# 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

This is a lot like trying to approximate the length of a hypotenuse of a right triangle by just adding up only the horizontal legs of small right triangles which fit together to make the larger right triangle. One clearly needs to add up the hypotenuses of the small right triangles in this case, not just the horizontal legs.

• Couldn't you make the same argument about regular scalar integrals? – Bobbie D Nov 12 '16 at 0:24
• @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
• I'll wait to see if someone else wants to give a more comprehensive answer before accepting, but this has definitely given me more insight. Thanks. +1 – Bobbie D Nov 12 '16 at 0:34
• Bobbie-- I agree you should wait, someone is likely to give a more compete answer. In the meanwhile, you might find good explanatory "pictures" for this in a reasonable calc 2 book in its section on surfaces of revolution, or even just google to get to a wiki page. – coffeemath Nov 12 '16 at 0:42
• No I just thought about it more and realized why approximating the area under a curve by rectangles can be used to bound the area from below and from above. But in the case of the surface area of a surface of revolution, using only the horizontal part of the length of a loop will bound it from below, but there's no reason to think that in the limit it will converge to the actual surface area. I've still got a bit more to think about it to see why it definitely won't converge to the correct answer, but at least I see the distinction now and can see that I shouldn't expect it to do so. – Bobbie D Nov 14 '16 at 0:55

You can think $ds$ as: $$ds=\sqrt{(dx)^2+(dy)^2)}\\ ds=\sqrt{(dx)^2(1+(\frac{dy}{dx})^2)}\\ ds=dx\sqrt{1+(\frac{dy}{dx})^2}\\ ds=dx\sqrt{1+(\frac{df}{dx})^2}$$ 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).