The area of the region \ (x, y): x ^ 2 ≤ y ≤ 8 - x ^ 2 , y ≤ 7\ is

The given curves are ${x}^{2} \le y, y \le 8 - {x}^{2}\text{and}y \le 7$

The point of intersection of the curves ${x}^{2} \le y\text{and}y \le 8 - {x}^{2}$ is obtained by,

$⇒ {x}^{2} = 8 - {x}^{2}$

$⇒ x = \pm 2$ and $y = 4$.

Hence the points are \left(2, 4\right), \left(- 2, 4\right)$.</p><p>The points of intersection of the curves$y \le 8 - {x}^{2}\text{and}y \le 7$is obtained by,</p><p>$⇒ 8 - {x}^{2} = 7$</p><p>$⇒ x = \pm 1$and$y = 7$</p><p>Hence the points are \left(1, 7\right), \left(- 1, 7\right)$.

The required graph is

The required area is symmetrical about $y -$axis.

Hence required area is $A = 2\left[{∫}_{0}^{4}\sqrt{y} d y + {∫}_{4}^{7}\sqrt{8 - y} d y\right]$

$A = 2\left[{\left[\dfrac{{y}^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]}_{0}^{4} - {\left[\dfrac{{\left(8 - y\right)}^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right]}_{4}^{7}\right]$

$= \dfrac{4}{3}\left[\left(4\right)\dfrac{3}{2} - \left(0\right)\dfrac{3}{2} - \left\{\left(8 - 7\right)\dfrac{3}{2} - \left(8 - 4\right)\dfrac{3}{2}\right\}\right]$

$= \dfrac{4}{3}\left(8 - 0 - 1 + 8\right)$

$= 20$ sq units.

Hence, the required area is $20$ sq units.