The disk method slices the solid into thin circular disks perpendicular to the axis of rotation and integrates their areas. The shell method wraps thin cylindrical shells around the axis and integrates their lateral surface areas times thickness. The disk method integrates parallel to the axis of rotation, while the shell method integrates perpendicular to it. Both give the same volume; the choice depends on which setup is easier for a given problem.
How do you find the radius and height of a cylindrical shell?
The radius is the distance from your strip to the axis of rotation. If the axis is x = c and you integrate with respect to x, then r(x) = |x − c|. The height is the length of the strip, determined by the function(s) bounding the region. For a region between y = f(x) and y = g(x) with f(x) ≥ g(x), the height is h(x) = f(x) − g(x).

Cylindrical Shell Method vs. Disk/Washer Method

Cylindrical Shell MethodDisk/Washer Method
Geometric ideaSums thin cylindrical shells wrapped around the axisSums thin circular disks (or washers) stacked along the axis
Integration directionPerpendicular to the axis of rotationParallel to the axis of rotation
Formula (revolving about y-axis)V=2π∫abx f(x) dxV = 2\pi \int_a^b x\, f(x)\, dxV=π∫cd[R(y)]2 dyV = \pi \int_c^d [R(y)]^2\, dy (or washer variant)
Best used whenThe function is hard to invert, or when disks require splitting into multiple integralsThe cross-sections are simple circles or washers and the function inverts easily
Typical pitfallForgetting the 2π factor or using the wrong radiusConfusing inner and outer radii in washers

Why It Matters

The shell method is essential in AP Calculus BC and college calculus courses whenever the disk or washer method leads to an awkward integral. Many real-world applications—such as computing the volume of a vase, a fuel tank, or a turbine component—involve solids whose cross-sections are easier to describe with shells. Mastering both the shell and disk methods gives you flexibility to choose the simpler setup for any volume-of-revolution problem.

Common Mistakes

Mistake: Using the wrong expression for the shell radius when the axis of rotation is not x = 0 or y = 0.
Correction: Always compute the radius as the distance from the strip to the axis. If the axis is x = c, the radius is |x − c|, not just x. Sketch the region and axis to visualize the distance.
Mistake: Forgetting the factor of 2π in the shell formula.
Correction: Each cylindrical shell has lateral surface area 2πr·h. The 2π comes from the circumference of the shell. If your answer seems too small by a factor of 2π, check that you included it.

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