Volume of Revolution: R around Y-Axis

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SUMMARY

The discussion focuses on calculating the volume of revolution around the y-axis for the region R bounded by the curves y=(x-1)^2 and y=2(x-1). Participants clarify that the integral must be set up with respect to y, requiring the inner and outer radii to be expressed as functions of y. The correct final volume is confirmed to be 64π/15, with a common mistake noted in miscalculating the radii and the integration process.

PREREQUISITES
  • Understanding of the washer method for volume calculation
  • Knowledge of curve equations and their transformations
  • Ability to perform integration with respect to y
  • Familiarity with solving equations for x in terms of y
NEXT STEPS
  • Study the washer method for volumes of revolution around the y-axis
  • Learn how to derive functions for inner and outer radii from given curves
  • Practice integration techniques involving variable substitutions
  • Explore examples of volume calculations for different axes of revolution
USEFUL FOR

Students and educators in calculus, particularly those focusing on volumes of revolution, as well as anyone seeking to improve their understanding of integration techniques related to geometric applications.

cmab
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R is bounded by the curves y=(x-1)^2 and y=2(x-1). Axis of revolution: y-axis.

How am i supposed to do this. I know how to do the washer method, I know how to apply it when it is revolved arround the x-axis, but I don't know how to do it when it is the y axis. Can anybody explain to me the steps to do ? :frown:
 
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cmab said:
R is bounded by the curves y=(x-1)^2 and y=2(x-1). Axis of revolution: y-axis.

How am i supposed to do this. I know how to do the washer method, I know how to apply it when it is revolved arround the x-axis, but I don't know how to do it when it is the y axis. Can anybody explain to me the steps to do ? :frown:

Your integral will be with respect to y, and you need to find the inner and outer radius as a function of y. That involves solving your equations for x.
 
OlderDan said:
Your integral will be with respect to y, and you need to find the inner and outer radius as a function of y. That involves solving your equations for x.

I did the reciprocals. so its x= sqrt(x)+1 and x=y/2 +1

I did everything and the result is 16pie/12, and in the answer sheet it says 64pie/12
 
cmab said:
I did the reciprocals. so its x= sqrt(x)+1 and x=y/2 +1

I did everything and the result is 16pie/12, and in the answer sheet it says 64pie/12

You mean x= sqrt(y)+1 and x=y/2 +1. Is that what you did? The answer given is correct (reduces to 16pi/3). When you take the difference between the lerger area and the smaller area of each washer, two terms cancel and two terms survive that need to be integrated.
 
Thx man, I gotted the answer but I was too tired to realize it :smile:
 
Answer sheet = 64pie/15
But I think the paper is wrong, cause it keeps giving me 64pi/12 (Unreduced)
 

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