Limits for Rho in Triple Integral for Volume of Solid Bounded by Two Surfaces

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Homework Help Overview

The discussion revolves around finding the volume of a solid defined by two surfaces in spherical coordinates, specifically bounded above by \(\rho=1+\cos\varphi\) and below by \(\rho=1\). Participants are examining the limits of integration for the variable \(\rho\) in a triple integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are comparing their integration limits for \(\rho\) and questioning why the professor's limits differ from their expectations. There is a focus on understanding the reasoning behind the limits of integration in the context of the problem.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about the professor's solution and their own results. Some guidance is suggested regarding potential errors in problem setup or interpretation, but no consensus has been reached.

Contextual Notes

There are indications of possible errors in the professor's answer key, and participants are considering the implications of these errors on their understanding of the problem.

HclGuy
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Homework Statement



Find the volume of the solid bounded above by [tex]\rho[/tex]=1+cos[tex]\varphi[/tex] and below by [tex]\rho[/tex]=1

Homework Equations




The Attempt at a Solution



I already solved it but was comparing my answer to my professor's solution, I was wondering why when he did the integration, his limits for [tex]\rho[/tex] were from 0 to 1, wouldn't the limits for rho be 1 to 1+cos[tex]\varphi[/tex]? Thanks
 
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There are any number of reasons why one integral might give the same result as another- since you did not post his complete integral I don't believe we can say why his method works for this particular problem.
 
The thing is, I got a different answer and was not sure if the professor had made an error on the answer key since I already spotted a few other errors on it..but let's see now
[tex]\int\int_0^\pi\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta[/tex]
outer limits are from 0 to 2pi
I was just wondering why his limits for [tex]\rho[/tex] were from 0 to 1. I thought it would be from 1 to 1+cos[tex]\varphi[/tex]
 
You may have copied the problem wrong or he may have set it up wrong himself. Have you asked your professor?
 

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