Question

The area of the region {(x, y) : 0 ≤ y ≤ 6 - x, y^2 ≥ 4x - 3, x ≥ 0} is:

• A. 8
• B. 9
• C. 12
• D. 15

Hint:

Identify the intersection points of the line and the parabola. Integrating with respect to y is simpler as it avoids dealing with radicals over most of the region.

Answer 1

The Correct Option is B

The region is defined by the following boundaries:
1. $y \ge 0$ and $x \ge 0$ (First quadrant).
2. $y \le 6 - x \implies x + y \le 6$ (Below the straight line).
3. $y^2 \ge 4x - 3 \implies x \le \frac{y^2 + 3}{4}$ (Inside/to the left of the parabola).

First, let's find the intersection of the line $x = 6 - y$ and the parabola $x = \frac{y^2 + 3}{4}$.
$$6 - y = \frac{y^2 + 3}{4} \implies 24 - 4y = y^2 + 3 \implies y^2 + 4y - 21 = 0$$
$$(y+7)(y-3) = 0$$
Since $y \ge 0$, we have $y = 3$. The corresponding $x$ value is $x = 6 - 3 = 3$. Point of intersection is $(3, 3)$.

It is easier to integrate along the $y$-axis. The horizontal width of the region for a fixed $y$ is from $x = 0$ to $x = \min(\frac{y^2+3}{4}, 6-y)$.
For $0 \le y \le 3$, $x$ is bounded by the parabola: $x = \frac{y^2+3}{4}$.
For $3 \le y \le 6$, $x$ is bounded by the line: $x = 6 - y$.

Area $A = \int_0^3 \frac{y^2+3}{4} dy + \int_3^6 (6-y) dy$
$$A_1 = \frac{1}{4} \left[ \frac{y^3}{3} + 3y \right]_0^3 = \frac{1}{4} \left[ \frac{27}{3} + 9 \right] = \frac{18}{4} = 4.5$$
$$A_2 = \left[ 6y - \frac{y^2}{2} \right]_3^6 = (36 - 18) - (18 - 4.5) = 18 - 13.5 = 4.5$$
Total Area = $4.5 + 4.5 = 9$.