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Unit 8.10 – Volume with Disc Method: Revolving Around Other Axes

AP® Calculus AB & BC | Advanced Solids of Revolution

Why This Matters: Now we extend the disc method beyond the x- and y-axes! When rotating around horizontal lines \(y = k\) or vertical lines \(x = h\), the radius function changes—it's the distance from the axis of rotation to the curve. This is where disc method problems become challenging and where AP® exams really test your understanding. Master this and you've conquered the full scope of volumes by rotation!

🎯 The Critical Difference: Finding Radius

THE FUNDAMENTAL PRINCIPLE

The Golden Rule:

Radius = Distance from axis of rotation to the curve

\[ R = |\text{curve} - \text{axis}| \]

📝 Critical Insight: For axes other than x-axis or y-axis, you MUST calculate the distance. Don't just use the function value!

↔️ Revolving Around Horizontal Line y = k

Horizontal Axis: y = k

THE FORMULA:

When rotating around horizontal line \(y = k\):

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

where

\[ R(x) = |f(x) - k| \]
Key Points:
  • Axis of rotation is \(y = k\) (horizontal line)
  • Radius = vertical distance from \(y = k\) to curve \(y = f(x)\)
  • \(R(x) = |f(x) - k|\) (usually absolute value not needed if curve above axis)
  • Still integrate with respect to \(x\) (using \(dx\))
  • When \(k = 0\), this reduces to x-axis rotation

Special Cases:

  • If \(f(x) > k\): \(R(x) = f(x) - k\)
  • If \(f(x) < k\): \(R(x) = k - f(x)\)

↕️ Revolving Around Vertical Line x = h

Vertical Axis: x = h

THE FORMULA:

When rotating around vertical line \(x = h\):

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

where

\[ R(y) = |g(y) - h| \]
Key Points:
  • Axis of rotation is \(x = h\) (vertical line)
  • Radius = horizontal distance from \(x = h\) to curve \(x = g(y)\)
  • \(R(y) = |g(y) - h|\)
  • Must integrate with respect to \(y\) (using \(dy\))
  • Must express curve as \(x = g(y)\)
  • When \(h = 0\), this reduces to y-axis rotation

📊 Complete Formula Reference

Disc Method for Different Axes
Axis of Rotation Radius Formula Volume Formula
x-axis (\(y = 0\)) \(R(x) = f(x)\) \(\pi\int_a^b [f(x)]^2\,dx\)
y-axis (\(x = 0\)) \(R(y) = g(y)\) \(\pi\int_c^d [g(y)]^2\,dy\)
\(y = k\) (horizontal) \(R(x) = |f(x) - k|\) \(\pi\int_a^b [f(x)-k]^2\,dx\)
\(x = h\) (vertical) \(R(y) = |g(y) - h|\) \(\pi\int_c^d [g(y)-h]^2\,dy\)

📖 Comprehensive Worked Examples

Example 1: Around Horizontal Line y = k

Problem: Find the volume when the region bounded by \(y = x^2\), \(y = 0\), and \(x = 2\) is rotated around the line \(y = -1\).

Solution:

Step 1: Identify setup

Rotating around \(y = -1\) (horizontal line below x-axis)

Bounds: \(x = 0\) to \(x = 2\)

Step 2: Find radius

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

\[ R(x) = x^2 - (-1) = x^2 + 1 \]

Step 3: Set up integral

\[ V = \pi \int_0^2 (x^2 + 1)^2 \, dx \]

Step 4: Expand and evaluate

\[ V = \pi \int_0^2 (x^4 + 2x^2 + 1) \, dx \]
\[ = \pi \left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^2 \]
\[ = \pi\left(\frac{32}{5} + \frac{16}{3} + 2\right) = \frac{206\pi}{15} \]

ANSWER: \(V = \frac{206\pi}{15}\) cubic units

Example 2: Around Vertical Line x = h

Problem: Rotate the region bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\) around the line \(x = -1\). Find the volume.

Step 1: Convert to x = g(y)

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

When \(x = 4\): \(y = 2\)

Bounds: \(y = 0\) to \(y = 2\)

Step 2: Find radius

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

\[ R(y) = y^2 - (-1) = y^2 + 1 \]

Step 3: Set up and evaluate

\[ V = \pi \int_0^2 (y^2 + 1)^2 \, dy = \pi \int_0^2 (y^4 + 2y^2 + 1) \, dy \]
\[ = \pi\left[\frac{y^5}{5} + \frac{2y^3}{3} + y\right]_0^2 = \pi\left(\frac{32}{5} + \frac{16}{3} + 2\right) = \frac{206\pi}{15} \]

Example 3: Around y = 2

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

Step 1: Find bounds

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

Step 2: Find radius

Distance from \(y = 2\) to outer curve \(y = x\):

\[ R(x) = 2 - x \]

(Note: both curves below \(y = 2\))

Step 3: Evaluate

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

Example 4: Around x = 3

Problem: Rotate region bounded by \(x = y^2\) and \(x = 4\) around \(x = 3\). Find volume.

Analysis:

Bounds: \(y = -2\) to \(y = 2\) (when \(x = 4\))

Axis \(x = 3\) is BETWEEN the curves!

Outer radius to \(x = 4\): \(R_{\text{out}} = 4 - 3 = 1\)

Inner radius to \(x = y^2\): \(R_{\text{in}} = |y^2 - 3|\)

This requires washer method (next topic)!

📋 Step-by-Step Process

Systematic Approach

The 7-Step Method:

  1. Identify axis of rotation: \(y = k\) or \(x = h\)?
  2. Sketch the region: Visualize rotation
  3. Determine integration variable:
    • Horizontal axis (\(y = k\)) → use \(dx\)
    • Vertical axis (\(x = h\)) → use \(dy\), convert to \(x = g(y)\)
  4. Find bounds: Limits of integration
  5. Calculate radius: \(R = |\text{curve} - \text{axis}|\)
  6. Set up integral: \(V = \pi\int [R]^2\,dx\) or \(dy\)
  7. Evaluate and simplify

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Always find distance: Radius = |curve - axis|
  • Sketch is crucial: Helps determine which is larger
  • Check position: Is curve above/below or left/right of axis?
  • Sign matters: Subtract axis value from curve (or vice versa)
  • For \(y = k\): Subtract k from y-coordinate
  • For \(x = h\): Subtract h from x-coordinate
  • Still square the radius! Don't forget this step
  • Include π: In the formula

🔥 Quick Decision Guide:

  • Horizontal line \(y = k\): Integrate with \(dx\), \(R = |f(x) - k|\)
  • Vertical line \(x = h\): Integrate with \(dy\), \(R = |g(y) - h|\)
  • Curve above axis: \(R = \text{curve} - \text{axis}\)
  • Curve below/left of axis: \(R = \text{axis} - \text{curve}\)

❌ Common Mistakes to Avoid

  • Mistake 1: Using function value as radius (forgetting to subtract axis value)
  • Mistake 2: Wrong subtraction order (curve - axis vs axis - curve)
  • Mistake 3: Forgetting to square the radius
  • Mistake 4: Using wrong variable (\(dx\) when should use \(dy\))
  • Mistake 5: Not converting to \(x = g(y)\) for vertical axes
  • Mistake 6: Wrong sign on axis value (e.g., using +k when axis is \(y = -k\))
  • Mistake 7: Expansion errors when squaring binomials
  • Mistake 8: Forgetting π in final answer
  • Mistake 9: Integration errors
  • Mistake 10: Not checking if axis is between curves (may need washer method)

📝 Practice Problems

Find the volume:

  1. Region bounded by \(y = x^2\), \(y = 0\), \(x = 2\) rotated around \(y = -2\).
  2. Region bounded by \(y = \sqrt{x}\), \(x = 0\), \(y = 2\) rotated around \(x = -1\).
  3. Region between \(y = x\) and \(y = x^2\) rotated around \(y = 1\).
  4. Region bounded by \(x = y^2\), \(x = 0\), \(y = 1\) rotated around \(x = 2\).

Answers:

  1. \(\frac{304\pi}{15}\) cubic units
  2. \(\frac{64\pi}{5}\) cubic units
  3. \(\frac{11\pi}{30}\) cubic units
  4. \(\frac{29\pi}{6}\) cubic units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Identify axis clearly: State which line you're rotating around
  • Show radius calculation: Write \(R = f(x) - k\) (or similar)
  • Explain distance: "Distance from \(y = k\) to curve is..."
  • Correct setup: \(V = \pi\int[R]^2\,dx\) with correct \(R\)
  • Show squaring: Expand the radius squared
  • Show integration: Find antiderivative
  • Evaluate at bounds: Show substitution
  • Include π: In final answer

💯 Exam Strategy:

  1. Identify the axis of rotation carefully
  2. Sketch region and axis if time permits
  3. Determine if curve is above/below or left/right of axis
  4. Calculate radius as distance: |curve - axis|
  5. Write setup showing radius calculation
  6. Square the radius expression
  7. Integrate carefully
  8. Simplify and include π

⚡ Quick Reference Guide

OTHER AXES ESSENTIALS

Around \(y = k\) (horizontal):

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

Radius = vertical distance from axis to curve

Around \(x = h\) (vertical):

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

Radius = horizontal distance from axis to curve

Remember:

  • Radius = |curve - axis value|
  • Calculate distance, not just function value!
  • Square the radius!
  • Include π!

Master Rotation Around Other Axes! The key principle: radius = distance from axis to curve. For horizontal line \(y = k\): \(V = \pi\int_a^b[f(x)-k]^2\,dx\) where radius \(R(x) = |f(x)-k|\) is vertical distance from axis to curve. For vertical line \(x = h\): \(V = \pi\int_c^d[g(y)-h]^2\,dy\) where radius \(R(y) = |g(y)-h|\) is horizontal distance (must convert to \(x = g(y)\)). Critical difference from x-axis/y-axis: you CANNOT just use function value—must calculate distance by subtracting axis value. Process: (1) identify axis (\(y = k\) or \(x = h\)), (2) determine which variable to integrate, (3) find radius = |curve - axis|, (4) square and integrate. Common error: using \(f(x)\) instead of \(f(x) - k\). Sketch helps determine if curve is above/below or left/right of axis. This is a major AP® challenge—requires careful radius calculation! 🎯✨