Question
The area of the region $R = \{(x, y) : xy \leq 8, 1 \leq y \leq x^2, x \geq 0\}$ is
The correct answer is
$\frac{1}{3}(40\log_e(2)+27)$
To solve the problem of finding the area of the region \( R = \{(x, y) : xy \leq 8, 1 \leq y \leq x^2, x \geq 0\} \), we will analyze each boundary condition and set up the double integral accordingly.
- Understand the boundary conditions:
- \(xy \leq 8\): This represents the region below the hyperbola \(y = \frac{8}{x}\).
- \(y \leq x^2\): This implies \(y\) is below the parabola \(y = x^2\).
- \(x \geq 0, 1 \leq y\): These are straightforward conditions restricting \(x\) to non-negative values and \(y\) to being at least 1.
- Find the area between boundaries:
The region is bounded by the curves \(y = x^2\) and \(y = \frac{8}{x}\). We first determine their intersection.
Set \(x^2 = \frac{8}{x}\). Solving this gives:
- \(x^3 = 8 \Rightarrow x = 2\)
Therefore, the intersection points are \((0,0)\) and x from \(0\) to \(2\):
- \(\int_{0}^{2} \left(\frac{8}{x} - x^2\right) dx\)
- Evaluate the integral:
- The integral of \(\frac{8}{x}\) is \(8\log_e|x|\).
- The integral of \(x^2\) is to = \left[ 8\log_e x - \frac{x^3}{3} \right]_{0}^{2}
Calculating this gives:
- \(= \left(8\log_e 2 - \frac{8}{3}\right) - (0 - 0)\)
= to in the lower bound, yielding the second integral:
- .
Thus, the total area of region R is
.
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