Answer
$\dfrac{2 \sqrt 2}{3}$
Work Step by Step
Here, we have: $y=1-x^2, y=x^{2}$
Suppose that $y=y \implies 1-x^2=x^2$ or, $x=\pm \dfrac{1}{\sqrt 2}$
The area is given by $A=\int_{-\frac{1}{\sqrt 2}}^{\frac{1}{\sqrt 2}} (1-2x^2) \ dx$
or, $= [x-\dfrac{2x^3}{3}]_{-\frac{1}{\sqrt 2}}^{\frac{1}{\sqrt 2}}$
or, $=[\dfrac{1}{\sqrt 2}-\dfrac{2(-\frac{1}{\sqrt 2})^3}{3}]-[-\dfrac{1}{\sqrt 2}-\dfrac{2(-\frac{1}{\sqrt 2})^3}{3}]$
or, $=2(\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{6})$
Therefore, the required area is: $Area=\dfrac{2 \sqrt 2}{3}$
