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Volume of Revolution

CSEC Additional Mathematics Essential Knowledge: Volume of revolution involves finding the volume of a 3D solid created by rotating a 2D region around an axis. This powerful application of integration allows us to calculate volumes of objects with curved surfaces that cannot be found using basic geometry formulas.

Key Concept: When a region in the xy-plane is rotated about a line (usually the x-axis or y-axis), it generates a three-dimensional solid. The volume of this solid can be found using integration with either the disk method or washer method.

Part 1: The Disk Method

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Revolution About the x-axis

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Disk Method Formula

When the region under \(y = f(x)\) from \(x = a\) to \(x = b\) is rotated about the x-axis:

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

Think of slicing the solid perpendicular to the x-axis: each slice is a circular disk with radius \(f(x)\) and thickness \(dx\).

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Visualization: Rotate curve about x-axis → Solid formed

Each vertical slice becomes a circular disk

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Revolution About the y-axis

When the region to the right of \(x = g(y)\) from \(y = c\) to \(y = d\) is rotated about the y-axis:

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

Key Idea: The radius is the horizontal distance from the curve to the y-axis.

About x-axis:
\(V = \pi \int_a^b y^2 \, dx\)

About y-axis:
\(V = \pi \int_c^d x^2 \, dy\)

📝 Example 1: Disk Method About x-axis

Find the volume generated when the region under \(y = \sqrt{x}\) from \(x = 0\) to \(x = 4\) is rotated about the x-axis.

Set up the integral: \(V = \pi \int_0^4 [\sqrt{x}]^2 \, dx\)

Simplify: \(V = \pi \int_0^4 x \, dx\)

Integrate: \(\pi \left[ \frac{x^2}{2} \right]_0^4\)

Evaluate: \(\pi \left( \frac{16}{2} – 0 \right) = \pi \times 8\)

Final answer: \(8\pi\) cubic units

Interpretation: The solid formed is a paraboloid (like a bowl).

📝 Example 2: Disk Method About y-axis

Find the volume generated when the region bounded by \(y = x^2\), \(y = 0\), and \(x = 2\) is rotated about the y-axis.

Visualize: We have \(y = x^2\) from \(x = 0\) to \(x = 2\), rotated about y-axis

Determine limits for y: When \(x = 0\), \(y = 0\); when \(x = 2\), \(y = 4\)

Express x in terms of y: \(x = \sqrt{y}\) (since \(x \geq 0\))

Set up integral: \(V = \pi \int_0^4 [\sqrt{y}]^2 \, dy\)

Simplify and integrate: \(\pi \int_0^4 y \, dy = \pi \left[ \frac{y^2}{2} \right]_0^4\)

Evaluate: \(\pi \left( \frac{16}{2} – 0 \right) = 8\pi\) cubic units

Part 2: The Washer Method

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Volume with a Hole

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Washer Method Formula

When the region between two curves \(y = f(x)\) (outer) and \(y = g(x)\) (inner) is rotated about the x-axis:

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

Each slice is a washer (disk with a hole). The volume is the difference between the outer disk and inner disk.

Outer Radius = f(x)
Area = π[f(x)]²

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Inner Radius = g(x)
Area = π[g(x)]²

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About the y-axis

For rotation about the y-axis with curves \(x = f(y)\) (right) and \(x = g(y)\) (left):

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

CSEC Strategy: Always sketch the region to identify which curve is “outer” and which is “inner” relative to the axis of rotation.

📝 Example 3: Washer Method About x-axis

Find the volume generated when the region between \(y = x^2\) and \(y = x\) (from \(x = 0\) to \(x = 1\)) is rotated about the x-axis.

Identify outer and inner curves: On [0,1], \(y = x\) is above \(y = x^2\)

Outer radius: \(R = x\)

Inner radius: \(r = x^2\)

Set up integral: \(V = \pi \int_0^1 \left( x^2 – (x^2)^2 \right) dx = \pi \int_0^1 (x^2 – x^4) dx\)

Integrate: \(\pi \left[ \frac{x^3}{3} – \frac{x^5}{5} \right]_0^1\)

Evaluate: \(\pi \left( \frac{1}{3} – \frac{1}{5} \right) = \pi \left( \frac{5}{15} – \frac{3}{15} \right) = \frac{2\pi}{15}\) cubic units

Visualization: The solid looks like a bowl with a conical hole.

📝 Example 4: Washer Method About y-axis (CSEC Past Paper Style)

The region bounded by \(y = x^2\), \(y = 4\), and the y-axis is rotated about the y-axis. Find the volume generated.

Sketch: We have parabola \(y = x^2\) from \(x = 0\) to where it meets \(y = 4\)

Find intersection: When \(y = 4\), \(x^2 = 4\) ⇒ \(x = 2\) (we take positive since right of y-axis)

For rotation about y-axis: We need x in terms of y: \(x = \sqrt{y}\)

Visualize washer: Outer radius is constant: \(R = 2\) (from y-axis to line x=2)

Inner radius varies: \(r = \sqrt{y}\) (from y-axis to curve)

y-limits: From \(y = 0\) to \(y = 4\)

Set up integral: \(V = \pi \int_0^4 \left( 2^2 – (\sqrt{y})^2 \right) dy = \pi \int_0^4 (4 – y) dy\)

Integrate: \(\pi \left[ 4y – \frac{y^2}{2} \right]_0^4\)

Evaluate: \(\pi \left( 16 – \frac{16}{2} \right) = \pi (16 – 8) = 8\pi\) cubic units

Part 3: Choosing the Right Method

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Disk vs Washer Method Decision Guide

Disk Method

Use when: The region touches the axis of rotation

Formula: \(V = \pi \int [radius]^2 \, d\)(axis)

Example: Region under \(y = \sqrt{x}\) from 0 to 4 rotated about x-axis

No hole in the solid

Washer Method

Use when: There’s a gap between region and axis of rotation

Formula: \(V = \pi \int ([R]^2 – [r]^2) \, d\)(axis)

Example: Region between \(y = x\) and \(y = x^2\) rotated about x-axis

Solid has a hole

General Strategy

1. Sketch the region

2. Identify axis of rotation

3. Draw a typical slice perpendicular to the axis

4. Determine if slice is disk (solid) or washer (has hole)

5. Set up integral with correct limits

Sketch the region and axis of rotation

Draw a typical slice perpendicular to the axis of rotation

Ask: Does the slice touch the axis?
• YES → Disk method (radius = distance to curve)
• NO → Washer method (R = outer distance, r = inner distance)

Set up integral: \(V = \pi \int (\text{radius expression})^2 \, d\)(axis variable)

📝 Example 5: Choosing the Method

The region bounded by \(y = x^3\), \(y = 8\), and \(x = 0\) is rotated about the y-axis. Find the volume.

Sketch: Cubic curve from (0,0) to (2,8), horizontal line y=8

Axis of rotation: y-axis

Typical slice: Horizontal slice (perpendicular to y-axis)

Does slice touch axis? YES – the region touches the y-axis at x=0

Method: Disk method (about y-axis)

Express x in terms of y: \(y = x^3\) ⇒ \(x = y^{1/3}\)

y-limits: From \(y = 0\) to \(y = 8\)

Set up: \(V = \pi \int_0^8 (y^{1/3})^2 dy = \pi \int_0^8 y^{2/3} dy\)

Integrate: \(\pi \left[ \frac{y^{5/3}}{5/3} \right]_0^8 = \pi \left[ \frac{3}{5} y^{5/3} \right]_0^8\)

Evaluate: \(\frac{3\pi}{5} (8^{5/3}) = \frac{3\pi}{5} (32) = \frac{96\pi}{5}\) cubic units

Part 4: Volume of Composite Solids

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Combining Methods and Regions

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Multiple Regions

Sometimes the solid is formed by rotating a region that needs to be split into parts:

\[V_{\text{total}} = V_1 + V_2 + \cdots\]

Each part might require a different method or have different radius expressions.

📝 Example 6: Composite Region (CSEC Level)

The region bounded by \(y = x^2\), \(y = 4\), and the y-axis is rotated about the x-axis. Find the volume.

Sketch: We have parabola from (0,0) to (2,4) and horizontal line y=4 from x=0 to x=2

When rotated about x-axis: Two different radius expressions

For x from 0 to 2: Outer radius = 4, inner radius = \(x^2\) (washer method)

Set up integral: \(V = \pi \int_0^2 (4^2 – (x^2)^2) dx = \pi \int_0^2 (16 – x^4) dx\)

Integrate: \(\pi \left[ 16x – \frac{x^5}{5} \right]_0^2\)

Evaluate: \(\pi \left( 32 – \frac{32}{5} \right) = \pi \left( \frac{160}{5} – \frac{32}{5} \right) = \frac{128\pi}{5}\) cubic units

Visualization: The solid is like a cylinder (radius 4) with a parabolic hole.

Common Mistake: Forgetting to square the radii! Remember: \(V = \pi \int (\text{radius})^2 \, d\)(axis), not \(\pi \int \text{radius} \, d\)(axis).

Part 5: Real-World Applications

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Practical Uses of Volume of Revolution

Everyday Objects as Volumes of Revolution

Bowls, glasses, and vases (rotated curves)

Bottles and containers (composite rotations)

Machine parts (precisely engineered solids)

Architectural domes (rotated parabolic curves)

Rocket nozzles and aerodynamic shapes

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Engineering Application

A tank is designed by rotating the curve \(y = \frac{1}{4}x^2\) from \(x = 0\) to \(x = 4\) about the y-axis. Find its capacity.

Curve: \(y = \frac{1}{4}x^2\), from \(x=0\) to \(x=4\)

Rotation about y-axis: Use disk method

Express x in terms of y: \(x = 2\sqrt{y}\)

y-limits: When \(x=0\), \(y=0\); when \(x=4\), \(y = \frac{1}{4}(16) = 4\)

Volume: \(V = \pi \int_0^4 (2\sqrt{y})^2 dy = \pi \int_0^4 4y \, dy\)

Integrate: \(4\pi \left[ \frac{y^2}{2} \right]_0^4 = 4\pi (8) = 32\pi\) cubic units

If units are meters: Capacity = \(32\pi \approx 100.5 \, \text{m}^3\)

Quiz: Test Your Understanding

Volume of Revolution Quiz

Question 1: Find the volume when the region under \(y = x^2 + 1\) from \(x = 0\) to \(x = 2\) is rotated about the x-axis.

Answer:
Disk method: \(V = \pi \int_0^2 (x^2 + 1)^2 dx = \pi \int_0^2 (x^4 + 2x^2 + 1) dx\)
\(= \pi \left[ \frac{x^5}{5} + \frac{2x^3}{3} + x \right]_0^2 = \pi \left( \frac{32}{5} + \frac{16}{3} + 2 \right)\)
\(= \pi \left( \frac{96}{15} + \frac{80}{15} + \frac{30}{15} \right) = \pi \times \frac{206}{15} = \frac{206\pi}{15}\) cubic units

Question 2: The region between \(y = \sqrt{x}\) and \(y = x\) (from \(x=0\) to \(x=1\)) is rotated about the x-axis. Find the volume.

Answer:
Washer method: Outer radius = \(\sqrt{x}\), inner radius = \(x\)
\(V = \pi \int_0^1 \left( (\sqrt{x})^2 – (x)^2 \right) dx = \pi \int_0^1 (x – x^2) dx\)
\(= \pi \left[ \frac{x^2}{2} – \frac{x^3}{3} \right]_0^1 = \pi \left( \frac{1}{2} – \frac{1}{3} \right) = \pi \left( \frac{3}{6} – \frac{2}{6} \right) = \frac{\pi}{6}\) cubic units

Question 3: Find the volume when the region bounded by \(y = 2x\), \(y = 0\), and \(x = 3\) is rotated about the y-axis.

Answer:
About y-axis: Express x in terms of y: \(y = 2x \Rightarrow x = \frac{y}{2}\)
y-limits: When \(x=0\), \(y=0\); when \(x=3\), \(y=6\)
Disk method: \(V = \pi \int_0^6 \left( \frac{y}{2} \right)^2 dy = \pi \int_0^6 \frac{y^2}{4} dy = \frac{\pi}{4} \left[ \frac{y^3}{3} \right]_0^6\)
\(= \frac{\pi}{4} \times \frac{216}{3} = \frac{\pi}{4} \times 72 = 18\pi\) cubic units

Question 4: The region bounded by \(y = 4 – x^2\) and \(y = 0\) is rotated about the x-axis. Find the volume.

Answer:
Find x-intercepts: \(4 – x^2 = 0 \Rightarrow x = \pm 2\)
Disk method: \(V = \pi \int_{-2}^2 (4 – x^2)^2 dx = \pi \int_{-2}^2 (16 – 8x^2 + x^4) dx\)
Even function, so can compute from 0 to 2 and double:
\(= 2\pi \int_0^2 (16 – 8x^2 + x^4) dx = 2\pi \left[ 16x – \frac{8x^3}{3} + \frac{x^5}{5} \right]_0^2\)
\(= 2\pi \left( 32 – \frac{64}{3} + \frac{32}{5} \right) = 2\pi \left( \frac{480}{15} – \frac{320}{15} + \frac{96}{15} \right) = 2\pi \times \frac{256}{15} = \frac{512\pi}{15}\) cubic units

Question 5: A solid is formed by rotating the region between \(y = x^2\) and \(y = 2x\) about the y-axis. Find the volume.

Answer:
Intersection: \(x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x = 0, 2\)
About y-axis: For washer method, need outer and inner radii in terms of y
Outer: from y-axis to line \(x = \frac{y}{2}\) (since \(y = 2x\))
Inner: from y-axis to curve \(x = \sqrt{y}\) (since \(y = x^2\))
y-limits: From \(y=0\) to \(y=4\) (when \(x=2\), both give y=4)
\(V = \pi \int_0^4 \left( \left(\frac{y}{2}\right)^2 – (\sqrt{y})^2 \right) dy = \pi \int_0^4 \left( \frac{y^2}{4} – y \right) dy\)
\(= \pi \left[ \frac{y^3}{12} – \frac{y^2}{2} \right]_0^4 = \pi \left( \frac{64}{12} – \frac{16}{2} \right) = \pi \left( \frac{16}{3} – 8 \right) = \pi \left( \frac{16}{3} – \frac{24}{3} \right) = -\frac{8\pi}{3}\)
Volume is positive: \(\frac{8\pi}{3}\) cubic units

🎯 Key Concepts Summary

Disk Method: \(V = \pi \int [radius]^2 \, d\)(axis) – when region touches axis
Washer Method: \(V = \pi \int ([R]^2 – [r]^2) \, d\)(axis) – when there’s a hole
About x-axis: Integrate with respect to x, radius = y-value
About y-axis: Integrate with respect to y, radius = x-value
Key Steps:
- 1. Sketch the region
- 2. Identify axis of rotation
- 3. Draw typical slice perpendicular to axis
- 4. Determine disk or washer
- 5. Set up integral with correct limits
- 6. Integrate and evaluate
Common CSEC Question Types:
- Rotate region under single curve (disk method)
- Rotate region between two curves (washer method)
- Rotate about x-axis or y-axis
- Find volume of practical objects (bowls, tanks)
Formula Memory Aid:
- DISK: \(V = \pi \int (\text{radius})^2 \, d\)(axis)
- WASHER: \(V = \pi \int (\text{big R}^2 – \text{small r}^2) \, d\)(axis)
- ALWAYS square the radius!

CSEC Exam Strategy: Volume of revolution questions typically appear in Paper 2. Common mistakes: 1) Forgetting to square the radius, 2) Using wrong limits, 3) Confusing disk vs washer method, 4) Integrating with respect to wrong variable. Always sketch the region, clearly identify your radius expression, and check that your final answer has the correct units (cubic units). Show all working for full marks.

Pro Tip: To check if your volume calculation is reasonable: 1) Volume should always be positive. 2) The volume should be less than the bounding cylinder. 3) For rotation about x-axis, if you swap the roles of x and y, you should get the same volume when rotating the inverse function about the y-axis (for one-to-one functions).