$V = 2\pi \int_a^b x \cdot f(x)\,dx$

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

Components:

• Radius = $x$ (distance from y-axis)
• Height = $f(x)$ (height of shell)
• Circumference = $2\pi x$

Where Does This Come From?

Volume of a Cylindrical Shell

Imagine unrolling a thin cylindrical shell:

• It becomes a rectangular sheet!
• Length = circumference = $2\pi r$
• Height = $h$
• Thickness = $\Delta x$

Volume = length × height × thickness $V = 2\pi r \cdot h \cdot \Delta x$

With Our Variables

At position $x$:

• Radius: $r = x$
• Height: $h = f(x)$
• Thickness: $\Delta x$

Shell volume: $2\pi x \cdot f(x) \cdot \Delta x$

Sum all shells and take limit → integral!

Example 1: Basic Shell Method

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

Step 1: Identify components

• Axis: y-axis (vertical)
• Radius: $r = x$ (distance from y-axis)
• Height: $h = f(x) = x^2$

Step 2: Set up the integral

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

$= 2\pi \int_0^2 x^3\,dx$

Step 3: Integrate

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

$= 2\pi \cdot \frac{16}{4} = 2\pi \cdot 4 = 8\pi$

Answer: $8\pi$ cubic units

Example 2: Region Between Curves

Find the volume when the region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ is rotated around the y-axis.

Step 1: Identify height

Height of shell = (top curve) - (bottom curve)

$h = x - x^2$

Step 2: Set up the integral

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

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

Step 3: Integrate

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

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

$= 2\pi \left(\frac{4-3}{12}\right) = \frac{2\pi}{12} = \frac{\pi}{6}$

Answer: $\frac{\pi}{6}$ cubic units

Rotating Around the x-axis with Shells

Formula

When rotating $x = g(y)$ from $y = c$ to $y = d$ around the x-axis:

$V = 2\pi \int_c^d y \cdot g(y)\,dy$

Components:

• Radius = $y$ (distance from x-axis)
• Height = $g(y)$ (horizontal extent)

Example 3: Horizontal Shells

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

Step 1: Set up with shells

• Radius: $r = y$
• Height: $h = y^2$

$V = 2\pi \int_0^2 y \cdot y^2\,dy$

$= 2\pi \int_0^2 y^3\,dy$

Step 2: Integrate

$= 2\pi \left[\frac{y^4}{4}\right]_0^2$

$= 2\pi \cdot \frac{16}{4} = 8\pi$

Answer: $8\pi$ cubic units

Rotating Around Other Lines

Around the line $x = k$

Radius = distance from line $x = k$ to the shell

If shell is at position $x$:

• If $x > k$: radius = $x - k$
• If $x < k$: radius = $k - x$

Generally: $r = |x - k|$

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

Example 4: Rotating Around $x = -1$

Find the volume when $y = \sqrt{x}$ from $x = 0$ to $x = 4$ is rotated around the line $x = -1$.

Step 1: Find radius

Shell at position $x$ is distance $x - (-1) = x + 1$ from the line.

• Radius: $r = x + 1$
• Height: $h = \sqrt{x}$

Step 2: Set up the integral

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

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

Step 3: Integrate

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

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

$= 2\pi \left(\frac{2(32)}{5} + \frac{2(8)}{3}\right)$

$= 2\pi \left(\frac{64}{5} + \frac{16}{3}\right)$

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

Answer: $\frac{544\pi}{15}$ cubic units

Shell Method vs Disk/Washer

When to Use Shell Method:

1. Rotating around y-axis but function is $y = f(x)$

   • Shell: easy! $2\pi \int x \cdot f(x)\,dx$
   • Disk: would need to solve for $x = g(y)$ (hard!)

2. Avoiding complicated algebra

   • Shell often simpler than squaring in washer method

3. Region naturally described in terms of x but rotating around y-axis

Example 5: Why Shell is Better

Find the volume when $y = e^{-x^2}$ from $x = 0$ to $x = 1$ is rotated around the y-axis.

With shells: $V = 2\pi \int_0^1 x \cdot e^{-x^2}\,dx$

Use substitution $u = -x^2$, $du = -2x\,dx$:

$= 2\pi \int_0^{-1} -\frac{1}{2}e^u\,du = \pi[e^u]_0^{-1}$

$= \pi(e^{-1} - 1) = \pi\left(\frac{1}{e} - 1\right)$

Easy!

With disks:

Need to solve $y = e^{-x^2}$ for $x$:

$x = \sqrt{-\ln y}$

This is much more complicated!

General Shell Method Formula

Around vertical line $x = k$

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

where $h(x)$ is the height of the shell.

Around horizontal line $y = k$

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

where $h(y)$ is the length of the shell.

⚠️ Common Mistakes

Mistake 1: Forgetting the $2\pi$

Shell method always has $2\pi$ (from circumference)!

$V = 2\pi \int (\text{radius})(\text{height})\,dx$

Mistake 2: Wrong Radius

Radius = distance from axis of rotation

Not just $x$! If rotating around $x = 3$, radius might be $|x - 3|$.

Mistake 3: Wrong Height

For region between two curves: $h = \text{top} - \text{bottom}$

Don't forget to subtract!

Mistake 4: Choosing Wrong Method

Sometimes both methods work, but one is much easier.

Quick check:

• Function is $y = f(x)$, rotating around y-axis? → Shell!
• Function is $x = g(y)$, rotating around x-axis? → Shell!
• Otherwise, compare which seems simpler.

Shell Method Formula Summary

Around y-axis (or $x = k$)

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

• Radius: distance from vertical axis
• Height: vertical extent of region
• Integrate with respect to $x$

Around x-axis (or $y = k$)

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

• Radius: distance from horizontal axis
• Height: horizontal extent of region
• Integrate with respect to $y$

Comparison Table

Choose based on which makes the integral simpler!

📝 Practice Strategy

1. Identify axis of rotation
2. Decide: Shell or Disk/Washer?
   • Shell if parallel slicing is easier
   • Disk/Washer if perpendicular slicing is easier
3. Find radius: distance from axis to shell
4. Find height: extent of shell (top - bottom)
5. Set up: $V = 2\pi \int (\text{radius})(\text{height})\,dx$
6. Remember $2\pi$!
7. Integrate and evaluate
8. Compare with other method as a check

So $x$ goes from 0 to 2.

Step 2: Find height of shell

At position $x$:

• Top: $y = 4$
• Bottom: $y = x^2$
• Height: $h = 4 - x^2$

Step 3: Set up with shell method

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

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

Step 4: Integrate

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

$= 2\pi \left(8 - 4\right) = 8\pi$

Answer: $8\pi$ cubic units

Step 2: Find height

$h = f(x) = \frac{1}{x}$

Step 3: Set up the integral

$V = 2\pi \int_1^3 (x+2) \cdot \frac{1}{x}\,dx$

$= 2\pi \int_1^3 \left(1 + \frac{2}{x}\right)\,dx$

Step 4: Integrate

$= 2\pi [x + 2\ln|x|]_1^3$

$= 2\pi [(3 + 2\ln 3) - (1 + 2\ln 1)]$

$= 2\pi [3 + 2\ln 3 - 1 - 0]$

$= 2\pi(2 + 2\ln 3)$

$= 4\pi(1 + \ln 3)$

Answer: $4\pi(1 + \ln 3)$ cubic units