Why is surface area not simply $2 \pi \int_{a}^{b} (y) dx$ instead of $2 \pi \int_a^b (y \cdot \sqrt{1 + y'^2}) dx$? Geometrically speaking, it seems to me that if you have for example $y^2=8x$ revolved around the x-axis, taking the limit of the sum of $n$ surfaces of cylinders as $n$ approaches infinity should give you the surface area of that surface of revolution.
This is how the author initially derives the formula for finding the volume of solids of revolution. Take a rectangle under the curve over $\Delta x$ and revolve it around the axis to get an approximation of the volume of the solid over that interval. Add up those rectangles over $n$ changes in $x$ and take the limit as $n$ approaches infinity, which is the integral of the function that gives you the $y$ value (radius of that approximating cylinder) for each $x$ value.
Following the same principle, why wouldn't we be able to take those same cylinders, but instead of taking their volume, taking their surface area and take the limit as the number of those cylinders approaches zero?
In other words, in this case each $y$ value is given by $y = \sqrt{8x}$, which is the radius of that cylinder of height $\Delta x$ and an approximation of the surface area over that interval.
Why doesn't that work? Why do we need to deal with arc length? I don't understand why it doesn't work in this case, it seems to me that you're still getting a better and better approximation of surface area as those cylinders get smaller and smaller, eventually getting the exact surface area with the limit as their number goes to infinity.     
PS: I saw this Surface area of a solid of revolution: Why does not $ \int_{b}^{a} 2\pi \,f(x) \,dx $ work? but it's still not making sense visually/geometrically.

In your link, it explains precisely why integrating $y$ doesn't work: Because the length $ds$ of a little piece of the arc is not $dx$, but $\sqrt{(dx)^2+(dy)^2}$, which can be written as
$${\sqrt{(dx)^2+(dy)^2}\over dx}\cdot dx = \sqrt{1 + \left(dy\over dx\right)^2} \cdot dx.$$ It's the same reasoning why the length of a hypotenuse is $\sqrt{a^2+b^2}$ and not $a+b$.
Or this "proof" that $pi=4$: Is value of $\pi = 4$?
