Volumes of Revolution (Shell Method)

Overview

The shell method calculates volume by using cylindrical shells parallel to the axis of rotation. It's often easier when the washer method requires solving for $x$ in terms of $y$ (or vice versa).

Basic Formula

Rotation around y-axis (shells parallel to y-axis)

$V = 2\pi \int_a^b x \cdot f(x) \, dx = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx$

Rotation around x-axis (shells parallel to x-axis)

$V = 2\pi \int_c^d y \cdot f(y) \, dy = 2\pi \int_c^d (\text{radius})(\text{height}) \, dy$

General Formulas

Rotation around $x = k$ (vertical line)

$V = 2\pi \int_a^b |x - k| \cdot h(x) \, dx$

Rotation around $y = k$ (horizontal line)

$V = 2\pi \int_c^d |y - k| \cdot h(y) \, dy$

Shell Components

• Radius: Distance from shell to axis of rotation
• Height: Length of the shell (difference between curves, or curve to axis)
• Thickness: $dx$ or $dy$

When to Use Shell Method

• When the region is easier to describe as a function of $x$, but you're rotating around a vertical line
• When solving for $x$ in terms of $y$ would be difficult
• When the washer method would require splitting the integral

Examples

Example 1: Basic shell method

Rotate the region under $y = x^2$ from $x = 0$ to $x = 1$ around the y-axis.

Radius $= x$ Height $= x^2$

$V = 2\pi \int_0^1 x \cdot x^2 \, dx = 2\pi \int_0^1 x^3 \, dx$

$= 2\pi \left[\frac{x^4}{4}\right]_0^1 = 2\pi \left(\frac{1}{4}\right) = \frac{\pi}{2}$

Example 2: Region between curves

Rotate the region between $y = x$ and $y = x^2$ around the y-axis.

Intersection: $x = 0, 1$ Height $= x - x^2$ (top minus bottom)

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

$= 2\pi \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 2\pi \left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \left(\frac{1}{12}\right) = \frac{\pi}{6}$

Example 3: Rotation around $x = 2$

Rotate the region under $y = \sqrt{x}$ from $x = 0$ to $x = 1$ around $x = 2$.

Radius $= 2 - x$ (distance from shell to $x = 2$) Height $= \sqrt{x}$

$V = 2\pi \int_0^1 (2-x)\sqrt{x} \, dx = 2\pi \int_0^1 (2x^{1/2} - x^{3/2}) \, dx$

$= 2\pi \left[2 \cdot \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2}\right]_0^1 = 2\pi \left[\frac{4}{3} - \frac{2}{5}\right]$

$= 2\pi \left(\frac{14}{15}\right) = \frac{28\pi}{15}$

Example 4: Rotation around x-axis using shells

Rotate the region bounded by $y = x^2$, $y = 4$, and $x = 0$ around the x-axis.

Use horizontal shells (integrate with respect to $y$).

Radius $= y$ Height $= \sqrt{y}$ (from $x = 0$ to $x = \sqrt{y}$) Limits: $y$ from $0$ to $4$

$V = 2\pi \int_0^4 y \cdot \sqrt{y} \, dy = 2\pi \int_0^4 y^{3/2} \, dy$

$= 2\pi \left[\frac{y^{5/2}}{5/2}\right]_0^4 = 2\pi \cdot \frac{2}{5} \cdot 32 = \frac{128\pi}{5}$

Example 5: Rotation around $y = -1$

Rotate the region under $y = x$ from $x = 0$ to $x = 1$ around $y = -1$.

Use horizontal shells.

Radius $= y - (-1) = y + 1$ Height $= 1 - y$ (from $x = y$ to $x = 1$) Limits: $y$ from $0$ to $1$

$V = 2\pi \int_0^1 (y + 1)(1 - y) \, dy = 2\pi \int_0^1 (1 - y^2 + y - y) \, dy$

$= 2\pi \int_0^1 (1 - y^2) \, dy = 2\pi \left[y - \frac{y^3}{3}\right]_0^1$

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

Example 6: Comparing methods

Rotate $y = \sqrt{x}$ from $x = 0$ to $x = 4$ around y-axis.

Shell method:

$V = 2\pi \int_0^4 x \cdot \sqrt{x} \, dx = 2\pi \int_0^4 x^{3/2} \, dx$

$= 2\pi \left[\frac{2x^{5/2}}{5}\right]_0^4 = 2\pi \cdot \frac{64}{5} = \frac{128\pi}{5}$

Washer method (integrate with respect to $y$): $x = y^2$, limits $y = 0$ to $2$

$V = \pi \int_0^2 (4^2 - (y^2)^2) \, dy = \pi \int_0^2 (16 - y^4) \, dy$

$= \pi \left[16y - \frac{y^5}{5}\right]_0^2 = \pi \left(32 - \frac{32}{5}\right) = \frac{128\pi}{5} \checkmark$