Key Concepts and Formulas

• Area Between Curves: The area between two curves $y=f(x)$ and $y=g(x)$, where $f(x) \ge g(x)$ on the interval $[a, b]$, is given by $A = \int_a^b [f(x) - g(x)] \, dx$.
• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. In particular, $x^2 + y^2 - 2ax = 0$ represents a circle with center $(a, 0)$ and radius $a$.
• Integration Formulas: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$.

Step-by-Step Solution

Step 1: Analyze the given inequalities

We are given the inequalities $y^2 \ge 2x$, $x^2 + y^2 \le 4x$, $x \ge 0$, and $y \ge 0$. The first inequality, $y^2 \ge 2x$, represents the region to the left of the parabola $y^2 = 2x$. The second inequality, $x^2 + y^2 \le 4x$, can be rewritten as $x^2 - 4x + y^2 \le 0$, which completing the square, gives $(x-2)^2 + y^2 \le 4$. This represents the region inside the circle with center $(2, 0)$ and radius 2. The conditions $x \ge 0$ and $y \ge 0$ restrict the region to the first quadrant.

Step 2: Find the intersection points

We need to find the intersection points of the parabola $y^2 = 2x$ and the circle $x^2 + y^2 = 4x$. Substituting $y^2 = 2x$ into the circle equation, we get $x^2 + 2x = 4x$, which simplifies to $x^2 - 2x = 0$. Factoring, we have $x(x - 2) = 0$, so $x = 0$ or $x = 2$.

When $x = 0$, $y^2 = 2(0) = 0$, so $y = 0$. When $x = 2$, $y^2 = 2(2) = 4$, so $y = \pm 2$. Since we are in the first quadrant, we take $y = 2$. Thus, the intersection points are $(0, 0)$ and $(2, 2)$.

Step 3: Set up the integral

The area we want to find is bounded by the circle $x^2 + y^2 = 4x$ (or $y = \sqrt{4x - x^2}$) and the parabola $y^2 = 2x$ (or $y = \sqrt{2x}$) in the first quadrant. The x-values range from 0 to 2. Thus, the area A is given by: $A = \int_0^2 (\sqrt{4x - x^2} - \sqrt{2x}) \, dx$

Step 4: Evaluate the integral

We can split the integral into two parts: $A = \int_0^2 \sqrt{4x - x^2} \, dx - \int_0^2 \sqrt{2x} \, dx$

First, let's evaluate $\int_0^2 \sqrt{4x - x^2} \, dx = \int_0^2 \sqrt{4 - (x - 2)^2} \, dx$. Let $x - 2 = 2\sin\theta$, so $dx = 2\cos\theta \, d\theta$. When $x = 0$, $2\sin\theta = -2$, so $\sin\theta = -1$, and $\theta = -\frac{\pi}{2}$. When $x = 2$, $2\sin\theta = 0$, so $\sin\theta = 0$, and $\theta = 0$. Then the integral becomes: $\int_{-\pi/2}^0 \sqrt{4 - 4\sin^2\theta} \cdot 2\cos\theta \, d\theta = \int_{-\pi/2}^0 2\cos\theta \cdot 2\cos\theta \, d\theta = 4 \int_{-\pi/2}^0 \cos^2\theta \, d\theta$ $= 4 \int_{-\pi/2}^0 \frac{1 + \cos(2\theta)}{2} \, d\theta = 2 \int_{-\pi/2}^0 (1 + \cos(2\theta)) \, d\theta = 2\left[\theta + \frac{\sin(2\theta)}{2}\right]_{-\pi/2}^0$ $= 2\left[(0 + 0) - \left(-\frac{\pi}{2} + 0\right)\right] = 2\left(\frac{\pi}{2}\right) = \pi$

Next, let's evaluate $\int_0^2 \sqrt{2x} \, dx = \sqrt{2} \int_0^2 \sqrt{x} \, dx = \sqrt{2} \left[\frac{2}{3}x^{3/2}\right]_0^2 = \sqrt{2} \cdot \frac{2}{3} (2)^{3/2} = \sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{8}{3}$.

Therefore, the area is $A = \pi - \frac{8}{3}$.

Common Mistakes & Tips

• Carefully sketch the region to avoid incorrect upper and lower bounds for the integrals.
• Remember to change the limits of integration when performing a substitution.
• Be careful with the signs when evaluating definite integrals.

Summary

The problem asks for the area of a region bounded by a parabola and a circle in the first quadrant. We first found the intersection points of the curves. Then, we set up a definite integral representing the area between the curves. Finally, we evaluated the integral, which involved a trigonometric substitution for the circle part. The final area is $\pi - \frac{8}{3}$.

Final Answer The final answer is $\boxed{\pi - {8 \over 3}}$, which corresponds to option (D).