The area of the region \ (x, y): xy ≤ 8, 1 ≤ y ≤ x ^ 2 \ is

To draw the inequality, let us draw the equation
$xy = 8$ and $y = 1$ and $y = {x}^{2}$
For point of intersection
(i) $xy = 8$ and $y = 1 ⇒ A ( 8,1 )$
(ii) $xy = 8$ and $y = {x}^{2} ⇒ {x}^{3} = 8 ⇒ x = 2 ⇒ B ( 2, 4 )$
(iii) $y = {x}^{2} and$y = 1 ⇒ $C\left(1, 1\right)$ and $D\left(1, 1\right)$
Now $xy \le 8 ⇒$ region which contains origin
$y \ge 1 ⇒$ region above line $y = 1$
$y \le {x}^{2} ⇒$ region outside the parabola
Now required area
Method I:
Using x-axis:
$A = \mathop{∫}_{1}^{2}\left({x}^{2} - 1\right) ⋅ dx + \mathop{∫}_{2}^{8}\left(\dfrac{8}{x} - 1\right) ⋅ dx$
$A = {\left[\dfrac{{x}^{3}}{3} - x\right]}_{1}^{2} + {\left[8\ln x - x\right]}_{2}^{8}$
$A = - \dfrac{14}{3} + 16\ln 2$
Method II:
Using y-axis:
$A = \mathop{∫}_{1}^{4}\left(\dfrac{8}{y} - \sqrt{y}\right) ⋅ dy$
$A = {\left[8\ln y - \dfrac{2}{3}{y}^{\dfrac{3}{2}}\right]}_{1}^{4} ⇒ A = - \dfrac{14}{3} + 16\ln 2$
Note: The question should include bounded area term as in ${2}^{nd}$ quadrant there exist a area which satisfy the inequality and is unbounded.