# Conceptual problem with surfaces of revolution When calculating the surface area of a surface of revolution, we derive the formula $$S = \int_a^b 2\pi f(x) ds$$ by approximating the surface by a set conical loops (for lack of a better term). Then we take the limit as the width of those loops goes to zero and get the surface area. That makes perfect sense.

But why couldn't we approximate the surface by cylindrical loops (I hope you guys understand what I mean by that)? To me this seems like the more Riemann style way to do it -- where the usual method looks a lot like the trapezoidal method for evaluating integrals. But in the limit as $\Delta x \to 0$, shouldn't the error of either way also go to zero?

Why would $\int_a^b 2\pi f(x) dx$ get you the wrong answer?

• I understand what you mean, I'm not able to provide some answer though; but note that while with the Riemann integral the limit will be $\Delta x \to 0$, with some surface integral the limit should be some kind of $\Delta V \to 0$, where V is some cylindrical volume volume – Eugenio Nov 12 '16 at 0:24
• – Hans Lundmark Nov 25 '19 at 15:20
• @ Bobbie D Consider an annular ring in a plane perpendicular to axis of symmetry. Taking $\int.. dx$ instead of $\int.. ds$ gives differential area as zero that is quite counter- intuitive. – Narasimham Nov 25 '19 at 16:47

• @BobbieD It depends what one is trying to compute. In a case like length of a hypotenuse the error (on replacing by only the horizontal parts) is fatal. Of course if one only wants the integral of a scalar function one would use $f(x)dx$ in the first place, and then $ds$ (distance along the curve $y=f(x)$) would be irrelevant for this. – coffeemath Nov 12 '16 at 0:30
You can think $ds$ as: \begin{equation} ds=\sqrt{(dx)^2+(dy)^2)}\\ ds=\sqrt{(dx)^2(1+(\frac{dy}{dx})^2)}\\ ds=dx\sqrt{1+(\frac{dy}{dx})^2}\\ ds=dx\sqrt{1+(\frac{df}{dx})^2} \end{equation} just applying Pythagoras's theorem.
The term inside the square root takes into account the slope of the curve; in the case $\frac{df}{dx}=0$, $ds$=$dx$ (it could be the case of a cylinder along x-axis).