1. Interpret the region

We need the area of $\{(x,y): xy \le 8,\; 1\le y \le x^2\}.$

This means:

• $y \ge 1$
• $y \le x^2$
• $xy \le 8 \Rightarrow y \le \dfrac{8}{x}$ (for relevant positive $x$ since $y\ge 1$)

So the region lies where $1 \le y \le \min\left(x^2,\frac{8}{x}\right).$

Since $y\ge 1$, we must also have:

• $x^2 \ge 1 \Rightarrow x\le -1$ or $x\ge 1$
• and $\dfrac{8}{x}\ge 1$; with positive $y$, this forces $x>0$, hence $x\le 8$

Also, because $y\le \frac{8}{x}$ with $y\ge 1$, negative $x$ are not possible. Thus the region is for $1\le x\le 8.$

2. Find where the upper curves switch

The two upper bounds are $y=x^2 \quad \text{and} \quad y=\frac{8}{x}.$

Their intersection is given by $x^2=\frac{8}{x} \Rightarrow x^3=8 \Rightarrow x=2.$

So:

• for $1\le x\le 2$, we have $x^2\le \dfrac{8}{x}$, so upper bound is $y=x^2$
• for $2\le x\le 8$, we have $\dfrac{8}{x}\le x^2$, so upper bound is $y=\dfrac{8}{x}$

3. Set up the area integral

Hence area is $A=\int_1^2 (x^2-1)\,dx+\int_2^8\left(\frac{8}{x}-1\right)dx.$

4. Evaluate 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. Evaluate the second integral

$\int_2^8\left(\frac{8}{x}-1\right)dx=\left[8\ln x-x\right]_2^8$ $=(8\ln 8-8)-(8\ln 2-2)$ $=8(\ln 8-\ln 2)-6$ $=8\ln 4-6=16\ln 2-6.$

6. Add both parts

$A=\frac{4}{3}+(16\ln 2-6)=16\ln 2-\frac{14}{3}.$

So the area is $\boxed{16\log_e 2-\frac{14}{3}}.$

7. Compare with options

This matches Option C.