Volume of Revolution: R around Y-Axis

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Homework Help Overview

The discussion revolves around finding the volume of a region R bounded by the curves y=(x-1)^2 and y=2(x-1) when revolved around the y-axis. Participants are exploring the application of the washer method in this context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to set up the integral with respect to y and to determine the inner and outer radii as functions of y. There is some confusion regarding the correct equations to use after solving for x.

Discussion Status

Some participants have provided guidance on the necessary steps to set up the integral, while others express uncertainty about the correctness of their results compared to the answer sheet. There is an ongoing exploration of the calculations involved.

Contextual Notes

Participants are working under the constraints of homework rules and are questioning the accuracy of the provided answer sheet, which presents conflicting results.

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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