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 are asked to find the area of a region R in the first quadrant bounded by the curves $y=1$ (a horizontal line), $y=x^2$ (a parabola), and $xy=27$ (a hyperbola). The first step is to visualize this region by finding the points where these curves intersect.
Step 2: Key Formula or Approach:
1. Find Intersection Points: Determine the coordinates where the boundary curves meet. This will define the limits of integration.
2. Choose Integration Variable: We can find the area by integrating with respect to $x$ or $y$. We should choose the variable that makes the calculation simpler. Integrating with respect to $y$ using the formula Area $= \int_{c}^{d} (x_{right} - x_{left}) dy$ is often easier if the left and right boundary curves are consistent over the range of $y$.
Step 3: Detailed Explanation:
Part A: Finding Intersection Points
- Intersection of $y=x^2$ and $y=1$: $x^2=1 \implies x=1$. Point: (1, 1).
- Intersection of $xy=27$ and $y=1$: $x(1)=27 \implies x=27$. Point: (27, 1).
- Intersection of $y=x^2$ and $xy=27$: Substitute $y=x^2$ into the second equation: $x(x^2)=27 \implies x^3=27 \implies x=3$. The corresponding y-value is $y=3^2=9$. Point: (3, 9).
Part B: Setting up the Integral
A sketch of the region reveals that integrating with respect to $y$ is more straightforward. The region extends from a lower bound of $y=1$ to an upper bound of $y=9$.
For any $y$ in this range, we need to identify the right and left boundaries in terms of $x$.
- The left boundary is the parabola $y=x^2$, which gives $x_{left} = \sqrt{y}$.
- The right boundary is the hyperbola $xy=27$, which gives $x_{right} = 27/y$.
The area $A$ is given by the integral of the horizontal width ($x_{right} - x_{left}$) over the vertical range of $y$.
\[ A = \int_{1}^{9} (x_{right} - x_{left}) dy = \int_{1}^{9} \left(\frac{27}{y} - \sqrt{y}\right) dy \] Part C: Evaluating the Integral
\[ A = \int_{1}^{9} \left(27y^{-1} - y^{1/2}\right) dy \] \[ A = \left[ 27\ln|y| - \frac{y^{3/2}}{3/2} \right]_{1}^{9} = \left[ 27\ln(y) - \frac{2}{3}y\sqrt{y} \right]_{1}^{9} \] Now, evaluate at the limits:
- At $y=9$: $27\ln(9) - \frac{2}{3}(9)\sqrt{9} = 27(2\ln 3) - \frac{2}{3}(27) = 54\ln 3 - 18$.
- At $y=1$: $27\ln(1) - \frac{2}{3}(1)\sqrt{1} = 0 - \frac{2}{3} = -\frac{2}{3}$.
The area is the difference between the values at the upper and lower limits:
\[ A = (54\ln 3 - 18) - \left(-\frac{2}{3}\right) = 54\ln 3 - 18 + \frac{2}{3} \] \[ A = 54\ln 3 - \frac{54}{3} + \frac{2}{3} = 54\ln 3 - \frac{52}{3} \] Step 4: Final Answer:
The area of the region is $54\log_e 3 - \frac{52}{3}$.