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Unit 8.11 – Volume with Washer Method: Revolving Around the x- or y-Axis

AP® Calculus AB & BC | Hollow Solids of Revolution

Why This Matters: The washer method extends the disc method to regions between TWO curves! When you rotate a region bounded by two functions around an axis, you create a hollow solid with a hole in the middle. Each cross section is a washer (annulus)—a disc with a hole. The volume is found by subtracting the inner disc from the outer disc. This is essential for AP® Calculus and appears frequently on exams!

🎯 The Washer Method Concept

FROM DISC TO WASHER

Disc Method: Region between ONE curve and an axis → solid disc cross sections

Washer Method: Region between TWO curves → washer (hollow disc) cross sections

The Fundamental Formula:

\[ \text{Washer Area} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \]

where \(R\) = outer radius, \(r\) = inner radius

📝 Key Insight: The volume of a washer is the volume of the outer disc MINUS the volume of the inner disc (the hole).

↔️ Washer Method: x-Axis

Revolving Around x-Axis

THE FORMULA:

When rotating region between \(y = f(x)\) (upper) and \(y = g(x)\) (lower) around the x-axis:

\[ V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right) \, dx \]
\[ V = \pi \int_a^b \left([f(x)]^2 - [g(x)]^2\right) \, dx \]
Where:
  • \(R(x) = f(x)\) = outer radius (distance from x-axis to outer curve)
  • \(r(x) = g(x)\) = inner radius (distance from x-axis to inner curve)
  • \(f(x) \geq g(x)\) on \([a, b]\) (upper curve minus lower curve)
  • \([a, b]\) = bounds of integration

⚠️ Critical: Outer radius is the curve FARTHER from the axis. Inner radius is the curve CLOSER to the axis. Order matters!

↕️ Washer Method: y-Axis

Revolving Around y-Axis

THE FORMULA:

When rotating region between \(x = f(y)\) (right) and \(x = g(y)\) (left) around the y-axis:

\[ V = \pi \int_c^d \left([R(y)]^2 - [r(y)]^2\right) \, dy \]
\[ V = \pi \int_c^d \left([f(y)]^2 - [g(y)]^2\right) \, dy \]
Where:
  • \(R(y) = f(y)\) = outer radius (distance from y-axis to outer curve)
  • \(r(y) = g(y)\) = inner radius (distance from y-axis to inner curve)
  • \(f(y) \geq g(y)\) on \([c, d]\) (right curve minus left curve)
  • Must express curves as \(x = f(y)\) and \(x = g(y)\)

🔍 Finding Outer and Inner Radius

How to Identify Outer vs Inner:

For x-Axis Rotation:
  • Outer radius: The function FARTHER from the x-axis (larger |y| value)
  • Inner radius: The function CLOSER to the x-axis (smaller |y| value)
  • If both positive: outer = upper curve, inner = lower curve
  • If one is negative: compare distances (absolute values)
For y-Axis Rotation:
  • Outer radius: The function FARTHER from the y-axis (larger |x| value)
  • Inner radius: The function CLOSER to the y-axis (smaller |x| value)
  • If both positive: outer = right curve, inner = left curve

💡 Pro Tip: Sketch the region! Visually identify which curve is farther from the axis of rotation.

📋 Step-by-Step Process

Complete Method

The 7-Step Approach:

  1. Sketch the region: Draw both curves
  2. Identify axis of rotation: x-axis or y-axis?
  3. Find intersection points: These are your bounds
  4. Determine outer and inner radius: Which curve is farther from axis?
  5. Set up integral: \(V = \pi\int_a^b([R]^2 - [r]^2)\,dx\) or \(dy\)
  6. Expand: \([R]^2 - [r]^2\) (DON'T factor as difference of squares!)
  7. Evaluate and simplify

📖 Comprehensive Worked Examples

Example 1: Around x-Axis (Basic)

Problem: Find the volume when the region bounded by \(y = x\) and \(y = x^2\) is rotated around the x-axis.

Solution:

Step 1: Find intersection points

\[ x = x^2 \quad \Rightarrow \quad x = 0, 1 \]

Bounds: \([0, 1]\)

Step 2: Determine outer and inner radius

On \([0, 1]\): \(x > x^2\), so \(y = x\) is above \(y = x^2\)

Outer radius: \(R(x) = x\) (farther from x-axis)

Inner radius: \(r(x) = x^2\) (closer to x-axis)

Step 3: Set up integral

\[ V = \pi \int_0^1 \left(x^2 - (x^2)^2\right) \, dx = \pi \int_0^1 (x^2 - x^4) \, dx \]

Step 4: Evaluate

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

ANSWER: \(V = \frac{2\pi}{15}\) cubic units

Example 2: Around y-Axis

Problem: Rotate the region bounded by \(y = x^2\) and \(y = 4\) around the y-axis. Find the volume.

Step 1: Convert to x = g(y)

From \(y = x^2\): \(x = \sqrt{y}\) (taking positive root)

Right boundary: \(x = 2\) (when \(y = 4\))

But actually the region extends from \(x = -2\) to \(x = 2\)

By symmetry, we can use \(x = \sqrt{y}\) on right side

Step 2: Setup

Bounds: \(y = 0\) to \(y = 4\)

Outer radius: \(R(y) = 2\) (constant, the boundary \(x = 2\))

Inner radius: \(r(y) = \sqrt{y}\) (the parabola)

Step 3: Integrate

\[ V = \pi \int_0^4 (2^2 - (\sqrt{y})^2) \, dy = \pi \int_0^4 (4 - y) \, dy \]
\[ = \pi\left[4y - \frac{y^2}{2}\right]_0^4 = \pi(16 - 8) = 8\pi \]

Example 3: Both Functions Non-Zero

Problem: Region between \(y = 2\sqrt{x}\) and \(y = x\) from \(x = 0\) to \(x = 4\), rotated around x-axis.

Determine which is outer:

Test \(x = 1\): \(2\sqrt{1} = 2\) and \(1 = 1\)

So \(2\sqrt{x} > x\) on this interval

Outer: \(R = 2\sqrt{x}\), Inner: \(r = x\)

Evaluate:

\[ V = \pi \int_0^4 \left[(2\sqrt{x})^2 - x^2\right] \, dx = \pi \int_0^4 (4x - x^2) \, dx \]
\[ = \pi\left[2x^2 - \frac{x^3}{3}\right]_0^4 = \pi\left(32 - \frac{64}{3}\right) = \frac{32\pi}{3} \]

Example 4: Around y-Axis with Two Curves

Problem: Region bounded by \(x = y^2\) and \(x = 2y\) rotated around y-axis.

Find intersections:

\[ y^2 = 2y \quad \Rightarrow \quad y = 0, 2 \]

Setup:

On \([0, 2]\): \(2y > y^2\)

Outer: \(R(y) = 2y\), Inner: \(r(y) = y^2\)

\[ V = \pi \int_0^2 [(2y)^2 - (y^2)^2] \, dy = \pi \int_0^2 (4y^2 - y^4) \, dy \]
\[ = \pi\left[\frac{4y^3}{3} - \frac{y^5}{5}\right]_0^2 = \pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{64\pi}{15} \]

📊 Disc vs Washer Method

Comparison of Methods
Feature Disc Method Washer Method
Region Between ONE curve and axis Between TWO curves
Cross Section Solid disc Washer (hollow disc)
Formula \(\pi\int [R]^2\,dx\) \(\pi\int([R]^2 - [r]^2)\,dx\)
Radius One radius \(R\) Outer \(R\) and inner \(r\)
Solid Type Solid (no hole) Hollow (has hole)

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Sketch the region: Critical for identifying outer/inner
  • Find intersections first: These are your bounds
  • Test a point: To determine which curve is farther from axis
  • Square EACH radius: \([R]^2 - [r]^2\), not \((R-r)^2\)!
  • Don't factor: \(R^2 - r^2 \neq (R-r)(R+r)\) in the integral
  • Keep π outside: Factor it out of the integral
  • Expand before integrating: Makes it easier
  • Check units: Cubic units for volume

🔥 Common Setups:

  • Region between curves: Use washer method
  • Region to axis: Use disc method (special case: inner radius = 0)
  • Outer is farther: Always farther from axis of rotation
  • For y-axis: Convert both curves to \(x = g(y)\)

❌ Common Mistakes to Avoid

  • Mistake 1: Squaring the difference: \((R-r)^2\) instead of \(R^2 - r^2\)
  • Mistake 2: Swapping outer and inner radius
  • Mistake 3: Using disc method when washer is needed
  • Mistake 4: Forgetting to square one or both radii
  • Mistake 5: Wrong bounds (not finding intersection points)
  • Mistake 6: Forgetting π in the formula
  • Mistake 7: Sign errors when expanding
  • Mistake 8: Integration errors
  • Mistake 9: Not converting to \(x = g(y)\) for y-axis problems
  • Mistake 10: Arithmetic errors in final simplification

📝 Practice Problems

Find the volume:

  1. Region between \(y = x^2\) and \(y = 4\) rotated around x-axis.
  2. Region between \(y = \sqrt{x}\) and \(y = x\) rotated around x-axis.
  3. Region between \(x = y^2\) and \(x = 4\) rotated around y-axis.
  4. Region between \(y = x\) and \(y = x^3\) from \(x = 0\) to \(x = 1\) around x-axis.

Answers:

  1. \(\frac{512\pi}{15}\) cubic units
  2. \(\frac{\pi}{6}\) cubic units
  3. \(\frac{128\pi}{5}\) cubic units
  4. \(\frac{4\pi}{21}\) cubic units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Identify the method: State "washer method"
  • Show intersection points: How you found bounds
  • Identify outer and inner: State which is which
  • Correct setup: \(V = \pi\int([R]^2 - [r]^2)\,dx\)
  • Show squaring: Square each radius separately
  • Show expansion: \(R^2 - r^2\) expanded
  • Show integration: Find antiderivative
  • Evaluate at bounds: Show substitution
  • Include π: In final answer

💯 Exam Strategy:

  1. Read carefully: One curve or two curves?
  2. Sketch if time permits
  3. Find intersection points (bounds)
  4. Determine which curve is farther from axis
  5. Write: Outer = ..., Inner = ...
  6. Set up: \(V = \pi\int([R]^2 - [r]^2)\)
  7. Square each radius separately
  8. Expand and integrate
  9. Simplify and include π

⚡ Quick Reference Guide

WASHER METHOD ESSENTIALS

The Formula:

\[ V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right) \, dx \]

or

\[ V = \pi \int_c^d \left([R(y)]^2 - [r(y)]^2\right) \, dy \]

Remember:

  • Outer radius = curve farther from axis
  • Inner radius = curve closer to axis
  • Square EACH radius: \(R^2 - r^2\)
  • NOT \((R-r)^2\)!
  • Include π!

Master the Washer Method! The fundamental formula: \(V = \pi\int_a^b([R]^2 - [r]^2)\,dx\) where \(R\) = outer radius (curve farther from axis) and \(r\) = inner radius (curve closer to axis). The washer method: region between TWO curves rotated creates hollow solid → each cross section is washer (disc with hole) → area = \(\pi R^2 - \pi r^2 = \pi(R^2 - r^2)\) → integrate. Critical: square EACH radius separately, then subtract: \(R^2 - r^2\), NOT \((R-r)^2\)! For x-axis: \(R\) and \(r\) are y-values (distances from x-axis). For y-axis: must convert to \(x = g(y)\), \(R\) and \(r\) are x-values. Process: find intersections (bounds), determine outer/inner (sketch helps!), set up integral, square each radius, expand and integrate. Disc method is special case of washer where inner radius = 0. This is a major AP® topic—appears every year! Practice until automatic! 🎯✨