TITLE: Why is area of a surface of revolution integral $2\pi y~ds$? not '$dx$'?
QUESTION [6 upvotes]: For me, intuitively, integral $2\pi y~dx$ make more sense. I know intuition can not be proof, but by far, most part of math I've learned does match with my intuition. So, I think this one should 'make sense' as well. Probably I didn't understand the way surface area is measured. It will be great if any one could tell me how 'integral $2\pi y~dx$' is wrong. (By the way, how to use mathematical symbols in texts?)

REPLY [2 votes]: OP, if I'm guessing right, the intuition point that you are puzzling over is "how come we can get away with integrating $\pi y^2 dx$ for volume (there too, we are approximating an infinitesimal frustrum by an infinitesimal cylinder) but we are not able to approximate the surface area of the infinitesimal frustrum $2 \pi y ds$ by that of the infinitesimal cylinder $2 \pi y dx$?"
To understand the difference consider the error of the approximation in each case as a fraction of the thing you are trying to calculate. The difference arises when the function is not a constant ($y' \neq 0$). In the case of volume, that error is negligible (because $\pi y^2 dx$ does capture most of the volume) while in the case of the surface area calculation, that error is of the same order as the surface area itself. 
To be more convinced, you can use the formula for the volume of that frustrum of thickness $dx$. It is $$dV=\pi y^2 dx + \pi y dy dx + \pi dy^2 dx$$ the error is $$dE=dV-\pi y^2 dx= \pi y dy dx + \pi dy^2 dx$$ which is tiny in relation to $dV$ (will vanish as we take $dx$ to zero and integrate) in relation to the true V.
The error in the surface area approximation on the other hand, is $$d\epsilon= dA - 2 \pi y dx = 2 \pi y (ds- dx) = 2 \pi y ds (1 - \frac{1}{\sqrt{1+y'^2}})$$which is of the same order as $ dA = 2 \pi y ds $ and upon taking the limit of dx to 0 to compute the integral, this results in a deviation from the true surface area.