The area (in sq. units) of the region bounded by the curves x^2+2y-1=0,y^2+4x-4=0 and y^2-4x-4=0 in the upper half plane is .

The given curves are ${x}^{2}+2y-1=0,{y}^{2}+4x-4=0$x2+2y−1=0,y2+4x−4=0 and ${y}^{2}-4x-4=0.$y2−4x−4=0.

Drawing their graphs, we get

The required area is shown by the shaded part and since the curves ${y}^{2}+4x-4=0$y2+4x−4=0 and ${y}^{2}-4x-4=0$y2−4x−4=0 are symmetric about the $y$y-axis, hence they have the same area.

Hence, the required area$\displaystyle =2[{\int }_{0}^{2}(\frac{4-{y}^{2}}{4})dy-{\int }_{0}^{1}(\frac{1-{x}^{2}}{2})dx]$=2[∫02​(44−y2​)dy−∫01​(21−x2​)dx]

$$=2[{\int }_{0}^{2}\frac{1}{4}(4-{y}^{2})dy-{\int }_{0}^{1}\frac{1}{2}(1-{x}^{2})dx]$$=2[∫02​41​(4−y2)dy−∫01​21​(1−x2)dx]

$$=2[\frac{1}{4}{(4y-\frac{{y}^{3}}{3})}_{0}^{2}-\frac{1}{2}{(x-\frac{{x}^{3}}{3})}_{0}^{1}]$$=2[41​(4y−3y3​)02​−21​(x−3x3​)01​]

$$=2[\frac{1}{4}(8-\frac{8}{3}-0)-\frac{1}{2}(1-\frac{1}{3}-0)]$$=2[41​(8−38​−0)−21​(1−31​−0)]

$=2[\frac{4}{3}-\frac{1}{3}]=2\mathrm{sq}.\mathrm{units}$=2[34​−31​]=2sq.units .