Find Area Bounded by Parabola and Lines

To find the area of the region $R = \{(x, y) : 0 \le y \le 6 - x, \, y^2 \ge 4x - 3, \, x \ge 0\}$, we can determine the boundaries and set up the integrals.

Step 1: Identify the Boundary Curves and Inequalities

1. $x \ge 0$ and $y \ge 0$ places the region in the first quadrant.
2. $y \le 6 - x \implies x + y \le 6$ (a straight line).
3. $y^2 \ge 4x - 3 \implies x \le \frac{y^2 + 3}{4}$ (a parabola opening to the right with vertex at $(3/4, 0)$).

Step 2: Find the Intersection Points

To find the intersection of the line $x = 6 - y$ and the parabola $x = \frac{y^2 + 3}{4}$, we equate the two expressions for $x$: $6 - y = \frac{y^2 + 3}{4}$

Multiply both sides by 4: $24 - 4y = y^2 + 3$ $y^2 + 4y - 21 = 0$

Factoring the quadratic equation: $(y + 7)(y - 3) = 0$

Since $y \ge 0$, we get $y = 3$.
Substituting $y = 3$ into $x = 6 - y$, we get $x = 3$. Thus, the intersection point in the first quadrant is $(3, 3)$.

Step 3: Integrate with Respect to $y$

It is simpler to integrate with respect to $y$. The limits for $y$ range from $0$ to $6$ (since $x \ge 0$ and $y \le 6 - x \implies y \le 6$).

The upper limit for $x$ as a function of $y$ is given by:

• For $y \in [0, 3]$: $x$ goes from $0$ to $\frac{y^2 + 3}{4}$.
• For $y \in [3, 6]$: $x$ goes from $0$ to $6 - y$.

Thus, the total area $A$ can be split into two integrals $A = A_1 + A_2$:

$A_1 = \int_{0}^{3} \left( \frac{y^2 + 3}{4} \right) dy$

$A_2 = \int_{3}^{6} (6 - y) dy$

Step 4: Evaluate the Integrals

Calculating $A_1$: $A_1 = \frac{1}{4} \left[ \frac{y^3}{3} + 3y \right]_{0}^{3} = \frac{1}{4} \left( \frac{27}{3} + 3(3) \right) = \frac{1}{4} (9 + 9) = \frac{18}{4} = \frac{9}{2}$

Calculating $A_2$: $A_2 = \left[ 6y - \frac{y^2}{2} \right]_{3}^{6} = \left( 36 - \frac{36}{2} \right) - \left( 18 - \frac{9}{2} \right) = 18 - \frac{27}{2} = \frac{9}{2}$

Step 5: Sum the Areas

$A = A_1 + A_2 = \frac{9}{2} + \frac{9}{2} = 9$

Thus, the area of the given region is 9.

Correct Answer: B