The typical way to compute the volume of a truncated cone is to slice into discs and calculate the volume of a differential cylinder. While doing that we first take the area of the disc $\pi f(x)^2$ and then multiply with thickness which we assume to be $dx$ and then get the final area is $\pi \int f(x)^2dx$. But why should $dx$ be the thickness. As I have indicated in the picture,we are basically taking a tiny step along the surface which is $dl$ which again can be written as $\sqrt{(dy)^2+(dx)^2}$. That changes the entire integration with a factor of $1+\sqrt{(\frac{dy}{dx})^2}$ and hence changes the volume? So which integral should be correct and why should $dx$ be the thickness and not $dL$? This issue is arising because if we were to calculate the surface area of this solid,we would have actually used $dL$ instead of $dx$. So why are we taking two different thickness while accounting for the same differential cylinder? An image or animation will be greatly helpful in pointing out the difference,
The way to think about it is that the truncated cone is first approximated by a union of thin right cylinders (of infinitesimal thickness), forming a ladder-like solid, like the Egyptian pyramids. It needs to be argued that the volume of the ladder-like solid approximates the volume of the original truncated cone, or more precisely the standard part of the volume of the ladder-like solid equals that of the truncated cone. There is certainly foundational work to be done here, which already preoccupied geometers in the 17th century!
Then $dx$ is precisely the thickness of each thin cylinder, and the square root factor you mentioned does not arise.
The calculation for the surface area of the truncated cone is certainly different because in that case the area of the ladder-like solid does not approximate the surface area of the truncated cone.