Answer
$$A \approx 0.87367$$
Work Step by Step
$$\eqalign{
& {\text{Let the graphs }}y = {x^2},\,\,\,x = 2{\sin ^2}y \cr
& y = {x^2} \Rightarrow x = \sqrt y {\text{ for }}y \geqslant 0 \cr
& \cr
& {\text{Find the intersection points between the graphs: }} \cr
& {\text{Let }}x = x \cr
& 2{\sin ^2}y = \sqrt y \cr
& {\text{Using a graphing utility we obtain}} \cr
& {y_1} \approx 0.705{\text{ and }}{y_2} \approx 2.12 \cr
& \cr
& {\text{For the interval }}\left( {0,0.705} \right){\text{: }}\sqrt y \geqslant 2{\sin ^2}y \cr
& {\text{For the interval }}\left( {0.705,2.12} \right){\text{: }}2{\sin ^2}y \geqslant \sqrt y \cr
& \cr
& {\text{Therefore}}{\text{, the enclosed area is given by}} \cr
& A \approx \int_0^{0.705} {\left( {\sqrt y - 2{{\sin }^2}y} \right)} dy + \int_{0.705}^{2.12} {\left( {2{{\sin }^2}y - \sqrt y } \right)} dy \cr
& {\text{Integrating by using a graphing utility we obtain}} \cr
& A \approx 0.1831 + 0.6905 \cr
& A \approx 0.87367 \cr} $$
