- #1
Fractal20
- 74
- 1
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 [itex]\int[/itex][itex]\int[/itex] f(x,y) dx dy. Similarly, if it was in polar coords, f(r,theta) then I would have [itex]\int[/itex][itex]\int[/itex] 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?
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 [itex]\int[/itex][itex]\int[/itex] f(x,y) dx dy. Similarly, if it was in polar coords, f(r,theta) then I would have [itex]\int[/itex][itex]\int[/itex] 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?