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

[HclGuy]

Homework Statement

Find the volume of the solid bounded above by $$\rho$$=1+cos$$\varphi$$ and below by $$\rho$$=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 $$\rho$$ were from 0 to 1, wouldn't the limits for rho be 1 to 1+cos$$\varphi$$? Thanks

 

[HallsofIvy]

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.

 

[HclGuy]

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$$\varphi$$

 

[HallsofIvy]

You may have copied the problem wrong or he may have set it up wrong himself. Have you asked your professor?