Rotating integrals and circles

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Homework Help Overview

The discussion revolves around the problem of finding the volume of a solid formed by rotating the curve defined by the equation x^2 + (y-1)^2 = 1 about the x-axis. This involves understanding the setup of integrals in the context of calculus and geometry.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss how to set up the integral for the volume calculation, with one participant expressing confusion about the correct approach. There is mention of a substitution method and the challenges of integrating the resulting expression.

Discussion Status

The conversation includes attempts to clarify the integral setup and explore substitution techniques. Some participants are sharing their struggles with the integration process, while others suggest potential methods to simplify the problem. There is no explicit consensus on the best approach yet.

Contextual Notes

One participant notes a hint received about considering the integral with respect to the y-axis, which may influence the setup and interpretation of the problem. There is also mention of a previous negative experience with the question, indicating a desire for deeper understanding.

lemurs
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kay had a question which i got zero on and now trying to figure out how the hell to do the question.

Given x^2 + (y-1)^2 =1 rotated about the x-axis.

if i could get the solution so i can review it so i can have a better idea how to do it on the mid term.

Hint:, I was told to do with respect to the y-axis.
 
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So how do you set up the integral?
 
well that part of the stuff that confused me.

I set it upas
x = sqrt(1- (y-1)^2)

so my intergral becomes

from 0-2 2(pi) y sqrt(2y - Y^2).

but given that. we end up with a nasty integration... I tryed to get help form soem one but we vouldn't figure it out..
 
That's right. Try the substitution u=y-1. You'll end up with the sum of two integrals, one you can do by another substitution and the other you can do just knowing the area of a circle.
 

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