Key Concepts and Formulas

• Area Between Curves: The area of a region bounded by curves $y=f(x)$ and $y=g(x)$, where $f(x) \ge g(x)$ on the interval $[a,b]$, is given by $\int_a^b (f(x) - g(x)) \, dx$.
• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
• Integration Techniques: Knowledge of basic integration, including trigonometric substitution, is required.

Step-by-Step Solution

Step 1: Understand the Region

We are given two inequalities: $x^2 + (y-2)^2 \leq 4$ and $x^2 \geq 2y$. The first inequality represents the region inside and on the circle with center $(0, 2)$ and radius $2$. The second inequality represents the region below and on the parabola $y = \frac{x^2}{2}$. We need to find the area of the region that satisfies both inequalities.

Step 2: Find the Points of Intersection

To find the points where the circle and parabola intersect, we solve their equations simultaneously: $x^2 + (y-2)^2 = 4$ $x^2 = 2y$ Substituting $x^2 = 2y$ into the circle's equation, we get: $2y + (y-2)^2 = 4$ $2y + y^2 - 4y + 4 = 4$ $y^2 - 2y = 0$ $y(y-2) = 0$ So, $y = 0$ or $y = 2$.

When $y = 0$, $x^2 = 2(0) = 0$, so $x = 0$. When $y = 2$, $x^2 = 2(2) = 4$, so $x = \pm 2$.

The points of intersection are $(0, 0)$, $(2, 2)$, and $(-2, 2)$.

Step 3: Set up the Integral

We want to find the area of the region where the circle is "above" the parabola. To express $y$ in terms of $x$ for the circle, we have: $x^2 + (y-2)^2 = 4$ $(y-2)^2 = 4 - x^2$ $y-2 = \pm \sqrt{4 - x^2}$ $y = 2 \pm \sqrt{4 - x^2}$ Since we are considering the upper part of the circle, we take the positive square root: $y_{circle} = 2 + \sqrt{4 - x^2}$ The parabola is given by $y_{parabola} = \frac{x^2}{2}$. The area of the region is given by the integral: $A = \int_{-2}^{2} \left( (2 + \sqrt{4 - x^2}) - \frac{x^2}{2} \right) dx$

Step 4: Evaluate the Integral

We can split the integral into three parts: $A = \int_{-2}^{2} 2 \, dx + \int_{-2}^{2} \sqrt{4 - x^2} \, dx - \int_{-2}^{2} \frac{x^2}{2} \, dx$

The first integral is: $\int_{-2}^{2} 2 \, dx = 2x \Big|_{-2}^{2} = 2(2) - 2(-2) = 4 + 4 = 8$

The second integral represents the area of a semicircle with radius 2. $\int_{-2}^{2} \sqrt{4 - x^2} \, dx = \frac{1}{2} \pi (2^2) = \frac{1}{2} \pi (4) = 2\pi$

The third integral is: $\int_{-2}^{2} \frac{x^2}{2} \, dx = \frac{1}{2} \int_{-2}^{2} x^2 \, dx = \frac{1}{2} \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{1}{2} \left( \frac{8}{3} - \frac{-8}{3} \right) = \frac{1}{2} \left( \frac{16}{3} \right) = \frac{8}{3}$

Therefore, the area is: $A = 8 + 2\pi - \frac{8}{3} = 2\pi + \frac{24 - 8}{3} = 2\pi + \frac{16}{3}$

Common Mistakes & Tips

• Incorrectly identifying the upper and lower curves: Carefully sketch the region to determine which curve is above the other within the interval of integration.
• Forgetting the limits of integration: The points of intersection determine the limits of integration. Ensure they are correctly identified.
• Difficulty integrating $\sqrt{a^2-x^2}$: Remember this integral represents the area under a circle or semicircle, and can be solved using trigonometric substitution ($x = a\sin\theta$).

Summary

We found the area of the region by identifying the curves, finding their points of intersection, setting up the integral representing the area between the curves, and then evaluating the integral. The area of the region is $2\pi + \frac{16}{3}$.

Final Answer

The final answer is \boxed{2 \pi+\frac{16}{3}}, which corresponds to option (A).