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Lesson 3: Volumes: Shell Method

Estimated time: 40-50 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Understand the idea of cylindrical shells
Apply the shell method formula for rotation about the y-axis
Apply the shell method for rotation about the x-axis
Decide when shells are preferable to disks/washers

The Idea of Cylindrical Shells

Instead of slicing perpendicular to the axis of rotation (as with disks/washers), the shell method uses slices parallel to the axis. Each slice generates a thin cylindrical shell when revolved.

Think of it like peeling layers off an onion: each layer is a thin cylinder. The volume of a thin cylindrical shell with radius r, height h, and thickness Δx is approximately 2πr · h · Δx.

Shell Method: Rotation About the y-Axis

Shell Method (rotation about the y-axis)

If the region under y = f(x) on [a, b] (where 0 ≤ a < b) is revolved about the y-axis:

V = 2π ∫_a^b x · f(x) dx

Here x is the shell radius (distance from x to the y-axis), and f(x) is the shell height.

Example 1: Basic Shell Method

Find the volume when y = x² on [0, 2] is revolved about the y-axis.

Solution: Shell radius = x, shell height = x².

V = 2π ∫₀² x · x² dx = 2π ∫₀² x³ dx = 2π [x⁴/4]₀² = 2π(4) = 8π.

Example 2: Region Between Two Curves

Find the volume when the region between y = x and y = x² on [0, 1] is revolved about the y-axis.

Solution: Shell radius = x, shell height = x − x².

V = 2π ∫₀¹ x(x − x²) dx = 2π ∫₀¹ (x² − x³) dx

= 2π [x³/3 − x⁴/4]₀¹ = 2π(1/3 − 1/4) = 2π(1/12) = π/6.

Shell Method: Rotation About the x-Axis

For rotation about the x-axis, the shells are horizontal. Express x as a function of y.

Shell Method (rotation about the x-axis)

V = 2π ∫_c^d y · g(y) dy

where g(y) is the shell height (horizontal width) and y is the shell radius.

Example 3: Shells About the x-Axis

Find the volume when x = √y on [0, 4] (for y) is revolved about the x-axis.

Solution: Shell radius = y, shell height = √y. Limits: y from 0 to 4.

V = 2π ∫₀⁴ y · √y dy = 2π ∫₀⁴ y^3/2 dy = 2π [(2/5)y^5/2]₀⁴ = 2π(2/5)(32) = 128π/5.

When to Use Shells vs. Disks/Washers

Choosing Your Method

Situation	Preferred Method
Rotate about x-axis, easy to write y = f(x)	Disk/Washer (dx)
Rotate about y-axis, easy to write y = f(x)	Shell (dx)
Rotate about y-axis, easy to write x = g(y)	Disk/Washer (dy)
Disk/washer would require multiple integrals	Try Shells

Example 4: When Shells Are Easier

Find the volume when y = x(2 − x) on [0, 2] is revolved about the y-axis.

Why shells? Using disks about the y-axis requires solving y = 2x − x² for x (quadratic formula), making the integral messy. Shells keep it simple.

Solution: V = 2π ∫₀² x · x(2 − x) dx = 2π ∫₀² (2x² − x³) dx

= 2π [2x³/3 − x⁴/4]₀² = 2π(16/3 − 4) = 2π(4/3) = 8π/3.

Shells with Non-Standard Axes

When rotating about a vertical line x = k (instead of x = 0), the shell radius becomes |x − k|.

Example 5: Rotation About x = 3

Find the volume when y = x² on [0, 2] is revolved about x = 3.

Solution: Shell radius = 3 − x (distance from x to x = 3). Shell height = x².

V = 2π ∫₀² (3 − x) · x² dx = 2π ∫₀² (3x² − x³) dx

= 2π [x³ − x⁴/4]₀² = 2π(8 − 4) = 8π.

Verifying with Both Methods

A good exercise is to compute the same volume using both methods to check your answer.

Example 6: Verification

Volume of y = √x on [0, 1] revolved about the y-axis.

Shells: V = 2π ∫₀¹ x · √x dx = 2π ∫₀¹ x^3/2 dx = 2π [(2/5)x^5/2]₀¹ = 4π/5.

Disks (dy): x = y², y from 0 to 1. V = π ∫₀¹ (1 − y⁴) dy ... this requires washer with outer R = 1, inner r = y². V = π ∫₀¹ (1 − y⁴) dy = π(1 − 1/5) = 4π/5. Same answer!

Key Takeaways

Shell method: V = 2π ∫ (radius)(height) dx (or dy).
Shells use slices parallel to the axis of rotation; disks use slices perpendicular.
Shells about the y-axis integrate with respect to x; shells about the x-axis integrate with respect to y.
Choose the method that avoids solving for the other variable and produces a single integral.

Check Your Understanding

1. Use shells to find the volume when y = 4 − x² on [0, 2] is revolved about the y-axis.

Answer: V = 2π ∫₀² x(4 − x²) dx = 2π ∫₀² (4x − x³) dx = 2π[2x² − x⁴/4]₀² = 2π(8 − 4) = 8π.

2. Use shells to find the volume when y = x³ on [0, 1] is revolved about the y-axis.

Answer: V = 2π ∫₀¹ x · x³ dx = 2π ∫₀¹ x⁴ dx = 2π[x⁵/5]₀¹ = 2π/5.

3. Which method (disk/washer or shell) would you choose for y = sin(x²) on [0, √π] revolved about the y-axis? Why?

Answer: Shells! You cannot easily solve y = sin(x²) for x, so using disks about the y-axis would be extremely difficult. Shells: V = 2π ∫₀^√π x sin(x²) dx, which can be evaluated with u-substitution (u = x²).

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