The area of the region $$R = \{(x, y) : xy \leq 27, \; 1 \leq y \leq x^2\}$$ is equal to:

To find the area of the region $$R = \{(x, y) : xy \le 27, 1 \le y \le x^2\}$$, we first need to identify the boundaries and their points of intersection.

The region is bounded by three curves:

1. $$y = x^2$$ (a parabola)
2. $$y = \frac{27}{x}$$ (a rectangular hyperbola)
3. $$y = 1$$ (a horizontal line)
4. Intersection of $$y = x^2$$ and $$y = 1$$:
5. Intersection of $$y = x^2$$ and $$y = \frac{27}{x}$$:
6. Intersection of $$y = \frac{27}{x}$$ and $$y = 1$$:
7. From $$x = 1$$ to $$x = 3$$: The upper curve is $$y = x^2$$ and the lower curve is $$y = 1$$.
8. From $$x = 3$$ to $$x = 27$$: The upper curve is $$y = \frac{27}{x}$$ and the lower curve is $$y = 1$$.

$$x^2 = 1 \implies x = 1$$ (since we are in the region where $$x > 0$$ to satisfy $$xy \le 27$$).

Point: $$(1, 1)$$

$$x^2 = \frac{27}{x} \implies x^3 = 27 \implies x = 3$$.

If $$x = 3$$, $$y = 3^2 = 9$$.

Point: $$(3, 9)$$

$$1 = \frac{27}{x} \implies x = 27$$.

Point: $$(27, 1)$$

The area is bounded above by two different functions as $$x$$ increases from $$1$$ to $$27$$. We split the integral at $$x = 3$$:

$$\text{Area} = \int_{1}^{3} (x^2 - 1) \, dx + \int_{3}^{27} \left( \frac{27}{x} - 1 \right) \, dx$$

Part 1:

$$\int_{1}^{3} (x^2 - 1) \, dx = \left[ \frac{x^3}{3} - x \right]_{1}^{3}$$

$$= \left( \frac{27}{3} - 3 \right) - \left( \frac{1}{3} - 1 \right) = (9 - 3) - \left( -\frac{2}{3} \right) = 6 + \frac{2}{3} = \frac{20}{3}$$

Part 2:

$$\int_{3}^{27} \left( \frac{27}{x} - 1 \right) \, dx = [27 \ln(x) - x]_{3}^{27}$$

$$= (27 \ln 27 - 27) - (27 \ln 3 - 3)$$

$$= 27 \ln(3^3) - 27 - 27 \ln 3 + 3$$

$$= 81 \ln 3 - 27 \ln 3 - 24 = 54 \ln 3 - 24$$

$$\text{Total Area} = \frac{20}{3} + 54 \ln 3 - 24$$

$$\text{Total Area} = 54 \ln 3 + \left( \frac{20 - 72}{3} \right)$$

$$\text{Total Area} = 54 \log_e 3 - \frac{52}{3}$$

Correct Option:

The correct answer is B: $$54 \log_e 3 - \frac{52}{3}$$.