@Aladdin This means that for a fixed $z$ you are inside the circle $x^2+y^2=3z$.

So $\int_0^3 dz \int_0^{\sqrt{3z}} dx \int_0^{\sqrt{3z-x^2}} dy$.

Or, using cylindrical coordinates, we would get $\int_0^3 dz \int_0^{\sqrt 3z} dr \int_0^{2\pi} d\varphi$.

So outer limit is z?

Then we fix x and y

@Aladdin Of course, there are many other ways to describe the same region, I went for z first - since that seems to be the simpler. (It has rotational symmetry around z-axis.)

I am not really sure what you mean by "outer limit is z". In any case, I checked the range for z and checked what happened for a fixed z.

I assumed you want to find bound for integration - similarly as in your previous questions.

If you're trying to do something else, you should explain what actually is the question.

If anyone is around I am doing some practice problems for an introductory real analysis class that I am in and am stuck on one problem in particular that is driving me nuts, and I would really appreciate the help.

@Matthew I think you have better chance of getting some response if you actually write what your questions is.

This is true for this room and for other rooms as well. There is a slogan: Don't ask to ask, just ask..

In fact, something along those lines is written also in Main Chatroom Guidelines.