Question

Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \] 

• A. $78\log_e 3 - \frac{52}{3}$
• B. $54\log_e 3 - \frac{52}{3}$
• C. $54\log_e 3 - \frac{26}{3}$
• D. $54\log_e 3 + \frac{26}{3}$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the area of a region in the xy-plane defined by a set of three inequalities: 
1. $xy \le 27$ 
2. $y \ge 1$ 
3. $y \le x^2$ 
This involves sketching the region bounded by the curves $y=1$, $y=x^2$, and $xy=27$, and then setting up a definite integral to calculate its area. 
Step 2: Key Formula or Approach: 
1. First, we identify the boundary curves and find their points of intersection. The curves are $y=1$, $y=x^2$ (a parabola), and $y=27/x$ (a hyperbola). The conditions imply the region is in the first quadrant, so we consider $x>0$. 
2. We can calculate the area by integrating either with respect to $x$ or with respect to $y$. Let's analyze both approaches. 
- Integrating with respect to $x$: The upper boundary of the region changes at an intersection point, requiring the sum of two separate integrals. 
- Integrating with respect to $y$: The left and right boundaries are consistent throughout the region, requiring only a single integral. This approach is simpler. The formula is Area $= \int_{c}^{d} (x_{right} - x_{left}) dy$. 
Step 3: Detailed Explanation: 
Finding Intersection Points: 
- Intersection of $y=x^2$ and $y=1$: $x^2=1 \implies x=1$ (since $x>0$). Point: (1, 1). 
- Intersection of $y=27/x$ and $y=1$: $1=27/x \implies x=27$. Point: (27, 1). 
- Intersection of $y=x^2$ and $y=27/x$: $x^2=27/x \implies x^3=27 \implies x=3$. Then $y=3^2=9$. Point: (3, 9). 
Setting up the Integral with respect to y: 
The region is defined by $1 \le y \le x^2$ and $xy \le 27$. We can rewrite these in terms of $x$: 
From $y \le x^2$, we get $x \ge \sqrt{y}$ (since $x>0$). This is the left boundary, $x_{left} = \sqrt{y}$. 
From $xy \le 27$, we get $x \le 27/y$. This is the right boundary, $x_{right} = 27/y$. 
The condition $x_{left} \le x_{right}$ must hold, so $\sqrt{y} \le 27/y \implies y^{3/2} \le 27 \implies y \le 9$. 
The condition $y \ge 1$ gives the lower bound for integration. 
Thus, the integration limits for $y$ are from 1 to 9. 
The area $A$ is given by: 
\[ A = \int_{1}^{9} (x_{right} - x_{left}) dy = \int_{1}^{9} \left(\frac{27}{y} - \sqrt{y}\right) dy \] Now, we perform the integration: 
\[ A = \int_{1}^{9} \left(27y^{-1} - y^{1/2}\right) dy \] \[ A = \left[ 27\ln|y| - \frac{y^{3/2}}{3/2} \right]_{1}^{9} = \left[ 27\ln(y) - \frac{2}{3}y^{3/2} \right]_{1}^{9} \] Evaluate at the upper limit ($y=9$): 
\[ 27\ln(9) - \frac{2}{3}(9)^{3/2} = 27\ln(3^2) - \frac{2}{3}(27) = 54\ln(3) - 18 \] Evaluate at the lower limit ($y=1$): 
\[ 27\ln(1) - \frac{2}{3}(1)^{3/2} = 0 - \frac{2}{3} = -\frac{2}{3} \] The area is the difference: 
\[ A = (54\ln 3 - 18) - \left(-\frac{2}{3}\right) = 54\ln 3 - 18 + \frac{2}{3} \] \[ A = 54\ln 3 - \frac{54}{3} + \frac{2}{3} = 54\ln 3 - \frac{52}{3} \] Verification (Integrating w.r.t. x): 
Area = $\int_1^3 (x^2-1)dx + \int_3^{27} (\frac{27}{x}-1)dx$ 
$= \left[\frac{x^3}{3}-x\right]_1^3 + \left[27\ln x - x\right]_3^{27}$ 
$= \left( (9-3) - (\frac{1}{3}-1) \right) + \left( (27\ln 27 - 27) - (27\ln 3 - 3) \right)$ 
$= \left( 6 - (-\frac{2}{3}) \right) + \left( 81\ln 3 - 27 - 27\ln 3 + 3 \right)$ 
$= \frac{20}{3} + 54\ln 3 - 24 = 54\ln 3 - \frac{52}{3}$. 
Both methods yield the same result. 
Step 4: Final Answer: 
The area of the region is $54\log_e 3 - \frac{52}{3}$.