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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?

Oddly Asymmetric
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  • Maybe ... because using disks for surface areas would give you the wrong answer? Try it out on some solid, e.g. a cone or a sphere, and see how it comes out. –  Jan 19 '18 at 00:28
  • One small point: the word is F-R-U-S-T-U-M. Only one "R". And if you want to sounds pompous, you can use the plural "Frusta". But "frustrum"? Not the right word. – John Hughes Jan 19 '18 at 00:45

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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.

heropup
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  • Oh I see, in the other cases of area under a curve and volume would using ds and dx come to the same result? I.e. If you use trapeziums with the arc length would it still get you the same formula for the area blinded by a curve? – Oddly Asymmetric Jan 19 '18 at 00:56