Find the area of the surface obtained by revolving $\sqrt{1-x^2}$ about the x-axis? So I began by choosing my formulas:
Since I know the curve is being rotated around the x-axis I choose $2\pi\int yds$
where $y=f(x)=\sqrt{1-x^2}$
$ds=\sqrt{1+[f'(x)]^2}$
When I compute ds, I find that $ds=\sqrt{x^6-2x^4+x^2+1}$
Therefore, my integral becomes: $2\pi\int(1-x^2)^{\frac{1}{2}}\sqrt{x^6-2x^4+x^2+1}dx$
Am I on the right track, because this integral itself seems very hard to solve?

No.  
$$f'(x) = -\frac{x}{\sqrt{1-x^2}} \implies 1+f'(x)^2 = \frac{1}{1-x^2}$$
