#Volume with Washer Method Revolving Around Other Axes

#Table of Contents

Introduction
Volume of Revolution Around a Line Parallel to the x-axis
- Worked Example
Volume of Revolution Around a Line Parallel to the y-axis
- Worked Example
Practice Questions
Glossary
Summary and Key Takeaways
Exam Strategy

#Introduction

The washer method is a technique used to find the volume of a solid of revolution when the solid is generated by rotating a region bounded by two curves around an axis. This method involves integrating the area of washers (or disks with holes) formed by the rotation.

#Volume of Revolution Around a Line Parallel to the x-axis

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

Identify the functions and the interval: Let $f(x)$ and $g(x)$ be continuous functions on the interval $[a, b]$ , with $f(x)$ closer to the horizontal line $y = k$ than $g(x)$ .
Set up the integral: The volume of revolution $V$ is given by: $V = \pi \int_{a}^{b} \left[ (g(x) - k)^2 - (f(x) - k)^2 \right] , dx$
Ensure proper boundaries: If the curves swap places over the interval, split the calculation into separate integrals.

#Worked Example

Let $R$ be the region enclosed by the graphs of $f(x) = \frac{1}{4}x^2$ and $g(x) = x$ . The region is rotated about the horizontal line $y = 5$ .

Step-by-Step Solution:

Find the points of intersection: $\begin{array}{rcl} \frac{1}{4}x^2 & = & x \\ x^2 - 4x & = & 0 \\ x(x - 4) & = & 0 \\ x & = & 0 \quad \text{or} \quad x = 4 \end{array}$ So, $a = 0$ and $b = 4$ .
Identify the functions closest to $y = 5$ : $g(x) = x$ is closer to $y = 5$ than $f(x) = \frac{1}{4}x^2$ . Thus, $y_1 = x$ and $y_2 = \frac{1}{4}x^2$ .
Set up and solve the integral: $\begin{array}{rcl} V & = & \pi \int_{0}^{4} \left[ \left(\frac{1}{4}x^2 - 5\right)^2 - (x - 5)^2 \right] , dx \\ & = & \pi \int_{0}^{4} \left( \frac{1}{16}x^4 - \frac{5}{2}x^2 + 25 - (x^2 - 10x + 25) \right) , dx \\ & = & \pi \int_{0}^{4} \left( \frac{1}{16}x^4 - \frac{7}{2}x^2 + 10x \right) , dx \\ & = & \pi \left[ \frac{1}{80}x^5 - \frac{7}{6}x^3 + 5x^2 \right]_{0}^{4} \\ & = & \pi \left( \left( \frac{1}{80}(4)^5 - \frac{7}{6}(4)^3 + 5(4)^2 \right) - 0 \right) \\ & = & \frac{272\pi}{15} \\ & \approx & 56.968 , \text{units}^3 , (\text{to 3 decimal places}) \end{array}$

#Volume of Revolution Around a Line Parallel to the y-axis

To calculate the volume of revolution around a line parallel to the y-axis, follow these steps:

Identify the functions and the interval: Let $f(y)$ and $g(y)$ be continuous functions on the interval $[a, b]$ , with $f(y)$ closer to the vertical line $x = k$ than $g(y)$ .
Set up the integral: The volume of revolution $V$ is given by: $V = \pi \int_{a}^{b} \left[ (g(y) - k)^2 - (f(y) - k)^2 \right] , dy$
Ensure proper boundaries: If the curves swap places over the interval, split the calculation into separate integrals.

#Worked Example

Let $R$ be the region enclosed by the graphs of $f(x) = \frac{1}{4}x^2$ and $g(x) = x$ . The region is rotated about the vertical line $x = -2$ .

Step-by-Step Solution:

Rewrite the functions as functions of $y$ : $\begin{array}{rcl} y & = & f(x) \\ y & = & \frac{1}{4}x^2 \\ x^2 & = & 4y \\ x & = & 2\sqrt{y} \\ x & = & y \end{array}$
Find the points of intersection: $\begin{array}{rcl} y & = & 2\sqrt{y} \\ y^2 & = & 4y \\ y(y - 4) & = & 0 \\ y & = & 0 \quad \text{or} \quad y = 4 \end{array}$ So, $a = 0$ and $b = 4$ .
Identify the functions closest to $x = -2$ : $g(x) = x$ is closer to $x = -2$ than $f(x) = \frac{1}{4}x^2$ . Thus, $x_1 = y$ and $x_2 = 2\sqrt{y}$ .
Set up and solve the integral: $\begin{array}{rcl} V & = & \pi \int_{0}^{4} \left[ \left(2\sqrt{y} - (-2)\right)^2 - (y - (-2))^2 \right] , dy \\ & = & \pi \int_{0}^{4} \left[ \left(2\sqrt{y} + 2\right)^2 - (y + 2)^2 \right] , dy \\ & = & \pi \int_{0}^{4} \left( 4y + 8\sqrt{y} + 4 - (y^2 + 4y + 4) \right) , dy \\ & = & \pi \int_{0}^{4} \left( 8\sqrt{y} - y^2 \right) , dy \\ & = & \pi \int_{0}^{4} \left( 8y^{1/2} - y^2 \right) , dy \\ & = & \pi \left[ \frac{16}{3}y^{3/2} - \frac{1}{3}y^3 \right]_{0}^{4} \\ & = & \pi \left( \left( \frac{16}{3}(4)^{3/2} - \frac{1}{3}(4)^3 \right) - 0 \right) \\ & = & \frac{64\pi}{3} \\ & \approx & 67.021 , \text{units}^3 , (\text{to 3 decimal places}) \end{array}$

#Practice Questions

Practice Question

#Question 1

Calculate the volume of the solid formed by rotating the region bounded by $f(x) = \sqrt{x}$ and $g(x) = x$ around the line $y = 3$ .

#Question 2

Find the volume of the solid formed by rotating the region bounded by $f(y) = y^2$ and $g(y) = y + 2$ around the line $x = -1$ .

#Glossary

Volume of Revolution: The volume of a solid formed by rotating a region around an axis.
Washer Method: A technique to find the volume of a solid of revolution when the solid has a hole in the middle.
Integral: A mathematical operation that sums the area under a curve.
Boundaries: The limits of integration, often determined by the intersection points of curves.

#Summary and Key Takeaways

The washer method involves integrating the area of washers formed by rotating a region around an axis.
Ensure the functions are properly identified and the correct integral setup is used.
Pay attention to the boundaries and whether the curves swap places within the interval.
Practice solving integrals to become proficient in applying the washer method.

#Exam Strategy

Read the question carefully: Identify the functions and the axis of rotation.
Sketch the region: Visualize the area being rotated to understand the setup.
Set up the integral correctly: Use the washer method formula and ensure the limits of integration are accurate.
Check your work: Verify the points of intersection and the integral setup before solving.

By following these steps and practicing regularly, you'll be well-prepared to tackle volume of revolution problems in your exams.