The area above the x-axis enclosed by the curves ${\mathrm{x}}^2-{\mathrm{y}}^2=0 \mathrm{\&} {\mathrm{x}}^2+\mathrm{y}-2=0$ is

The curve ${\mathrm{x}}^2-{\mathrm{y}}^2=0\Rightarrow \mathrm{y}=\pm \mathrm{x}$ represents a pair of straight lines with slopes 1 and -1 passing origin. The second curve ${\mathrm{x}}^2+\mathrm{y}-2=0\Rightarrow {\mathrm{x}}^2=-\mathrm{y}+2\Rightarrow {\mathrm{x}}^2=-(\mathrm{y}-2)$

Solving two equations we get $\left.{\mathrm{y}}^2+\mathrm{y}-2=0\text{ [putting }{\mathrm{x}}^2={\mathrm{y}}^2\right]$

$(\mathrm{y}+2)(\mathrm{y}-1)=0$

Giving y = -2, y = 1, but y = -2 is discarded as the required area is above the x-axis

$\therefore \mathrm{y}=1\Rightarrow \mathrm{x}=\pm 1$

The points A and C are respectively $\left(-1, 1\right)$ and $\left(1, 1\right)$ now due to symmetry

Area of the bounded region$=2{\int }_0^1 \left[\left(2-{\mathrm{x}}^2\right)-\mathrm{x}\right]\mathrm{dx}$

Since $\mathrm{y}=2-{\mathrm{x}}^2$ is the upper curve and y = x is the lower curve

$2{\left[2\mathrm{x}-\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}\right]}_0^1=2\left[2-\frac{1}{3}-\frac{1}{2}\right]=\frac{7}{3}$