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Shells vs. Washers: Choosing the Right Method

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Reference: Stewart §5.3

Quick Decision Guide

Axis Direction Shells Use Washers Use
Vertical ($y$-axis, $x = k$) Vertical strips, $dx$ Horizontal slices, $dy$
Horizontal ($x$-axis, $y = k$) Horizontal strips, $dy$ Vertical slices, $dx$

The Key Question: Which variable describes the region more naturally?

  • If $y = f(x)$ is given and easy to use → favor $dx$ integrals
  • If $x = g(y)$ is given or $f(x)$ is easy to invert → favor $dy$ integrals

Before You Start

1. Can you set up a shell integral for rotation about a vertical axis?

For rotation about $x = k$: $$V = \int 2\pi \lvert x - k \rvert \cdot (\text{height}) \, dx$$

If unclear, review Shell Method: Other Axes.

2. Can you set up a washer integral for rotation about the $x$-axis?

$$V = \int \pi [R(x)^2 - r(x)^2] \, dx$$

where $R$ is the outer radius and $r$ is the inner radius.

If unclear, review Disk/Washer Method.

3. Can you solve $y = x^2$ for $x$ in terms of $y$?

Answer: $x = \sqrt{y}$ (for $x \geq 0$) or $x = -\sqrt{y}$ (for $x \leq 0$).

Inverting functions is often the deciding factor in which method to choose.

The Core Question

You have a region. You need to rotate it about an axis. You have two methods: shells and washers. Which one should you use?

The answer depends on three factors:

  1. How the region is described (functions of $x$ or $y$?)
  2. Which axis you’re rotating around
  3. Which integral is easier to evaluate

Making that choice quickly and confidently is the goal: read the region and the axis before computing.

Prerequisite Map

This skillShells vs. Washers: Choosing the Right Method
Leads tono further branch yet

Quick Reference

Property Value
Chapter 5 - Applications of Integration
Section 5.3
Difficulty Intermediate
Time ~25 minutes

Key Concepts

The Fundamental Difference

Aspect Washers/Disks Shells
Slice direction Perpendicular to axis Parallel to axis
Shape of slice Circular cross-section Cylindrical tube
Integration variable Same direction as axis Opposite direction to axis
Formula structure $\pi R^2 - \pi r^2$ $2\pi r h$

The Decision Flowchart

              ┌─────────────────────────────┐
              │ What axis of rotation?      │
              └─────────────┬───────────────┘
                            │
          ┌─────────────────┴─────────────────┐
          │                                   │
  ┌───────▼───────┐                   ┌───────▼───────┐
  │ Vertical axis │                   │ Horizontal    │
  │ (y-axis, x=k) │                   │ axis (x-axis, │
  │               │                   │ y=k)          │
  └───────┬───────┘                   └───────┬───────┘
          │                                   │
  ┌───────┴───────┐                   ┌───────┴───────┐
  │               │                   │               │
  ▼               ▼                   ▼               ▼
┌─────────┐  ┌─────────┐        ┌─────────┐  ┌─────────┐
│ Washers │  │ Shells  │        │ Washers │  │ Shells  │
│ use dy  │  │ use dx  │        │ use dx  │  │ use dy  │
└─────────┘  └─────────┘        └─────────┘  └─────────┘

The Quick Decision Rules

Rule 1: Match the natural description

Region Description Favored Variable Reason
$y = f(x)$ given, $f$ is complicated $dx$ Avoid solving for $x$
$x = g(y)$ given $dy$ Natural form
$y = f(x)$ is easy to invert Either Choose based on axis

Rule 2: Match variable to method

Axis Direction To use $dx$ To use $dy$
Vertical ($x = k$) Shells Washers
Horizontal ($y = k$) Washers Shells

Rule 3: Combine Rules 1 and 2

Situation Best Choice
Vertical axis + $y = f(x)$ natural Shells (both favor $dx$)
Vertical axis + $x = g(y)$ natural Washers (both favor $dy$)
Horizontal axis + $y = f(x)$ natural Washers (both favor $dx$)
Horizontal axis + $x = g(y)$ natural Shells (both favor $dy$)

When Shells Clearly Win

Use shells when:

Situation Example Why Shells Win
$y = f(x)$ is hard to invert $y = x^3 - x$ Avoid cubic formula
Function has multiple branches $y^2 = x$ gives $y = \pm\sqrt{x}$ One integral vs. two
Rotating about vertical axis with $y = f(x)$ Standard setup Natural match

When Washers Clearly Win

Use washers when:

Situation Example Why Washers Win
Region has obvious inner/outer radii Annular region Direct application
Rotating about $x$-axis with simple $f(x)$ $y = x^2$ Just square the function
$x = g(y)$ is the given form $x = y^2$ Natural match

The Strip Direction Rule

Visual test: Draw a thin strip in the region. Which direction is it?

      Vertical axis                    Horizontal axis
           │                                 ═══
     ┌─────┼─────┐                     ┌─────────────┐
     │  │  │  │  │   ← vertical        │ ─ ─ ─ ─ ─ │   ← horizontal
     │  │  │  │  │     strips          │ ─ ─ ─ ─ ─ │     strips
     │  │  │  │  │     = SHELLS        │ ─ ─ ─ ─ ─ │     = SHELLS
     └─────┴─────┘                     └─────────────┘

     ┌───────────┐                     ┌─────────────┐
     │ ───────── │   ← horizontal      │  │  │  │  │ │   ← vertical
     │ ───────── │     strips          │  │  │  │  │ │     strips
     │ ───────── │     = WASHERS       │  │  │  │  │ │     = WASHERS
     └───────────┘                     └─────────────┘

Side-by-Side Comparisons

Comparison 1: $y = x^3$, $y = 0$, $x = 2$ rotated about $y$-axis

Shells (use $dx$): $$V = \int_0^2 2\pi x \cdot x^3 \, dx = 2\pi \int_0^2 x^4 \, dx = 2\pi \cdot \frac{32}{5} = \frac{64\pi}{5}$$

Washers (use $dy$, need $x = y^{1/3}$):

Bounds: $y \in [0, 8]$ $$V = \int_0^8 \pi(4 - y^{2/3}) \, dy = \pi\left[4y - \frac{3y^{5/3}}{5}\right]_0^8 = \pi\left(32 - \frac{96}{5}\right) = \frac{64\pi}{5}$$

Verdict: Both work. Shells required slightly fewer steps (no cube root inversion needed).

Comparison 2: $y = 4x^2 - x^3$, $y = 0$ rotated about $y$-axis

Shells (use $dx$): $$V = \int_0^4 2\pi x(4x^2 - x^3) \, dx = 2\pi \int_0^4 (4x^3 - x^4) \, dx$$

Straightforward polynomial integration gives $V = \frac{512\pi}{5}$.

Washers (use $dy$):

Would need to solve $y = 4x^2 - x^3$ for $x$ in terms of $y$. This cubic equation has no nice closed form: you’d need the cubic formula.

Verdict: Shells win decisively. This type of problem is exactly why the shell method was developed.

Comparison 3: $y = x^{1/3}$, $y = 0$, $x = 8$ rotated about $x$-axis

Washers (use $dx$): $$V = \int_0^8 \pi(x^{1/3})^2 \, dx = \pi \int_0^8 x^{2/3} \, dx = \pi \cdot \frac{3(32)}{5} = \frac{96\pi}{5}$$

Shells (use $dy$, need $x = y^3$): $$V = \int_0^2 2\pi y(8 - y^3) \, dy = 2\pi \int_0^2 (8y - y^4) \, dy = \frac{96\pi}{5}$$

Verdict: Both work. Washers slightly simpler (just square the function).

📜 Why Both Methods Give the Same Answer

This isn’t coincidence: it’s a theorem. Both methods compute the same volume by summing up pieces differently:

  • Washers sum circular cross-sections perpendicular to the axis
  • Shells sum cylindrical tubes parallel to the axis

It’s like computing the area of a rectangle as (width × height) vs. (height × width), two different approaches giving the same answer.

The fact that both methods agree provides a powerful verification technique: if you’re unsure of your answer, try the other method!

Practice Problems

Level 1 Method Identification

For each situation, state whether shells or washers would be simpler (don’t solve):

  1. $y = \sin x$, $y = 0$, $0 \le x \le \pi$, rotated about the $y$-axis
  2. $x = y^2$, $x = 4$, rotated about the $x$-axis
  3. $y = e^x$, $y = 1$, $y = 2$, $x = 0$, rotated about the $x$-axis
Thought Process

Strategy for each:

  1. Note the axis of rotation
  2. Identify how the region is naturally described ($x$ or $y$?)
  3. Check if inversion would be required for the other method
  4. Choose the method that avoids complications

Key questions:

  • Is the function easy to invert?
  • Would one method require multiple integrals?
  • Is the natural variable compatible with the axis?
Show Answer

(a) $y = \sin x$ rotated about $y$-axis → Shells

  • Axis: vertical ($y$-axis)
  • Natural form: $y = \sin x$ (function of $x$)
  • To use washers: would need to solve $y = \sin x$ for $x$, giving $x = \arcsin y$. This is valid only for $y \in [0,1]$ and complicated.
  • Shells use $dx$ and keep the simple form $\sin x$.

(b) $x = y^2$, $x = 4$ rotated about $x$-axis → Washers (slightly better)

  • Axis: horizontal ($x$-axis)
  • Natural form: $x = y^2$ (function of $y$)
  • Washers: horizontal slices give outer radius 4, inner radius $y^2$
  • Shells: would also work with horizontal strips

Both methods are reasonable. Washers are slightly more direct since we see the “hole” clearly.

(c) $y = e^x$ region rotated about $x$-axis → Shells

  • Axis: horizontal ($x$-axis)
  • The region is bounded by horizontal lines $y = 1$ and $y = 2$
  • Shells: horizontal strips with radius $y$ and width $\ln y$ (from $x = \ln y$)
  • Washers: would require splitting into parts

Shells give a single integral; washers would need careful handling of the region.

Level 2 Both Methods, Same Answer

The region bounded by $y = \sqrt{x}$ and $y = x^2$ is rotated about the $x$-axis.

  1. Set up and evaluate the integral using washers.
  2. Set up and evaluate the integral using shells.
  3. Verify both give the same answer.
Thought Process

For washers (about $x$-axis): Use vertical slices. The outer radius is the curve farther from the axis, inner radius is the curve closer.

For shells (about $x$-axis): Use horizontal strips. We need to express boundaries as functions of $y$: from $y = \sqrt{x}$ get $x = y^2$; from $y = x^2$ get $x = \sqrt{y}$.

Finding bounds:

  • Intersection: $\sqrt{x} = x^2 \Rightarrow x = x^4 \Rightarrow x(x^3 - 1) = 0 \Rightarrow x = 0$ or $x = 1$
  • For washers: $x \in [0, 1]$
  • For shells: $y \in [0, 1]$
Show Answer

Intersection points: $\sqrt{x} = x^2 \Rightarrow x = x^4 \Rightarrow x = 0$ or $x = 1$

Which curve is farther from the $x$-axis? At $x = 0.5$: $\sqrt{0.5} \approx 0.71$ and $(0.5)^2 = 0.25$. So $y = \sqrt{x}$ is farther.

(a) Washers:

Vertical slices perpendicular to $x$-axis:

  • Outer radius: $R = \sqrt{x}$ (farther from $x$-axis)
  • Inner radius: $r = x^2$ (closer to $x$-axis)
  • Bounds: $x \in [0, 1]$

$$V = \int_0^1 \pi(x - x^4) \, dx = \pi\left[\frac{x^2}{2} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{2} - \frac{1}{5}\right) = \frac{3\pi}{10}$$

(b) Shells:

Horizontal strips parallel to $x$-axis. First, solve for $x$ in terms of $y$:

  • From $y = \sqrt{x}$: $x = y^2$
  • From $y = x^2$: $x = \sqrt{y}$

At $y = 0.5$: $y^2 = 0.25$ and $\sqrt{y} \approx 0.71$. So right boundary is $x = \sqrt{y}$, left is $x = y^2$.

  • Radius: $y$ (distance to $x$-axis)
  • Width: $\sqrt{y} - y^2$
  • Bounds: $y \in [0, 1]$

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

$$= 2\pi\left[\frac{2y^{5/2}}{5} - \frac{y^4}{4}\right]_0^1 = 2\pi\left(\frac{2}{5} - \frac{1}{4}\right) = 2\pi \cdot \frac{3}{20} = \frac{3\pi}{10}$$

(c) Verification: Both methods give $V = \frac{3\pi}{10}$ ✓

Level 3 Strategic Choice

Find the volume of the solid obtained by rotating the region bounded by $y = 4 - x^2$ and $y = 3$ about the line $y = 3$.

Thought Process

Axis analysis: The axis $y = 3$ is horizontal.

Region analysis: The parabola $y = 4 - x^2$ intersects $y = 3$ when $4 - x^2 = 3$, giving $x = \pm 1$. The region is above $y = 3$ (touching it at $x = \pm 1$).

Method comparison:

  • Washers: Vertical slices give disks (no hole since region touches axis). Radius = $(4 - x^2) - 3 = 1 - x^2$.
  • Shells: Horizontal strips. Would need to express $x$ as function of $y$: $x = \pm\sqrt{4-y}$, giving two branches.

Decision: Washers are simpler: one integral, no square roots.

Show Answer

Find bounds: $4 - x^2 = 3 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$

Choose method:

Since the region touches the axis $y = 3$ at its boundary, vertical slices become disks (not washers, no hole).

Washers (disks) setup:

  • Radius: $(4 - x^2) - 3 = 1 - x^2$
  • Bounds: $x \in [-1, 1]$

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

Use symmetry (integrand is even): $$= 2\pi \int_0^1 (1 - 2x^2 + x^4) \, dx = 2\pi\left[x - \frac{2x^3}{3} + \frac{x^5}{5}\right]_0^1$$

$$= 2\pi\left(1 - \frac{2}{3} + \frac{1}{5}\right) = 2\pi \cdot \frac{15 - 10 + 3}{15} = 2\pi \cdot \frac{8}{15} = \frac{16\pi}{15}$$

Why washers won: The region is bounded by $y = 4 - x^2$ (function of $x$), and the axis is horizontal. Using shells would require solving for $x = \pm\sqrt{4-y}$ and handling two branches.

Level 4 Shells Necessary

Find the volume of the solid obtained by rotating the region bounded by $y = x^3 - x$ and $y = 0$ (for $x \geq 0$) about the $y$-axis.

Hint: Solving $y = x^3 - x$ for $x$ in terms of $y$ leads to the cubic formula.

Thought Process

Axis: Vertical ($y$-axis)

Natural form: $y = x^3 - x = x(x^2 - 1) = x(x-1)(x+1)$

Roots for $x \geq 0$: $x = 0$ and $x = 1$

Method comparison:

  • Shells: Use $dx$, keep the polynomial form. Simple!
  • Washers: Would need $x$ as a function of $y$, requiring the cubic formula. Nightmare!

Careful: For $x \in [0, 1]$, check the sign of $y = x^3 - x = x(x-1)(x+1)$.

  • At $x = 0.5$: $y = 0.125 - 0.5 = -0.375 < 0$

The curve is below the $x$-axis! So the “height” is $\vert y\vert = \vert x^3 - x\vert = x - x^3$ (since $x^3 - x < 0$).

Show Answer

Analyze the curve: $y = x^3 - x = x(x-1)(x+1)$

For $x \geq 0$:

  • Roots at $x = 0$ and $x = 1$
  • For $x \in (0, 1)$: $y < 0$ (curve below $x$-axis)

The bounded region for $x \geq 0$ is between the curve and the $x$-axis on $[0, 1]$.

Shell setup:

The “height” of each shell is the vertical distance from $y = 0$ to the curve: $$\text{height} = \vert x^3 - x\vert = x - x^3 \quad \text{(since } x^3 - x < 0 \text{ on } (0,1)\text{)}$$

  • Radius: $x$
  • Height: $x - x^3$
  • Bounds: $[0, 1]$

$$V = \int_0^1 2\pi x(x - x^3) \, dx = 2\pi \int_0^1 (x^2 - x^4) \, dx$$

$$= 2\pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = 2\pi\left(\frac{1}{3} - \frac{1}{5}\right) = 2\pi \cdot \frac{2}{15} = \frac{4\pi}{15}$$

Why shells were essential: Solving $y = x^3 - x$ for $x$ requires the cubic formula. For each $y$-value in $\left(-\frac{2}{3\sqrt{3}}, 0\right)$, there are multiple $x$-values, making washers impractical.

Level 5 Complete Method Analysis

Consider the region $R$ bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$.

  1. Set up volume integrals for rotating $R$ about each of these four axes using the most efficient method: $x$-axis, $y$-axis, $x = 4$, and $y = 2$.
  2. For which axis does the choice of method matter most? Explain.
  3. Evaluate all four integrals.
Thought Process

For each axis:

  1. Determine if it’s horizontal or vertical
  2. Consider which variable describes the region naturally
  3. Match method to minimize inversion and complexity

The region: Under $y = \sqrt{x}$ from $x = 0$ to $x = 4$.

  • As $y = f(x)$: $y = \sqrt{x}$
  • As $x = g(y)$: $x = y^2$

Both forms are simple, so we have flexibility.

Systematic analysis:

  • $x$-axis (horizontal): Washers use $dx$ (natural), shells use $dy$
  • $y$-axis (vertical): Shells use $dx$ (natural), washers use $dy$
  • $x = 4$ (vertical): Shells use $dx$, washers use $dy$
  • $y = 2$ (horizontal): Washers use $dx$, shells use $dy$
Show Answer

(a) Integrals:

About $x$-axis (horizontal):

Best: Washers (vertical slices, use $dx$)

  • Radius: $\sqrt{x}$
  • Bounds: $x \in [0, 4]$

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

About $y$-axis (vertical):

Best: Shells (vertical strips, use $dx$)

  • Radius: $x$
  • Height: $\sqrt{x}$
  • Bounds: $x \in [0, 4]$

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

About $x = 4$ (vertical):

Best: Washers (horizontal slices, use $dy$)

  • Outer radius: $4 - y^2$ (from $x = y^2$ to $x = 4$)
  • Inner radius: $0$ (disk, not washer)
  • Bounds: $y \in [0, 2]$

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

About $y = 2$ (horizontal):

Best: Washers (vertical slices, use $dx$)

  • Radius: $2 - \sqrt{x}$
  • Bounds: $x \in [0, 4]$

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

(b) Where method matters most:

For the $y$-axis rotation, the choice matters most:

  • Shells: $\int 2\pi x^{3/2} \, dx$ (single term, easy)
  • Washers: $\int \pi(16 - y^4) \, dy$ (requires converting bounds and inverting)

If the function were harder to invert (like $y = x + \sin x$), shells would be the only practical option. The $y$-axis case best illustrates why shells were invented.

(c) Evaluate all four:

$x$-axis: $$V = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi$$

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

$x = 4$: $$V = \pi \int_0^2 (16 - 8y^2 + y^4) \, dy = \pi\left[16y - \frac{8y^3}{3} + \frac{y^5}{5}\right]_0^2$$ $$= \pi\left(32 - \frac{64}{3} + \frac{32}{5}\right) = \pi \cdot \frac{480 - 320 + 96}{15} = \frac{256\pi}{15}$$

$y = 2$: $$V = \pi \int_0^4 (4 - 4\sqrt{x} + x) \, dx = \pi\left[4x - \frac{8x^{3/2}}{3} + \frac{x^2}{2}\right]_0^4$$ $$= \pi\left(16 - \frac{64}{3} + 8\right) = \pi\left(24 - \frac{64}{3}\right) = \pi \cdot \frac{72 - 64}{3} = \frac{8\pi}{3}$$

Common Mistakes

Mistake Why It Happens Correction
Always using the method learned first Habit, not strategy Ask: “Which variable describes the region more naturally?”
Forgetting to change bounds when switching methods Mixing $x$ and $y$ bounds If switching $dx \to dy$, convert all bounds to $y$-values.
Not checking if inversion is feasible Jumping into washers without thinking Before choosing washers for vertical axis, verify that $y = f(x)$ can be solved for $x$.
Using shells when washers are simpler “Shells are always better” mindset For simple functions rotated about the $x$-axis, washers often win.

Still Confused?

  • Shell setup unclear? → Review Shell Method: Other Axes
  • Washer setup unclear? → Review Disk/Washer Method
  • Can’t decide which method? → Default rule: use the variable that matches how the region is described
  • Both methods seem hard? → Try both! Comparing partial setups often reveals which is simpler

Common Misconceptions

Common misconception

shells always require integrating with respect to $x$, and washers always require integrating with respect to $y$.

This is the iconic-graph error. The integration variable is determined by the direction of the strips, not by the method label. Shells use strips parallel to the axis of rotation: for a vertical axis, the strips are vertical and integration is with respect to $x$; for a horizontal axis, the strips are horizontal and integration is with respect to $y$. Washers use slices perpendicular to the axis, so the integration variable is opposite. For the region under $y = \sqrt{x}$ rotated about the $x$-axis, the washer method uses vertical slices and integrates with $dx$, while the shell method uses horizontal strips and integrates with $dy$. The method and the integration variable are linked to the strip direction, not fixed.


Mastery Checklist

Looking Ahead

With both methods mastered, you’re ready for:

Mental Model

The Strip Direction Rule:

Draw your region. Draw a thin strip. Rotate it mentally:

Pick the strip direction that makes the boundaries easiest to describe.

If boundaries are given as $y = f(x)$, vertical strips (which preserve that form) are usually easier. If boundaries are given as $x = g(y)$, horizontal strips are usually easier.


Last updated: 2026-01-23