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Unit 8.12 – Volume with Washer Method: Revolving Around Other Axes

AP® Calculus AB & BC | Advanced Hollow Solids of Revolution

Why This Matters: This is the ULTIMATE challenge in volumes of revolution! Combining the washer method with rotation around lines like \(y = k\) or \(x = h\) requires careful calculation of BOTH outer and inner radii as distances from the axis. This is where AP® exams test your complete mastery of solids of revolution. Master this and you've conquered the entire topic!

🎯 The Critical Concept: Double Distance Calculation

THE GOLDEN RULE

Both Radii Are Distances:

  • Outer radius: Distance from axis to outer curve
  • Inner radius: Distance from axis to inner curve
\[ R = |\text{outer curve} - \text{axis}| \]
\[ r = |\text{inner curve} - \text{axis}| \]

⚠️ CRITICAL: You CANNOT just use function values. You MUST subtract the axis value from BOTH curves!

↔️ Revolving Around Horizontal Line y = k

Washer Method: y = k

THE FORMULA:

When rotating region between \(y = f(x)\) and \(y = g(x)\) around \(y = k\):

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

where

\[ R(x) = |\text{curve farther from } y=k - k| \]
\[ r(x) = |\text{curve closer to } y=k - k| \]
Common Scenario: Both curves on same side of axis

If \(f(x) > g(x) > k\) (both above the axis):

  • Outer radius: \(R(x) = f(x) - k\)
  • Inner radius: \(r(x) = g(x) - k\)

If \(k > f(x) > g(x)\) (both below the axis):

  • Outer radius: \(R(x) = k - g(x)\)
  • Inner radius: \(r(x) = k - f(x)\)

↕️ Revolving Around Vertical Line x = h

Washer Method: x = h

THE FORMULA:

When rotating region between \(x = f(y)\) and \(x = g(y)\) around \(x = h\):

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

where

\[ R(y) = |\text{curve farther from } x=h - h| \]
\[ r(y) = |\text{curve closer to } x=h - h| \]
Common Scenario:

If \(f(y) > g(y) > h\) (both right of axis):

  • Outer radius: \(R(y) = f(y) - h\)
  • Inner radius: \(r(y) = g(y) - h\)

If \(h > f(y) > g(y)\) (both left of axis):

  • Outer radius: \(R(y) = h - g(y)\)
  • Inner radius: \(r(y) = h - f(y)\)

🔍 Determining Outer and Inner Radii

Step-by-Step Process:

  1. Sketch the region and axis: Visualize which curve is farther
  2. For each curve, calculate distance from axis:
    • For \(y = k\): distance = \(|f(x) - k|\) and \(|g(x) - k|\)
    • For \(x = h\): distance = \(|f(y) - h|\) and \(|g(y) - h|\)
  3. Compare distances: Larger distance = outer radius
  4. Set up integral: \(V = \pi\int([R]^2 - [r]^2)\)

📖 Comprehensive Worked Examples

Example 1: Around y = -1 (Horizontal Line)

Problem: Find the volume when the region bounded by \(y = x\) and \(y = x^2\) is rotated around \(y = -1\).

Solution:

Step 1: Find intersection points

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

Bounds: \([0, 1]\)

Step 2: Sketch and analyze

Both curves are ABOVE \(y = -1\)

On \([0, 1]\): \(x > x^2\)

Step 3: Calculate distances from \(y = -1\)

Distance from \(y = x\) to \(y = -1\): \(x - (-1) = x + 1\)

Distance from \(y = x^2\) to \(y = -1\): \(x^2 - (-1) = x^2 + 1\)

Since \(x + 1 > x^2 + 1\) when \(x < x^2\) is false, we need to check:

At \(x = 0.5\): \(0.5 + 1 = 1.5\) and \(0.25 + 1 = 1.25\)

So \(y = x\) is farther!

Step 4: Set up and evaluate

Outer: \(R = x + 1\), Inner: \(r = x^2 + 1\)

\[ V = \pi \int_0^1 [(x+1)^2 - (x^2+1)^2] \, dx \]
\[ = \pi \int_0^1 [(x^2 + 2x + 1) - (x^4 + 2x^2 + 1)] \, dx \]
\[ = \pi \int_0^1 (-x^4 - x^2 + 2x) \, dx \]
\[ = \pi\left[-\frac{x^5}{5} - \frac{x^3}{3} + x^2\right]_0^1 = \pi\left(-\frac{1}{5} - \frac{1}{3} + 1\right) = \frac{7\pi}{15} \]

ANSWER: \(V = \frac{7\pi}{15}\) cubic units

Example 2: Around y = 2 (Above the Region)

Problem: Region between \(y = x^2\) and \(y = 0\) from \(x = 0\) to \(x = 1\), rotated around \(y = 2\).

Analysis:

Both curves BELOW \(y = 2\)

Distance from \(y = x^2\) to \(y = 2\): \(2 - x^2\)

Distance from \(y = 0\) to \(y = 2\): \(2 - 0 = 2\)

\(y = 0\) is farther from axis (constant distance of 2)

Setup and evaluate:

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

\[ V = \pi \int_0^1 [2^2 - (2-x^2)^2] \, dx \]
\[ = \pi \int_0^1 [4 - (4 - 4x^2 + x^4)] \, dx = \pi \int_0^1 (4x^2 - x^4) \, dx \]
\[ = \pi\left[\frac{4x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{4}{3} - \frac{1}{5}\right) = \frac{17\pi}{15} \]

Example 3: Around x = -1 (Vertical Line)

Problem: Region bounded by \(x = y^2\) and \(x = 4\) rotated around \(x = -1\).

Setup:

When \(x = 4\): \(y = \pm 2\), so bounds: \(y \in [-2, 2]\)

Both curves RIGHT of \(x = -1\)

Distance from \(x = 4\) to \(x = -1\): \(4 - (-1) = 5\)

Distance from \(x = y^2\) to \(x = -1\): \(y^2 - (-1) = y^2 + 1\)

\(x = 4\) is farther (constant distance 5 > \(y^2 + 1\) for all \(y \in [-2,2]\))

Evaluate:

Outer: \(R = 5\), Inner: \(r = y^2 + 1\)

\[ V = \pi \int_{-2}^2 [25 - (y^2+1)^2] \, dy \]
\[ = \pi \int_{-2}^2 [25 - y^4 - 2y^2 - 1] \, dy = \pi \int_{-2}^2 (24 - 2y^2 - y^4) \, dy \]

By symmetry:

\[ = 2\pi \int_0^2 (24 - 2y^2 - y^4) \, dy = 2\pi\left[24y - \frac{2y^3}{3} - \frac{y^5}{5}\right]_0^2 \]
\[ = 2\pi\left(48 - \frac{16}{3} - \frac{32}{5}\right) = \frac{1088\pi}{15} \]

Example 4: Around x = 2 (Between the Curves)

Problem: Region between \(x = y^2\) and \(x = 2y\) rotated around \(x = 2\).

Find intersections and analyze:

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

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

Distance from \(x = 2y\) to \(x = 2\): \(|2y - 2| = 2|y - 1|\)

Distance from \(x = y^2\) to \(x = 2\): \(|y^2 - 2|\)

Need to check which is farther...

For \(y \in [0, \sqrt{2}]\): \(y^2 < 2\), so distance = \(2 - y^2\)

For \(y \in [\sqrt{2}, 2]\): \(y^2 > 2\), so distance = \(y^2 - 2\)

This problem requires splitting the integral!

📋 Complete Step-by-Step Process

Systematic Approach

The 8-Step Method:

  1. Identify axis of rotation: \(y = k\) or \(x = h\)?
  2. Sketch region and axis: Critical for visualization!
  3. Find intersection points: Determine bounds
  4. For each curve, calculate distance from axis:
    • Subtract axis value from curve
    • Use absolute value if needed
  5. Compare distances: Identify outer (larger) and inner (smaller)
  6. Set up integral: \(V = \pi\int([R]^2 - [r]^2)\)
  7. Expand and simplify
  8. Evaluate and check answer

📊 Complete Comparison Table

Washer Method: All Axes
Axis Outer Radius Inner Radius Variable
x-axis \(R = f(x)\) (upper) \(r = g(x)\) (lower) \(dx\)
y-axis \(R = f(y)\) (right) \(r = g(y)\) (left) \(dy\)
\(y = k\) \(R = |f(x) - k|\) (farther) \(r = |g(x) - k|\) (closer) \(dx\)
\(x = h\) \(R = |f(y) - h|\) (farther) \(r = |g(y) - h|\) (closer) \(dy\)

💡 Essential Tips & Strategies

✅ Success Strategies:

  • ALWAYS sketch: Cannot stress this enough for other axes!
  • Calculate BOTH distances: From axis to each curve
  • Compare distances at a test point: To determine outer/inner
  • Subtract axis value from BOTH curves
  • Watch the signs: If curve below/left of axis, axis - curve
  • Square each radius separately: Then subtract
  • Expand carefully: Many terms to keep track of
  • Include π in answer

🔥 Quick Decision Guide:

  • Both curves above \(y = k\): Subtract k from both
  • Both curves below \(y = k\): Subtract both from k
  • Both curves right of \(x = h\): Subtract h from both
  • Both curves left of \(x = h\): Subtract both from h
  • Curves on opposite sides: Calculate each distance separately

❌ Common Mistakes to Avoid

  • Mistake 1: Using function values instead of distances (forgetting to subtract axis value)
  • Mistake 2: Only subtracting axis from one curve, not both
  • Mistake 3: Wrong subtraction order (curve - axis vs axis - curve)
  • Mistake 4: Not determining which is outer and which is inner
  • Mistake 5: Squaring \((R - r)\) instead of \(R^2 - r^2\)
  • Mistake 6: Sign errors when expanding
  • Mistake 7: Not sketching the region (huge mistake for this topic!)
  • Mistake 8: Wrong variable (\(dx\) vs \(dy\))
  • Mistake 9: Integration errors
  • Mistake 10: Forgetting π or wrong final simplification

📝 Practice Problems

Find the volume:

  1. Region between \(y = x\) and \(y = x^2\) rotated around \(y = 2\).
  2. Region between \(y = x^2\) and \(y = 4\) rotated around \(y = -1\).
  3. Region between \(x = y^2\) and \(x = 4\) rotated around \(x = 5\).
  4. Region between \(y = \sqrt{x}\) and \(y = x\) rotated around \(y = -1\).

Answers:

  1. \(\frac{13\pi}{30}\) cubic units
  2. \(\frac{608\pi}{15}\) cubic units
  3. \(\frac{192\pi}{5}\) cubic units
  4. \(\frac{13\pi}{6}\) cubic units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Clearly identify axis: State which line you're rotating around
  • Show distance calculations: For BOTH curves
  • Explain outer vs inner: "Distance from \(y = k\) to curve is..."
  • Correct setup: \(V = \pi\int([R]^2 - [r]^2)\) with correct distances
  • Show expansion: \([f(x)-k]^2 - [g(x)-k]^2\)
  • Show integration work
  • Evaluate at bounds
  • Include π in final answer

💯 Exam Strategy:

  1. Read axis of rotation VERY carefully
  2. Sketch region AND axis of rotation
  3. Calculate distance from axis to EACH curve
  4. Compare distances to determine outer/inner
  5. Write: "Outer radius = [formula], Inner radius = [formula]"
  6. Set up integral showing both squared terms
  7. Expand completely before integrating
  8. Evaluate and simplify
  9. Check: Does answer make sense?

⚡ Quick Reference Guide

WASHER - OTHER AXES ESSENTIALS

Around \(y = k\):

\[ V = \pi \int_a^b \left([f(x)-k]^2 - [g(x)-k]^2\right) \, dx \]

Calculate distance from \(y = k\) to EACH curve

Around \(x = h\):

\[ V = \pi \int_c^d \left([f(y)-h]^2 - [g(y)-h]^2\right) \, dy \]

Calculate distance from \(x = h\) to EACH curve

THE GOLDEN RULE:

  • BOTH radii are distances from axis!
  • Outer = curve FARTHER from axis
  • Inner = curve CLOSER to axis
  • Subtract axis value from BOTH curves!
  • SKETCH IS MANDATORY!

Master Washer Method Around Other Axes! The ultimate formula: \(V = \pi\int_a^b([R]^2-[r]^2)\,dx\) where BOTH \(R\) and \(r\) are DISTANCES from axis of rotation. For \(y = k\): \(R = |f(x)-k|\) and \(r = |g(x)-k|\) where farther curve gives outer radius. For \(x = h\): \(R = |f(y)-h|\) and \(r = |g(y)-h|\). Critical concept: you CANNOT use function values directly—must calculate distance by subtracting axis value from BOTH curves. Common scenario: both curves same side of axis → if above/right, subtract axis from curves; if below/left, subtract curves from axis. Outer is always curve FARTHER from axis. Process: (1) identify axis, (2) SKETCH, (3) calculate BOTH distances, (4) compare to determine outer/inner, (5) set up integral with both distances, (6) expand and integrate. Most common error: forgetting to subtract axis from both curves. This is the HARDEST volume topic—requires complete mastery of all concepts. Practice extensively! 🎯✨