area computations

The circle ( $x^2 + y^2 = 1$ ) and the parabola intersect in the points:

$\large y + y^2 = 1 \Rightarrow y^2 + y + \frac{1}{4} = 1 + \frac{1}{4} \Rightarrow (y+\frac{1}{2})^2 = \frac{5}{4} \Rightarrow y = -\frac{1}{2} + \frac{\sqrt{5}}{2}$

since $y > 0$ per the above figure. This in turn yields $\large x^2 = \frac{-1+\sqrt{5}}{2} \Rightarrow x = \pm \sqrt{\frac{\sqrt{5}-1}{2}}.$

By exploiting symmetry, we have:

$\Large 2\int_{0}^{ \sqrt{\frac{\sqrt{5}-1}{2}}} \sqrt{1-x^2} - x^2 dx = 2[\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\arcsin(x)] - \frac{2x^3}{3}|_{0}^{ \sqrt{\frac{\sqrt{5}-1}{2}}} \approx \boxed{1.0666}.$