Area of the region $$\{(x, y): x^2 + (y-2)^2 \leq 4, x^2 \geq 2y\}$$ is

We need to find the area of the region satisfying both $$x^2 + (y-2)^2 \leq 4$$ and $$x^2 \geq 2y$$.

To begin,

Circle: center $$(0, 2)$$, radius 2. Its equation: $$x^2 + (y-2)^2 = 4$$.

Parabola: $$x^2 = 2y$$ or $$y = \frac{x^2}{2}$$. The region $$x^2 \geq 2y$$ means $$y \leq \frac{x^2}{2}$$.

Next,

Substituting $$x^2 = 2y$$ in the circle equation:

$$2y + (y-2)^2 = 4$$

$$2y + y^2 - 4y + 4 = 4$$

$$y^2 - 2y = 0$$

$$y(y-2) = 0$$

So $$y = 0$$ (giving $$x = 0$$) and $$y = 2$$ (giving $$x = \pm 2$$).

Intersection points: $$(0, 0)$$, $$(2, 2)$$, and $$(-2, 2)$$.

From this,

The region is bounded above by the parabola $$y = \frac{x^2}{2}$$ and below by the lower arc of the circle $$y = 2 - \sqrt{4 - x^2}$$.

$$\text{Area} = \int_{-2}^{2} \left[\frac{x^2}{2} - \left(2 - \sqrt{4 - x^2}\right)\right] dx$$

Since the integrand is an even function:

$$= 2\int_0^2 \left[\frac{x^2}{2} - 2 + \sqrt{4 - x^2}\right] dx$$

Continuing,

$$\int_0^2 \frac{x^2}{2} dx = \frac{1}{2} \cdot \frac{x^3}{3}\bigg|_0^2 = \frac{4}{3}$$

$$\int_0^2 (-2) dx = -4$$

$$\int_0^2 \sqrt{4 - x^2} \, dx = \frac{\pi(2)^2}{4} = \pi$$ (quarter circle area)

$$\text{Area} = 2\left[\frac{4}{3} - 4 + \pi\right] = 2\left[\pi - \frac{8}{3}\right] = 2\pi - \frac{16}{3}$$