Question

Area bounded by \(\{(x, y) : xy \le 8 ; y \le x^2, y \ge 1\}\) is :

• A. \(16 \log_e 2 - \frac{14}{3}\)
• B. \(16 \log_e 2 - \frac{13}{3}\)
• C. \(16 \log_e 2 - \frac{17}{3}\)
• D. \(16 \log_e 2 - \frac{19}{3}\)

Hint:

Sketching the region helps identify when the upper curve changes, which indicates where the integration interval needs to be split.

Answer 1

The Correct Option is A

To find the area bounded by the given conditions \(xy \le 8\), \(y \le x^2\), and \(y \ge 1\), we need to determine the intersections of these curves and integrate accordingly. Let's break this down step-by-step:

1.  Identify the Curves:
   • The region \(xy \le 8\) represents an area under the hyperbola \(xy = 8\).
   • The region \(y \le x^2\) represents an area below the parabola \(y = x^2\).
   • The region \(y \ge 1\) starts at \(y = 1\), a horizontal line.
2. Find the Intersection Points:
   • Intersect \(xy = 8\) and \(y = x^2\):
   • From \(y = x^2\), substitute in the hyperbola equation:
   • \(x(x^2) = 8 \Rightarrow x^3 = 8 \Rightarrow x = 2\).
   • Thus, \(y = x^2\) gives \(y = 4\).
   • Intersect \(y = x^2\) and \(y = 1\):
   • Substitute \(y = 1\) in \(y = x^2\):
   • \(x^2 = 1 \Rightarrow x = \pm 1\).
3. Set Up the Integrals:
   • For the region above \(y = 1\) and at/below \(y = x^2\), between \(x = -1\) and \(x = 2\), we calculate:
   • \(\int_{1}^{4} x^2 \, dy - \int_{1}^{4} \frac{8}{y} \, dy\).
4. Calculate the Integrals:
   • Find \(\int_{1}^{4} x^2 \, dy = \left[x^3/3\right]_{x=1}^{x=2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\).
   • Find \(\int_{1}^{4} \frac{8}{y} \, dy = 8[\log_e(y)]_{1}^{4} = 8 (\log_e 4 - \log_e 1) = 8 \log_e 4\).
   • Substituting values: \(8 \log_e 4 - \frac{7}{3}\).
5. Recognize \(\log_e 4 = 2 \log_e 2\):
6. Thus, we have \(16 \log_e 2 - \frac{7}{3}\).

Conclusion: The total area bounded by the given conditions is \(16 \log_e 2 - \frac{14}{3}\). Therefore, the correct answer is \(16 \log_e 2 - \frac{14}{3}\).