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I don't quite understand why the formula to find the surface area of a revolution is what it is:

$$A = 2\pi \int_a^b x\ \sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2}\ \text{d} x.$$

I would've though that A could equal: $A = 2\pi \int_a^b y\ \text{d} x$.

ΔS / | Δy
  /  |

I've looked at proofs but I don't really understand why $\Delta s\to 0$ and not $\Delta x\to 0$. I would've thought that as $\Delta x\to 0$ (or $\Delta y\to 0$) you'd end up with a slice infinitelly small, and it wouldn't have a slope to it, then I would've though that you could add up the perimeter of these 'slopeless' slices, and get the surface area.

I'd would help if I could geometrically see what is going wrong with letting $\Delta x\to 0$ (or $\Delta y\to 0$).

More specifically, if I just go $2\pi y$, what am I not taking into account (graphically).

I know this is like the arc length (length of a curve) formula, I have the same grievance with that.

Thank you.

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Since you say you have the same question about the arc length formula, have you looked at this previous question? Intuition behind arc length formula – Rahul Mar 9 '12 at 13:10
Well, the point really is that even infinitesimal segments have a slope. IOW $\Delta s\to0$ and $\Delta x\to0$ at the same time, but their ratio $$\frac{\Delta s}{\Delta x}\to \sqrt{1+\left(\frac{dy}{dx}\right)^2}.$$ And this limit is $>1$ more often than not. – Jyrki Lahtonen Mar 9 '12 at 13:32
I think that your problem (shared many a student around the globe) is that the kind of "infinitesimal approximations" that lead us to correct formulas for surface area (resp. volume of the solid of revolution) fail here. A key difference is that when approximating area by rectangles (resp. volume by disks) the relative error is negligible, i.e. the ratio of (estimation error)/(true area, resp. volume) tends to zero. That does not happen with arc length (resp. area of surface of revolution) unless you take the slope into account. – Jyrki Lahtonen Mar 9 '12 at 14:29

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