Question

The area enclosed between the region given by \( xy \le 27 \) and \( 1 \le y \le x^2 \) is:

• A. \( 54 \ln 3 - \frac{52}{3} \)
• B. \( 52 \ln 3 - \frac{52}{3} \)
• C. \( 54 \ln 2 - \frac{54}{3} \)
• D. \( 52 \ln 2 - \frac{52}{3} \)

Hint:

When a region has multiple "upper" functions, split the integral at the x-coordinate where the functions intersect. Visualizing the boundaries helps prevent integration over the wrong interval.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The region is bounded by the rectangular hyperbola \( y = 27/x \), the parabola \( y = x^2 \), and the horizontal line \( y = 1 \). We need to find the intersection points to set up the definite integral.

Step 2: Key Formula or Approach:
1. Intersection of \( y = x^2 \) and \( y = 1 \): \( x^2 = 1 \implies x = 1 \). 2. Intersection of \( y = x^2 \) and \( xy = 27 \): \( x(x^2) = 27 \implies x^3 = 27 \implies x = 3 \). 3. Intersection of \( xy = 27 \) and \( y = 1 \): \( x(1) = 27 \implies x = 27 \).

Step 3: Detailed Explanation:
1. The upper boundary is \( y = x^2 \) from \( x = 1 \) to \( x = 3 \), and \( y = 27/x \) from \( x = 3 \) to \( x = 27 \). 2. The lower boundary is \( y = 1 \) from \( x = 1 \) to \( x = 27 \). 3. Total Area \( A = \int_{1}^{3} (x^2 - 1) \, dx + \int_{3}^{27} (\frac{27}{x} - 1) \, dx \). 4. Evaluate first part: \( [ \frac{x^3}{3} - x ]_1^3 = (9 - 3) - (\frac{1}{3} - 1) = 6 + \frac{2}{3} = \frac{20}{3} \). 5. Evaluate second part: \( [ 27 \ln x - x ]_3^{27} = (27 \ln 27 - 27) - (27 \ln 3 - 3) \). \( = 27 \ln(3^3) - 27 - 27 \ln 3 + 3 = 81 \ln 3 - 27 \ln 3 - 24 = 54 \ln 3 - 24 \). 6. Sum: \( \frac{20}{3} + 54 \ln 3 - 24 = 54 \ln 3 - \frac{72 - 20}{3} = 54 \ln 3 - \frac{52}{3} \).

Step 4: Final Answer:
The area is \( 54 \ln 3 - \frac{52}{3} \).