I have to calculate the surface area of the solid of revolution which is produced from rotating $f: (-1,1) \rightarrow \mathbb{R}$, $f(x) = 1-x^2$ about the $x$-axis. I do know there is a formula: $$S=2 \pi \int_{a}^b f(x) \sqrt{1+f'(x)^2}\, \mathrm dx$$ Which will work very well. However, I am not very comfortable with the integral $$\int_{-1}^1 (1-x^2)\sqrt{1+4x^2}\, \mathrm dx$$ which I would have to calculate in order to get to the surface area (I tried to substitute $x=\frac{1}{2} \sinh(u)$, but it did not work out too well). Thus, I had the idea to apply Pappus' centroid theorem. I first found the centroid of the area between the parabola and the x-axis to be at $y=\frac{2}{5}$, hence the surface area of the solid of revolution would be: $$S = 2 \pi \frac{2}{5} \int_{-1}^1 \sqrt{1+4x^2}\, \mathrm dx$$ But this leads me to a different result than I should get (I calculated the value of the first integral with the help of wolframalpha, it's about ~11...).
What did I do wrong? My best guess is that I misunderstood Pappus' centroid theorem, but what's the mistake? How can I fix it?