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

HclGuy
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Homework Statement



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

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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 \rho were from 0 to 1, wouldn't the limits for rho be 1 to 1+cos\varphi? 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
\int\int_0^\pi\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta
outer limits are from 0 to 2pi
I was just wondering why his limits for \rho were from 0 to 1. I thought it would be from 1 to 1+cos\varphi
 
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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