#Volume with Washer Method Revolving Around the x-axis

#Table of Contents

Introduction
When to Use the Washer Method
Calculating Volume Using the Washer Method
Exam Tips
Worked Example
Practice Questions
Glossary
Summary and Key Takeaways

#Introduction

The washer method is a powerful tool for calculating the volume of a solid of revolution when there is a gap between the region being rotated and the axis of rotation. This method extends the disk method by accounting for the inner radius, which is subtracted from the outer radius.

#When to Use the Washer Method

The washer method should be used when there is a gap between the region being rotated and the axis of rotation (x-axis in this case).

For example, consider rotating the shaded area between two curves around the x-axis. If there is a gap between the x-axis and the curve

y\_1 = f(x)

, then the cross-section of the solid of revolution will resemble a washer.

#Cross Section

For $x$ between $a$ and $b$ , the cross-section of the solid of revolution will have the shape of a washer, with: $\text{Area} = \pi \left( \left( g(x) \right)^2 - \left( f(x) \right)^2 \right)$

#Calculating Volume Using the Washer Method

To calculate the volume of revolution around the x-axis using the washer method, follow these steps:

Identify the Curves: Let $f$ and $g$ be continuous functions such that $|f(x)| < |g(x)|$ on the interval $[a, b]$ .
Set Up the Integral: The volume $V$ of the region bounded by $y_1 = f(x)$ and $y_2 = g(x)$ between $x = a$ and $x = b$ , when rotated around the x-axis, is given by: $V = \pi \int_{a}^{b} \left( \left( y_2 \right)^2 - \left( y_1 \right)^2 \right) , dx$ where $y_1$ and $y_2$ are both functions of $x$ .
Evaluate the Integral: Make sure to carefully evaluate the integral, considering if the curves swap places over the interval.

Exam Tip

Be careful not to confuse $\left( y_2 \right)^2 - \left( y_1 \right)^2$ with $\left( y_2 - y_1 \right)^2$ . These are not equal!

Key Concept

The washer method uses the concept of definite integrals to calculate an accumulation of change. The volume is derived from the area of the washer cross-sections.

#Exam Tips

Exam Tip

Be careful not to confuse $\left( y_2 \right)^2 - \left( y_1 \right)^2$ with $\left( y_2 - y_1 \right)^2$ . These are not equal! The correct formula is: $\left( y_2 \right)^2 - \left( y_1 \right)^2 \neq \left( y_2 - y_1 \right)^2$

Common Mistake

Common Mistake: Confusing the square of the difference of functions with the difference of their squares. $\left( y_2 - y_1 \right)^2 = \left( y_2 \right)^2 - 2 y_1 y_2 + \left( y_1 \right)^2$

#Worked Example

Let

R

be the region enclosed by the graphs of

f(x) = \frac{1}{4} x^2

and

g(x) = x

To find the volume of $R$ when rotated around the x-axis:

Find Points of Intersection: $\frac{1}{4} x^2 = x$ $x^2 - 4x = 0$ $x(x - 4) = 0$ $x = 0 \text{ or } x = 4$ Thus, $a = 0$ and $b = 4$ .
Set Up the Integral: $V = \pi \int_{0}^{4} \left( x^2 - \left( \frac{1}{4} x^2 \right)^2 \right) , dx$ $V = \pi \int_{0}^{4} \left( x^2 - \frac{1}{16} x^4 \right) , dx$
Evaluate the Integral: $V = \pi \left[ \frac{1}{3} x^3 - \frac{1}{80} x^5 \right]_{0}^{4}$ $V = \pi \left[ \left( \frac{1}{3} (4)^3 - \frac{1}{80} (4)^5 \right) - 0 \right]$ $V = \pi \left( \frac{64}{3} - \frac{1024}{80} \right)$ $V = \pi \left( \frac{64}{3} - \frac{128}{10} \right)$ $V = \pi \left( \frac{64}{3} - \frac{128}{10} \right)$ $V \approx 26.808 \ \text{units}^3 \ (\text{to 3 decimal places})$

#Practice Questions

Practice Question

Question 1: Find the volume of the solid obtained by rotating the region bounded by $y = x^2$ and $y = 2x$ around the x-axis from $x = 0$ to $x = 2$ .

Answer:

Find points of intersection: $x^2 = 2x$ $x(x - 2) = 0$ $x = 0 \text{ or } x = 2$
Set up the integral: $V = \pi \int_{0}^{2} \left( (2x)^2 - (x^2)^2 \right) , dx$ $V = \pi \int_{0}^{2} \left( 4x^2 - x^4 \right) , dx$
Evaluate the integral: $V = \pi \left[ \frac{4}{3} x^3 - \frac{1}{5} x^5 \right]_{0}^{2}$ $V = \pi \left( \frac{4}{3} (2)^3 - \frac{1}{5} (2)^5 \right)$ $V = \pi \left( \frac{32}{3} - \frac{32}{5} \right)$ $V \approx 14.612 \ \text{units}^3 \ (\text{to 3 decimal places})$

Practice Question

Question 2: Determine the volume of the solid obtained by rotating the region bounded by $y = \sqrt{x}$ and $y = x$ around the x-axis from $x = 0$ to $x = 1$ .

Answer:

Find points of intersection: $\sqrt{x} = x$ $x = 0 \text{ or } x = 1$
Set up the integral: $V = \pi \int_{0}^{1} \left( x^2 - (\sqrt{x})^2 \right) , dx$ $V = \pi \int_{0}^{1} \left( x^2 - x \right) , dx$
Evaluate the integral: $V = \pi \left[ \frac{1}{3} x^3 - \frac{1}{2} x^2 \right]_{0}^{1}$ $V = \pi \left( \frac{1}{3} - \frac{1}{2} \right)$ $V = \pi \left( \frac{2}{6} - \frac{3}{6} \right)$ $V = \pi \left( -\frac{1}{6} \right)$ $V = -\frac{\pi}{6} \ \text{units}^3$

#Glossary

Washer Method: A method for finding the volume of a solid of revolution when there is a gap between the region and the axis of rotation.
Volume of Revolution: The volume of a solid formed by rotating a region around an axis.
Definite Integral: A mathematical expression representing the accumulation of quantities.

#Summary and Key Takeaways

The washer method is used for calculating the volume of a solid of revolution when there is a gap between the region and the axis of rotation.
The formula for the volume using the washer method around the x-axis is: $V = \pi \int_{a}^{b} \left( \left( y_2 \right)^2 - \left( y_1 \right)^2 \right) , dx$
Remember to correctly set up and evaluate the integral, ensuring that the functions and bounds are accurately represented.
Be cautious of common mistakes, such as confusing the square of the difference with the difference of the squares.

Key Concept

The washer method extends the disk method by subtracting the inner radius from the outer radius, accounting for the gap between the region and the axis of rotation.

Volumes of Revolution

Emily Davis

7 min read

Next Topic - Washer Method Around the y-Axis

Listen to this study note

Study Guide Overview

This study guide covers the washer method for finding volumes of revolution around the x-axis. It explains when to use this method (when a gap exists between the region and the x-axis), how to calculate volume using the washer method formula with definite integrals, and common mistakes to avoid. It includes a worked example and practice questions, and defines key terms like volume of revolution.

#Volume with Washer Method Revolving Around the x-axis

#Table of Contents

Introduction
When to Use the Washer Method
Calculating Volume Using the Washer Method
Exam Tips
Worked Example
Practice Questions
Glossary
Summary and Key Takeaways

#Introduction

#When to Use the Washer Method

The washer method should be used when there is a gap between the region being rotated and the axis of rotation (x-axis in this case).

For example, consider rotating the shaded area between two curves around the x-axis. If there is a gap between the x-axis and the curve

y\_1 = f(x)

, then the cross-section of the solid of revolution will resemble a washer.

#Cross Section

For $x$ between $a$ and $b$ , the cross-section of the solid of revolution will have the shape of a washer, with: $\text{Area} = \pi \left( \left( g(x) \right)^2 - \left( f(x) \right)^2 \right)$

#Calculating Volume Using the Washer Method

To calculate the volume of revolution around the x-axis using the washer method, follow these steps:

Identify the Curves: Let $f$ and $g$ be continuous functions such that $|f(x)| < |g(x)|$ on the interval $[a, b]$ .
Set Up the Integral: The volume $V$ of the region bounded by $y_1 = f(x)$ and $y_2 = g(x)$ between $x = a$ and $x = b$ , when rotated around the x-axis, is given by: $V = \pi \int_{a}^{b} \left( \left( y_2 \right)^2 - \left( y_1 \right)^2 \right) , dx$ where $y_1$ and $y_2$ are both functions of $x$ .
Evaluate the Integral: Make sure to carefully evaluate the integral, considering if the curves swap places over the interval.

Exam Tip

Be careful not to confuse $\left( y_2 \right)^2 - \left( y_1 \right)^2$ with $\left( y_2 - y_1 \right)^2$ . These are not equal!

Key Concept

The washer method uses the concept of definite integrals to calculate an accumulation of change. The volume is derived from the area of the washer cross-sections.

#Exam Tips

Exam Tip

Common Mistake

Common Mistake: Confusing the square of the difference of functions with the difference of their squares. $\left( y_2 - y_1 \right)^2 = \left( y_2 \right)^2 - 2 y_1 y_2 + \left( y_1 \right)^2$

#Worked Example

Let

R

be the region enclosed by the graphs of

f(x) = \frac{1}{4} x^2

and

g(x) = x

To find the volume of $R$ when rotated around the x-axis:

Find Points of Intersection: $\frac{1}{4} x^2 = x$ $x^2 - 4x = 0$ $x(x - 4) = 0$ $x = 0 \text{ or } x = 4$ Thus, $a = 0$ and $b = 4$ .
Set Up the Integral: $V = \pi \int_{0}^{4} \left( x^2 - \left( \frac{1}{4} x^2 \right)^2 \right) , dx$ $V = \pi \int_{0}^{4} \left( x^2 - \frac{1}{16} x^4 \right) , dx$
Evaluate the Integral: $V = \pi \left[ \frac{1}{3} x^3 - \frac{1}{80} x^5 \right]_{0}^{4}$ $V = \pi \left[ \left( \frac{1}{3} (4)^3 - \frac{1}{80} (4)^5 \right) - 0 \right]$ $V = \pi \left( \frac{64}{3} - \frac{1024}{80} \right)$ $V = \pi \left( \frac{64}{3} - \frac{128}{10} \right)$ $V = \pi \left( \frac{64}{3} - \frac{128}{10} \right)$ $V \approx 26.808 \ \text{units}^3 \ (\text{to 3 decimal places})$

#Practice Questions

Practice Question

Question 1: Find the volume of the solid obtained by rotating the region bounded by $y = x^2$ and $y = 2x$ around the x-axis from $x = 0$ to $x = 2$ .

Answer:

Find points of intersection: $x^2 = 2x$ $x(x - 2) = 0$ $x = 0 \text{ or } x = 2$
Set up the integral: $V = \pi \int_{0}^{2} \left( (2x)^2 - (x^2)^2 \right) , dx$ $V = \pi \int_{0}^{2} \left( 4x^2 - x^4 \right) , dx$
Evaluate the integral: $V = \pi \left[ \frac{4}{3} x^3 - \frac{1}{5} x^5 \right]_{0}^{2}$ $V = \pi \left( \frac{4}{3} (2)^3 - \frac{1}{5} (2)^5 \right)$ $V = \pi \left( \frac{32}{3} - \frac{32}{5} \right)$ $V \approx 14.612 \ \text{units}^3 \ (\text{to 3 decimal places})$

Practice Question

Question 2: Determine the volume of the solid obtained by rotating the region bounded by $y = \sqrt{x}$ and $y = x$ around the x-axis from $x = 0$ to $x = 1$ .

Answer:

Find points of intersection: $\sqrt{x} = x$ $x = 0 \text{ or } x = 1$
Set up the integral: $V = \pi \int_{0}^{1} \left( x^2 - (\sqrt{x})^2 \right) , dx$ $V = \pi \int_{0}^{1} \left( x^2 - x \right) , dx$
Evaluate the integral: $V = \pi \left[ \frac{1}{3} x^3 - \frac{1}{2} x^2 \right]_{0}^{1}$ $V = \pi \left( \frac{1}{3} - \frac{1}{2} \right)$ $V = \pi \left( \frac{2}{6} - \frac{3}{6} \right)$ $V = \pi \left( -\frac{1}{6} \right)$ $V = -\frac{\pi}{6} \ \text{units}^3$

#Glossary

Washer Method: A method for finding the volume of a solid of revolution when there is a gap between the region and the axis of rotation.
Volume of Revolution: The volume of a solid formed by rotating a region around an axis.
Definite Integral: A mathematical expression representing the accumulation of quantities.

#Summary and Key Takeaways

The washer method is used for calculating the volume of a solid of revolution when there is a gap between the region and the axis of rotation.
The formula for the volume using the washer method around the x-axis is: $V = \pi \int_{a}^{b} \left( \left( y_2 \right)^2 - \left( y_1 \right)^2 \right) , dx$
Remember to correctly set up and evaluate the integral, ensuring that the functions and bounds are accurately represented.
Be cautious of common mistakes, such as confusing the square of the difference with the difference of the squares.

Key Concept

The washer method extends the disk method by subtracting the inner radius from the outer radius, accounting for the gap between the region and the axis of rotation.

Continue your learning journey

Flashcard

Continute to Flashcard

Question Bank

Continute to Question Bank

Mock Exam

Continute to Mock Exam

Previous Topic - Disc Method Around Other Axes Next Topic - Washer Method Around the y-Axis

How are we doing?

Give us your feedback and let us know how we can improve

Question 1 of 11

When should you use the washer method to find the volume of a solid of revolution? 🤔

When the region is rotated around the y-axis

When there is no gap between the region and the axis of rotation

When there is a gap between the region and the axis of rotation

When the region is bounded by a single curve