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

We can also find the area by integrating with respect to $x$ by splitting the region. The parabola $y^2 = 4x - 3$ has its vertex at $(3/4, 0)$.
The condition $y^2 \ge 4x - 3$ implies that for $x \ge 3/4$, the $y$-coordinate must be outside the interior of the parabola arms, i.e., $y \ge \sqrt{4x-3}$ or $y \le -\sqrt{4x-3}$. Since $y \ge 0$, we have $y \ge \sqrt{4x-3}$ for $x \ge 3/4$. For $x<3/4$, any $y \ge 0$ satisfies $y^2 \ge 4x-3$ as the right side is negative.

Area = $\int_0^{3/4} (6-x) dx + \int_{3/4}^3 (6-x - \sqrt{4x-3}) dx$
Integrating the first part:
$$I_1 = [6x - \frac{x^2}{2}]_0^{3/4} = 6(\frac{3}{4}) - \frac{1}{2}(\frac{9}{16}) = 4.5 - \frac{9}{32} = \frac{144-9}{32} = \frac{135}{32}$$
Integrating the second part:
$$I_2 = [6x - \frac{x^2}{2}]_{3/4}^3 - \int_{3/4}^3 (4x-3)^{1/2} dx = (18 - 4.5) - \frac{135}{32} - [\frac{2}{3} \cdot \frac{(4x-3)^{3/2}}{4}]_{3/4}^3$$
$$I_2 = 13.5 - \frac{135}{32} - \frac{1}{6} [(12-3)^{3/2} - 0] = 13.5 - \frac{135}{32} - \frac{27}{6} = 13.5 - \frac{135}{32} - 4.5 = 9 - \frac{135}{32}$$
Total Area = $I_1 + I_2 = \frac{135}{32} + 9 - \frac{135}{32} = 9$.