Conceptual problem with surfaces of revolution

Question

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?

Answer 1

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.

Answer 2

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).