Can someone please explain to me why the surface area of a solid of revolution formula was derived in the way it was? After learning more about disc integration, that formula made perfect since to me. That is
$$\pi \int_{a}^{b} [R(x)]^2 \space dx$$
where $R(x)$ is the distance between the function and the axis of rotation (in this case the horizontal axis).
I understood this intuitively as viewing the integral as the summation of an infinite number of "slices" with infinitesimal width that can be interpreted as radius's. The formula above can then be viewed as the sum of a lot of circle areas to get the volume of the object generated using the function $R(x)$.
This thinking lead me to think that the surface area of a solid of revolution could be thought of in a similar way and I thought that the formula would have the following form:
$$ 2\pi \int_{a}^{b} [R(x)] \space dx$$
Why is this not the case? I understand the derivation explanations online that use frustums and the arc length as a motivation to arrive at the following formula:
$$S = \int_{a}^{b} 2 \pi \space f(x) \space \sqrt{1 + [f'(x)]^2} \space dx$$
I'm just trying to understand why the way I originally thought about the problem is incorrect.
