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Volumes of Revolution

Emily Davis

Emily Davis

7 min read

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Study Guide Overview

This study guide covers the washer method for finding the volume of revolution around axes other than the x or y-axis. It explains how to set up and solve integrals for rotations around lines parallel to both the x-axis and y-axis. Examples, practice questions, and exam strategies are included.

Volume with Washer Method Revolving Around Other Axes

Table of Contents

  1. Introduction
  2. Volume of Revolution Around a Line Parallel to the x-axis
  3. Volume of Revolution Around a Line Parallel to the y-axis
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways
  7. 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:

  1. Identify the functions and the interval: Let f(x)f(x) and g(x)g(x) be continuous functions on the interval [a,b][a, b], with f(x)f(x) closer to the horizontal line y=ky = k than g(x)g(x).

  2. Set up the integral: The volume of revolution VV is given by: V=π∫ab[(g(x)−k)2−(f(x)−k)2],dxV = \pi \int_{a}^{b} \left[ (g(x) - k)^2 - (f(x) - k)^2 \right] , dx

  3. Ensure proper boundaries: If the curves swap places over the interval, split the calculation into separate integrals.

Worked Example

Let RR be the region enclosed by the graphs of f(x)=14x2f(x) = \frac{1}{4}x^2 and g(x)=xg(x) = x. The region is rotated about the horizontal line y=5y = 5.

Step-by-Step Solution:

  1. Find the points of intersection: 14x2=xx2−4x=0x(x−4)=0x=0orx=4\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=0a = 0 and b=4b = 4.

  2. Identify the functions closest to y=5y = 5: g(x)=xg(x) = x is closer to y=5y = 5 than f(x)=14x2f(x) = \frac{1}{4}x^2. Thus, y1=xy_1 = x and y2=14x2y_2 = \frac{1}{4}x^2.

  3. Set up and solve the integral: V=π∫04[(14x2−5)2−(x−5)2],dx=π∫04(116x4−52x2+25−(x2−10x+25)),dx=π∫04(116x4−72x2+10x),dx=π[180x5−76x3+5x2]04=π((180(4)5−76(4)3+5(4)2)−0)=272π15≈56.968,units3,(to 3 decimal places)\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:

  1. Identify the functions and the interval: Let f(y)f(y) and g(y)g(y) be continuous functions on the interval [a,b][a, b], with f(y)f(y) closer to the vertical line x=kx = k than g(y)g(y).

  2. Set up the integral: The volume of revolution VV is given by: V=π∫ab[(g(y)−k)2−(f(y)−k)2],dyV = \pi \int_{a}^{b} \left[ (g(y) - k)^2 - (f(y) - k)^2 \right] , dy

  3. Ensure proper boundaries: If the curves swap places over the interval, split the calculation into separate integrals.

Worked Example

Let RR be the region enclosed by the graphs of f(x)=14x2f(x) = \frac{1}{4}x^2 and g(x)=xg(x) = x. The region is rotated about the vertical line x=−2x = -2.

Step-by-Step Solution:

  1. Rewrite the functions as functions of yy: y=f(x)y=14x2x2=4yx=2yx=y\begin{array}{rcl} y & = & f(x) \\ y & = & \frac{1}{4}x^2 \\ x^2 & = & 4y \\ x & = & 2\sqrt{y} \\ x & = & y \end{array}

  2. Find the points of intersection: y=2yy2=4yy(y−4)=0y=0ory=4\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=0a = 0 and b=4b = 4.

  3. Identify the functions closest to x=−2x = -2: g(x)=xg(x) = x is closer to x=−2x = -2 than f(x)=14x2f(x) = \frac{1}{4}x^2. Thus, x1=yx_1 = y and x2=2yx_2 = 2\sqrt{y}.

  4. Set up and solve the integral: V=π∫04[(2y−(−2))2−(y−(−2))2],dy=π∫04[(2y+2)2−(y+2)2],dy=π∫04(4y+8y+4−(y2+4y+4)),dy=π∫04(8y−y2),dy=π∫04(8y1/2−y2),dy=π[163y3/2−13y3]04=π((163(4)3/2−13(4)3)−0)=64π3≈67.021,units3,(to 3 decimal places)\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)=xf(x) = \sqrt{x} and g(x)=xg(x) = x around the line y=3y = 3.

Question 2

Find the volume of the solid formed by rotating the region bounded by f(y)=y2f(y) = y^2 and g(y)=y+2g(y) = y + 2 around the line x=−1x = -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

  1. Read the question carefully: Identify the functions and the axis of rotation.
  2. Sketch the region: Visualize the area being rotated to understand the setup.
  3. Set up the integral correctly: Use the washer method formula and ensure the limits of integration are accurate.
  4. 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.

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