In integrating to find a volume when is the function included?

1. Aug 20, 2012

Fractal20

I just want to clarify this. I have had trouble before with including the function in the integral when I am trying to find volume. I have come to think that this is necessarily only when it is a double integral. But I had a specific question I posted earlier and I feel like somebody told me that the function is never included.

More specifically, if I have z = f(x,y) and I want to integrate over some domain in the xy plane then I want to say the integral would look like $\int$$\int$ f(x,y) dx dy. Similarly, if it was in polar coords, f(r,theta) then I would have $\int$$\int$ f(r,theta)r dr d theta. I think this is true since the integral is "adding" little boxes of volume z dx dz or z r dr dtheta respectively.

However, if it is a triple integral, whether in spherical, cylindrical or Cartesian, the function is not included in the integral because dV is the little units of volume. Is this correct?

2. Aug 21, 2012

Muphrid

Yes, this comes from the way one decides to construct the volume from buildling blocks, so to speak. Whne you find the volume as a double integral, you're dividing the xy plane into a bunch of tiny squares and then stacking squares up to a height based on the given function f. Obviously this is reasonable to find the volume beneath a surface.

When you're finding volume as a triple integral, you're constructing tiny cubes and, to find volume, you need only sum them up. There is no need for a weighting function like f. All cubes are created equal, so to speak.

3. Aug 21, 2012

Fractal20

Thanks for the verification!

4. Aug 21, 2012

haruspex

Here's how to see the equivalence algebraically:
Volume = $\int_V dv$
= $\int \int \int_{z = 0}^{f(x,y)} dz dx dy$
= $\int \int [z]_{z = 0}^{f(x,y)} dx dy$
= $\int \int f(x,y) dx dy$