1. Interpret the region

We need the area of

$$\{(x,y): xy\le 8,\; 1\le y\le x^2\}.$$

The conditions mean:

• $xy\le 8 \Rightarrow y\le \dfrac{8}{x}$ if $x>0$.
• Also $1\le y\le x^2$.

Since $y\ge 1>0$, we must have $x>0$ for the inequality $xy\le 8$ to give a bounded upper condition on $y$.

So the region lies in the first quadrant and satisfies

$$1\le y\le \min\left(x^2,\frac{8}{x}\right).$$

For the region to exist, we need

$$\min\left(x^2,\frac{8}{x}\right)\ge 1.$$

2. Find intersection of the curves

The upper boundary changes where

$$x^2 = \frac{8}{x}$$

which gives

$$x^3=8 \Rightarrow x=2.$$

Then the corresponding $y$-value is

$$y=2^2=4.$$

So the curves $y=x^2$ and $y=\dfrac{8}{x}$ intersect at $(2,4)$.

3. Determine the relevant $x$-intervals

From $1\le y\le x^2$, we need

$$x^2\ge 1 \Rightarrow x\ge 1$$

(since $x>0$).

From $1\le y\le \dfrac{8}{x}$, we need

$$\frac{8}{x}\ge 1 \Rightarrow x\le 8.$$

Hence the region exists for

$$1\le x\le 8.$$

Now compare the two upper bounds:

• 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 $y=\dfrac{8}{x}$.

Thus the area is

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

4. Evaluate the first integral

$$\int_1^2 (x^2-1)\,dx = \left[\frac{x^3}{3}-x\right]_1^2 = \left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-1\right).$$

Simplify:

$$\frac{8}{3}-2 = \frac{2}{3}, \qquad \frac{1}{3}-1 = -\frac{2}{3}.$$

So,

$$\int_1^2 (x^2-1)\,dx = \frac{2}{3}-\left(-\frac{2}{3}\right)=\frac{4}{3}.$$

5. Evaluate the second integral

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

Since $\ln 4 = 2\ln 2$,

$$8\ln 4 = 16\ln 2.$$

Therefore,

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

6. Add both parts

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

7. Match with options

This is exactly

$$16\log_e 2 - \frac{14}{3},$$

which corresponds to Option A.