The area of the region R = \ (x, y): xy ≤ 27, 1 ≤ y ≤ x^2\ is equal...

The given region is bounded by the curves $y = 1$, $y = x^2$, and $xy = 27$ in the first quadrant.

Let us find the points of intersection of these curves:
Intersection of $y = 1$ and $y = x^2$ gives $x = 1$.
Intersection of $y = x^2$ and $xy = 27$ gives $x(x^2) = 27 \Rightarrow x^3 = 27 \Rightarrow x = 3$, so $y = 9$.
Intersection of $y = 1$ and $xy = 27$ gives $x = 27$.

We can find the area by integrating with respect to $y$ from $y = 1$ to $y = 9$. For a given $y$, the value of $x$ ranges from the parabola $x = \sqrt{y}$ to the hyperbola $x = \dfrac{27}{y}$.

The area $A$ is given by:
$A = \int_{1}^{9} \left( \dfrac{27}{y} - \sqrt{y} \right) dy$

Integrating the terms:
$A = \left[ 27 \ln y - \dfrac{2}{3} y^{3/2} \right]_{1}^{9}$

Substituting the limits:
$A = \left( 27 \ln 9 - \dfrac{2}{3} (9)^{3/2} \right) - \left( 27 \ln 1 - \dfrac{2}{3} (1)^{3/2} \right)$

$A = \left( 27 \ln(3^2) - \dfrac{2}{3} (27) \right) - \left( 0 - \dfrac{2}{3} \right)$

$A = 54 \ln 3 - 18 + \dfrac{2}{3}$

$A = 54 \ln 3 - \dfrac{54}{3} + \dfrac{2}{3}$

$A = 54 \ln 3 - \dfrac{52}{3}$

Answer: $54\log_e 3 - \dfrac{52}{3}$