Volumes of Solids of Revolution

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SUMMARY

The discussion focuses on calculating the volumes of solids of revolution, specifically for the region R bounded by the curves y = x² and y = x + 2. The area of R is determined to be 4.5 using the integral ∫((x + 2) - (x²))dx over the interval (-1, 2). The volume when R is rotated about the x-axis is calculated as (72π/5) using the integral ∏∫((x + 2)² - (x²))dx. The participant expresses uncertainty regarding the shell method for calculating the volume when R is rotated about the line x = 4.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with the disk and shell methods for volume calculation
  • Knowledge of the curves y = x² and y = x + 2
  • Basic proficiency in calculus
NEXT STEPS
  • Study the shell method for calculating volumes of solids of revolution
  • Practice finding areas between curves using definite integrals
  • Explore applications of the disk method in volume calculations
  • Review integration techniques for polynomial functions
USEFUL FOR

Students studying calculus, particularly those focusing on integration and volumes of solids of revolution, as well as educators seeking to clarify these concepts for learners.

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I'm really having issues grasping the Volumes of Solids of revolution. I could use some help solving this question, it isn't very hard.

1. Let R be the region bounded by y = x2 and y = x + 2. Find:

a) the area of R

b) the volume of the solid if R is rotated about the x-axis

c) the volume of the solid if R is rotated about the the line x = 4





Homework Equations





The Attempt at a Solution



Basically this is what I've gotten.



a)
Area = ∫((x+2)-(x2))dx
= (x2/2 + 2x) - (x3/3), evaluated over the interval (-1,2)
= 4.5


b) this is what I did but I don't know if it is right

∏∫((x+2)2 - (x2))dx
= ∏∫((x2+4x+4) - x4) dx
= ∏(x5/5 - x3/3 -2x2 -4x) evaluated from (-1,2)
= 72∏/5

c) I'm not sure how to start this one.
 
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a and b look right. For c, have you learned the shell method for integration?
 

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