Summary: Essentially I don't understand this inconsistency between using frustrums or disks in derivations (or trapeziums and rectangular strips) I also don't understand why the limit of a sum of disks doesn't work, and shouldn't give you the surface area of a solid of revolution
Full question: In the derivation for the surface area of a Solid of revolution they take the sum and limit of the surface area of frustums to give you $\int(2\pi y ds)$ where $s$ is the arc length of the curve.
When deriving the volume of this shape you use disks of width $dx$ and radius $y$ so the volume of each is $\pi y^2dx$ and then take the sum and then take the limit. Why not use disks for the surface area of a solid of revolution, giving you $\int(2\pi y dx)$ instead, why should the limit of the surface area of a sum of disks of width $dx$ and height $y$ not give you the surface area?
Surely using frustrums is like using trapeziums instead of rectangular strips when deriving the area under a general curve, so why don't we use this method to give you $\int(y ds)$ instead of $dx$? Furthermore why wouldn't we similarly use frustrums when deriving the volume of a solid of revolution?
The reason is essentially the same as the reason why, for a right triangle with legs $a$ and $b$, and hypotenuse $c$, $$a^2 + b^2 = c^2,$$ and not $$a + b = c.$$ The same fallacy occurs when trying to argue the latter for a right triangle by incorrectly reasoning that the limit of a sum of the lengths of stepwise horizontal/vertical lines parallel to the legs will converge to the length of the hypotenuse.