It's known that the volume of revolution of the function $f(x)$ (assuming it's real, continuous...) is $$V=\pi\int_a^b f(x)^2dx$$
This can be modelized as if we add together all the infinitesimal cylinders of radius $|f(x)|$ integrating over all the domain $x \in [a,b]$
However, what's the geometric interpretation of the surface
$$S=2\pi\int_a^b f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx$$
And this isn't the sum of the perimeter of all the small circumference with radius $|f(x)|$ (at least I can figure out why) ?
It will be $$S=2\pi\int_a^b ydx$$ Can anyone explain the geometric meaning of $\sqrt{1+\bigg(\dfrac{dy}{dx}\bigg)^2}$
Thanks in advance.
I moved this answer to the more active: Why is surface area not simply $2 \pi \int_{a}^{b} (y) dx$ instead of $2 \pi \int_a^b (y \cdot \sqrt{1 + y'^2}) dx$?