I have recently learned the principle of finding the surface area of revolution, but I don't understand why it works.
It was asserted that
$\delta A\approx 2\pi y\delta s$
I would have thought that you are just summing the circumferences of circles, so instead of the arc length tending to zero, a change in x would tend to zero, giving you a formula of:
$A=2\pi \int y\mathrm{dx}$
James Thimont

Let's simplify the situation to a cone, generated by a right angled triangle of height h and base b. The curved surface area we can work out as

$A=\pi bl$
where l is the slant height. (If you don't see that, try to 'unfold' the curved bit to a sector of circle, and work out its area.) That is why we integrate with respect to s, rather than x. What you have asserted there would only have worked for a cylinder, as a special case ds=dx.

Or...

By doing the integral with respect to the arc length what you are doing is to chop the surface up into lots of almost cylinders. These have radius y and 'height' ds so the total area is what you wrote down. So it isn't quite the same as adding up lots of circles but adding cylinders.

The other reason for doing things with respect to ds is that is ds is small we can guarantee dy is small; but if dy is small there is no reason for ds to be small.

AlexB.