Question

Find the area bounded by \( 0 \le y \le 6 - x \), \( y^2 + 3 \le 4x \) and \( x>0 \):

Answer 1

Correct Answer: 9

Step 1: Understanding the Concept:
The region is bounded by the line \( x + y = 6 \), the parabola \( y^2 = 4x - 3 \), and the x-axis (\( y=0 \)). We find the area by integrating with respect to \( y \).
Step 2: Key Formula or Approach:
1. Intersection points: \( y^2 + 3 = 4(6 - y) \implies y^2 + 4y - 21 = 0 \).
2. Roots: \( (y+7)(y-3) = 0 \). Since \( y \ge 0 \), \( y = 3 \).
3. Area \( = \int_0^3 (x_{right} - x_{left}) dy \).
Step 3: Detailed Explanation:
1. Identify the boundaries: \( x_{right} = 6 - y \) and \( x_{left} = \frac{y^2 + 3}{4} \).
2. Set up the integral: \[ \text{Area} = \int_0^3 \left( 6 - y - \frac{y^2 + 3}{4} \right) dy \] \[ \text{Area} = \int_0^3 \left( \frac{24 - 4y - y^2 - 3}{4} \right) dy = \frac{1}{4} \int_0^3 (21 - 4y - y^2) dy \] 3. Evaluate: \[ \frac{1}{4} \left[ 21y - 2y^2 - \frac{y^3}{3} \right]_0^3 = \frac{1}{4} \left( 63 - 18 - 9 \right) = \frac{36}{4} = 9 \]
Step 4: Final Answer:
The area bounded by the curves is 9 square units.