Calculate the surface area of a solid of revolution

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?

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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.

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If you want to try the original integral again, you can use the trig substitution x = 0.5 tan(t). This will give you an integral involving tan and sec which can be worked out.

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The problem is that we did not define the sec during our lectures yet... –  Huy Dec 5 '10 at 8:01
I'm not too sure, but if I recall correctly, one way you can make this a little easier is to note that $g(x)=(1-x^2)\sqrt{1+4x^2}$ is even $g(-x) = g(x)$ and its range is from $-a$ to $a$.
So you can do $2\int_0^a g(x)\, dx$