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Volumes of Revolution

Emily Davis

Emily Davis

6 min read

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Study Guide Overview

This study guide covers the disc method for finding volumes of solids of revolution. It focuses on rotations around axes parallel to the x-axis and y-axis. Key topics include setting up integrals using (y−k)2(y-k)^2(y−k)2 or (x−k)2(x-k)^2(x−k)2, rewriting equations, and applying the method through worked examples. Practice questions, a glossary, and exam strategies are also provided.

#Volume with Disc Method Revolving Around Other Axes

#Table of Contents

  1. Introduction
  2. Volume of Revolution Around Axes Parallel to the x-axis
    • Key Concepts
    • Worked Example
  3. Volume of Revolution Around Axes Parallel to the y-axis
    • Key Concepts
    • Worked Example
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways
  7. Exam Strategy

#Introduction

The disc method is a powerful technique used in calculus to find the volume of a solid of revolution. When a region in the plane is revolved around a line that is parallel to either the x-axis or the y-axis, we can use the disc method to calculate the volume of the resulting solid.

#Volume of Revolution Around Axes Parallel to the x-axis

#Key Concepts

  • For a continuous function fff, if the region bounded by:
    • The curve y=f(x)y=f(x)y=f(x) and the line y=ky=ky=k
    • Between x=ax=ax=a and x=bx=bx=b
  • is rotated around the line y=ky=ky=k, then the volume of revolution is:

V=π∫ab(y−k)2,dxV = \pi \int_{a}^{b} (y - k)^{2} , dxV=π∫ab​(y−k)2,dx

Note that xxx here is a function of yyy.
  • Thinking of this as an accumulation of change:
    • π(y−k)2Δx\pi (y - k)^{2} \Delta xπ(y−k)2Δx is the volume of a disc with:
      • Circular cross section of radius ∣y−k∣\left|y - k\right|∣y−k∣
      • Length Δx\Delta xΔx
    • π(y−k)2,dx\pi (y - k)^{2} , dxπ(y−k)2,dx...
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Question 1 of 7

What is the volume of the solid generated when the region bounded by y=xy=xy=x, y=0y=0y=0, x=0x=0x=0 and x=1x=1x=1 is rotated about the x-axis? 🤔

π4\frac{\pi}{4}4π​

π3\frac{\pi}{3}3π​

π2\frac{\pi}{2}2π​

π\piπ