Area of Region Bounded by Hyperbola and Parabola

To find the area of the region defined by the set $R = \{(x,y) : xy \le 8, 1 \le y \le x^2\}$, we analyze the boundaries of the region in the first quadrant (where $y \ge 1 > 0$):

1. Curves and Intersections:

   • The lower boundary of the region is given by the line $y = 1$.
   • The upper boundary is bounded by the parabola $y = x^2$ and the hyperbola $y = \frac{8}{x}$.
   • The intersection of $y = x^2$ and $y = 1$ occurs at $x = 1$ (for $x > 0$).
   • The intersection of $y = x^2$ and $y = \frac{8}{x}$ gives $x^3 = 8 \implies x = 2$.
   • The intersection of $y = \frac{8}{x}$ and $y = 1$ occurs at $x = 8$.

2. Splitting the Region into Two Parts:

   • For $x \in [1, 2]$: The region is bounded above by $y = x^2$ and below by $y = 1$.
   • For $x \in [2, 8]$: The region is bounded above by $y = \frac{8}{x}$ and below by $y = 1$.

3. Setting up the Definite Integrals: $\text{Area } A = \int_{1}^{2} (x^2 - 1) \, dx + \int_{2}^{8} \left(\frac{8}{x} - 1\right) \, dx$

4. Evaluating the First Integral: $\int_{1}^{2} (x^2 - 1) \, dx = \left[ \frac{x^3}{3} - x \right]_{1}^{2} = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right) = \frac{2}{3} - \left( -\frac{2}{3} \right) = \frac{4}{3}$

5. Evaluating the Second Integral: $\int_{2}^{8} \left(\frac{8}{x} - 1\right) \, dx = \left[ 8 \log_e x - x \right]_{2}^{8} = (8 \log_e 8 - 8) - (8 \log_e 2 - 2)$ $= 8(\log_e 8 - \log_e 2) - 6 = 8 \log_e \left(\frac{8}{2}\right) - 6 = 8 \log_e 4 - 6 = 16 \log_e 2 - 6$

6. Total Area: $A = \frac{4}{3} + 16 \log_e 2 - 6 = 16 \log_e 2 - \frac{14}{3}$

Thus, the correct option is (A).