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Solution

The region lies between the circle $x^2 + y^2 = 1$ and the parabola $y^2 = 1 - x$. Converting parabola into standard form: $x = 1 - y^2$. To find points of intersection, substitute $x = 1 - y^2$ in $x^2 + y^2 = 1$: $(1 - y^2)^2 + y^2 = 1 \Rightarrow 1 - 2y^2 + y^4 + y^2 = 1 \Rightarrow y^2(y^2 - 1) = 0.$ $\Rightarrow y = 0, \pm 1.$ Required area (using symmetry about x-axis): $A = 2 \int_{0}^{1} [\sqrt{1 - y^2} - (1 - y^2)] dy.$ Compute separately: $\int_0^1 \sqrt{1 - y^2}dy = \dfrac{\pi}{4}$, $\int_0^1 (1 - y^2)dy = \dfrac{2}{3}.$ Hence $A = 2\left(\dfrac{\pi}{4} - \dfrac{2}{3}\right) = \dfrac{\pi}{2} - \dfrac{4}{3}.$