You did misinterpret Pappus' Theorem. You used the geometric centroid of the region between $1-x^2$ and the $x$-axis, whereas Pappus' Theorem wants you to use the geometric centroid of the curve $1-x^2$. The geometric centroid of the latter is not $\frac{2}{5}$ but (by definition)
$$\frac{\int_{-1}^1 (1-x^2) \sqrt{1+4x^2} dx}{\int_{-1}^1 \sqrt{1+4x^2} dx} \approx 0.59002.$$
Unfortunately, multiplying this by $2\pi$ times the arc length just gives you the integral you started with. So it doesn't appear that Pappus' Theorem is an easier route to take. You could also try switching to an integral in $dy$, but I doubt that will be any better. I would try Joe's suggestion on your first integral.
For more on finding geometric centroids of curves, see this.
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:
- each slice is a cylinder
- each slice is a section of cone cut by two parallel planes (a frustum of a cone)
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).
Best Answer
The parametric equation of the cycloid is $$x(t)=r(t-\sin t) \, \quad y(t)=r(1-\cos t) \quad \mbox{for $t\in[0,2\pi]$.}$$ Its surface of revolution around the $x$-axis is given by $$S:=2\pi \int_0^{2\pi}y(t) \sqrt{x'(t)^2+y'(t)^2}dt.$$ Then $$x'(t)=r(1-\cos t) \ , \ y'(t)=r\sin t \implies x'(t)^2+y'(t)^2=2r^2(1-\cos t)=4r^2\sin^2(t/2) $$ and we find that $$ \begin{align} S&=2\pi \int_0^{2\pi}r(1-\cos t) \cdot 2r\sin(t/2)dt =8\pi r^2\int_0^{2\pi}\sin^3(t/2)dt\\ &=16\pi r^2\int_0^{\pi}\sin^3(s)ds =16\pi r^2\int_0^{\pi}(1-\cos^2(s))d(-\cos(s))\\ &=16\pi r^2\left[\cos(s)-\frac{1}{3}\cos^3(s)\right]_{\pi}^0=\frac{64\pi r^2}{3}. \end{align}$$