The area of the region given by $$\{(x, y) : xy \leq 8, 1 \leq y \leq x^2\}$$ is:

We need to find the area of the region $$\{(x, y) : xy \leq 8, 1 \leq y \leq x^2\}$$.

The boundary curves are $$y = x^2$$, $$y = 1$$, and $$xy = 8$$ (i.e., $$y = 8/x$$). The curves $$y = 1$$ and $$y = x^2$$ intersect when $$x^2 = 1$$, giving $$x = 1$$ (with $$x > 0$$); $$y = x^2$$ and $$y = 8/x$$ meet when $$x^3 = 8$$, so $$x = 2$$; and $$y = 1$$ meets $$y = 8/x$$ when $$1 = 8/x$$, giving $$x = 8$$.

The region is therefore defined by $$y \geq 1$$, $$y \leq x^2$$, and $$y \leq \frac{8}{x}$$. For $$1 \leq x \leq 2$$, one has $$x^2 \leq 8/x$$, so $$y$$ ranges from $$1$$ to $$x^2$$. For $$2 \leq x \leq 8$$, one has $$8/x \leq x^2$$ (in particular at $$x=8$$, $$8/x=1$$ and $$x^2=64$$), so $$y$$ ranges from $$1$$ to $$8/x$$.

Thus the area is given by $$A = \int_1^2 (x^2 - 1)\,dx + \int_2^8 \Bigl(\frac{8}{x} - 1\Bigr)\,dx.$$

Evaluating the first integral gives $$\int_1^2 (x^2 - 1)\,dx = \Bigl[\tfrac{x^3}{3} - x\Bigr]_1^2 = \tfrac{4}{3},$$ and the second integral yields $$\int_2^8 \Bigl(\frac{8}{x} - 1\Bigr)\,dx = \bigl[8\ln x - x\bigr]_2^8 = 16\ln 2 - 6.$$ Therefore $$A = \frac{4}{3} + 16\ln 2 - 6 = 16\ln 2 - \frac{14}{3}.$$

The correct answer is Option B: $$16\log_e 2 - \frac{14}{3}$$.