Area of Region Defined by Inequalities Involving Logarithm

To find the area of the region $R = \{(x, y): xy \le 27, 1 \le y \le x^2\}$, we analyze the boundaries of the region in the first quadrant (since $y \ge 1 > 0$ and $y \le x^2$, which implies $x^2 \ge 1 \implies x \ge 1$ for a bounded region):

1. Upper boundary:

   • $y = x^2$ for smaller values of $x$
   • $y = \frac{27}{x}$ for larger values of $x$

2. Lower boundary:

   • $y = 1$

Now, let's find the points of intersection of these curves:

• Intersection of $y = x^2$ and $y = 1$: $x^2 = 1 \implies x = 1 \quad (\text{since } x > 0)$

• Intersection of $y = x^2$ and $y = \frac{27}{x}$: $x^2 = \frac{27}{x} \implies x^3 = 27 \implies x = 3$

• Intersection of $y = \frac{27}{x}$ and $y = 1$: $\frac{27}{x} = 1 \implies x = 27$

Thus, the region $R$ is split into two sub-regions along the $x$-axis:

1. For $x \in [1, 3]$, $y$ varies from $1$ to $x^2$.
2. For $x \in [3, 27]$, $y$ varies from $1$ to $\frac{27}{x}$.

The total area $A$ of region $R$ is given by the integral: $A = \int_{1}^{3} (x^2 - 1) \, dx + \int_{3}^{27} \left( \frac{27}{x} - 1 \right) \, dx$

Evaluating the first integral $I_1$: $I_1 = \int_{1}^{3} (x^2 - 1) \, dx = \left[ \frac{x^3}{3} - x \right]_{1}^{3}$ $I_1 = \left( \frac{3^3}{3} - 3 \right) - \left( \frac{1^3}{3} - 1 \right) = (9 - 3) - \left( \frac{1}{3} - 1 \right) = 6 - \left( -\frac{2}{3} \right) = \frac{20}{3}$

Evaluating the second integral $I_2$: $I_2 = \int_{3}^{27} \left( \frac{27}{x} - 1 \right) \, dx = \left[ 27 \log_e x - x \right]_{3}^{27}$ $I_2 = (27 \log_e 27 - 27) - (27 \log_e 3 - 3)$

Since $27 = 3^3$, we have $\log_e 27 = 3 \log_e 3$: $I_2 = (81 \log_e 3 - 27) - (27 \log_e 3 - 3)$ $I_2 = (81 \log_e 3 - 27 \log_e 3) - (27 - 3) = 54 \log_e 3 - 24$

Adding $I_1$ and $I_2$ to find the total area $A$: $A = \frac{20}{3} + 54 \log_e 3 - 24$ $A = 54 \log_e 3 + \frac{20 - 72}{3}$ $A = 54 \log_e 3 - \frac{52}{3}$

Thus, the correct option is B.