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.
Best Answer
We have an option to cut the solid of revolution (obtained by revolution of $y = f(x)$ between $x = a$ and $x = b$) into multiple slices in the following manner:
Let the desired slicing be done via partition $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ of $[a, b]$. We will apply both the approaches mentioned earlier to calculate the surface area as well as volume of the solid of revolution.
First we deal with volume which has an easier analysis. If we slice the solid as cylinders then the approximation of volume is given by $$V(P) = \pi\sum_{i = 1}^{n}\{f(x_{i})\}^{2}(x_{i} - x_{i - 1})\tag{1}$$ which is a Riemann sum for the integral $\pi\int_{a}^{b}\{f(x)\}^{2}\,dx$ and this is the desired volume.
If we slice the solid into frustums of cone we get the approximation of volume as $$V(P) = \frac{\pi}{3}\sum_{i = 1}^{n}\left[\{f(x_{i - 1})\}^{2} + f(x_{i - 1})f(x_{i}) + \{f(x_{i})\}^{2}\right](x_{i} - x_{i -1})\tag{2}$$ which is split into 3 terms and each term is a Riemann sum for $(\pi/3)\int_{a}^{b}\{f(x)\}^{2}\,dx$ so that the desired volume is again $\pi\int_{a}^{b}\{f(x)\}^{2}\,dx$
Let's now come to surface area of the solid of revolution. If we slice the solid into cylinders then the surface area is approximated by $$S(P) = 2\pi\sum_{i = 1}^{n}f(x_{i})(x_{i} - x_{i - 1})\tag{3}$$ which tends to $2\pi\int_{a}^{b}f(x)\, dx$.
If we slice the solid into frustums we get the approximation for surface area as $$S(P) = \pi\sum_{i = 1}^{n}\{f(x_{i - 1}) + f(x_{i})\}\sqrt{(x_{i} - x_{i - 1})^{2} + ((f(x_{i}) - f(x_{i - 1}))^{2}}\tag{4}$$ which can be simplified by the use of mean value theorem as $$S(P) = \pi\sum_{i = 1}^{n}\{f(x_{i - 1}) + f(x_{i})\}\sqrt{1 + \{f'(t_{i})\}^{2}}\cdot(x_{i} - x_{i - 1})\tag{5}$$ for some points $t_{i} \in (x_{i - 1}, x_{i})$. This can be split into two sums each of which is a Riemann sum for $\pi\int_{a}^{b}f(x)\sqrt{1 + \{f'(x)\}^{2}}\,dx$ so that the desired surface area is $2\pi\int_{a}^{b}f(x)\sqrt{1 + \{f'(x)\}^{2}}\,dx$.
We see that in case of volume both the approaches give the same answer. But in case of surface area the answers obtained by both the methods are different. Further note that out of the two answers we can easily verify which one is correct by using $y = x, a = 0, b = 1$ so that the solid of revolution is a circular cone. This verification shows that the technique used in equation $(4), (5)$ gives the correct surface area.
The question which OP is asking is this:
Why do both the approaches (using cylinders and frustums) give the same result for volume but different results for surface area?
The reason is simple. Both the sums in $(3)$ and $(4)$ are trying to approximate the surface area of the solid, but there is a huge difference between them namely $$\Delta = 2\pi\sum_{i = 1}^{n}f(x_{i})\left[\sqrt{1 + \{f'(t_{i})\}^{2}} - 1\right](x_{i} - x_{i - 1})$$ and this itself is a non-zero sum unless $f'(x)$ is identically zero. So the approximation $(4)$ is trying to take into account some additional surface area which is left out by sum $(3)$ and this additional part is significant unless $f'(x) = 0$ identically. Hence $(4)$ is a better and correct approximation.
In case of volume both the approximations $(1), (2)$ are Riemann sums for the same integral (but are expressed in slightly different ways).