Animated Solution for Mathematics - Definite Integration: The area of the region $R=\{(x,y):xy\le 8,1\le y\le x^2,x\ge 0\}$ is

Before we touch a single integral, we must visualize the region. The constraints $y\le x^2$, $y\le 8/x$, and $y\ge 1$ define an enclosed shape in the first quadrant.

Notice that the "roof" of this shape changes. From the left, the parabola $y=x^2$ acts as the ceiling, but it eventually intersects the hyperbola $y=8/x$. At that point, the hyperbola becomes the new ceiling.

To define our limits, we calculate the intersection points of the boundary curves:

1. The parabola $y=x^2$ meets the line $y=1$ at $x^2=1$, which gives $x=1$. Our region starts at $(1,1)$.

2. The parabola $y=x^2$ meets the hyperbola $y=8/x$ at $x^2=8/x$, leading to $x^3=8$, or $x=2$. At $x=2$, $y=4$. This is our transition point $(2,4)$.

3. The hyperbola $y=8/x$ meets the line $y=1$ at $8/x=1$, which gives $x=8$. Our region ends at $(8,1)$.

Because the upper boundary changes at $x=2$, we must partition the region into two distinct parts:

Part 1 ($A_1$) spans from $x=1$ to $x=2$, where the roof is $y=x^2$ and the floor is $y=1$.

Part 2 ($A_2$) spans from $x=2$ to $x=8$, where the roof is $y=8/x$ and the floor is $y=1$. The total area is $A=A_1+A_2$.

First, we calculate $A_1$:

Evaluating this expression:

Next, we calculate $A_2$:

Evaluating this expression:

Using the property $\ln 8=3\ln 2$, this simplifies to:

We combine our results to find the total area $A$:

Combining the constants:

Thus, the final area is:

Factoring out $\frac{2}{3}$ for an elegant form, we obtain: