Let $$A = \left\{(x,y) \in \mathbb{R}^2 : y \geq 0, 2x \leq y \leq \sqrt{4-(x-1)^2}\right\}$$ and
$$B = \left\{(x,y) \in \mathbb{R} \times \mathbb{R} : 0 \leq y \leq \min\left\{2x, \sqrt{4-(x-1)^2}\right\}\right\}$$. Then the ratio of the area of $$A$$ to the area of $$B$$ is

To find the ratio of the area of region $$A$$to the area of region$$B$$, let's analyze the curves that bound these regions.

The two primary curves involved are:

1. A straight line: $$y = 2x$$

2. A curve: $$y = \sqrt{4 - (x-1)^2}$$

Squaring both sides and rearranging, we get $$(x-1)^2 + y^2 = 4$$. This is an equation of a circle with its center at $$(1, 0)$$and a radius of$$r = 2$$. Since $$y \geq 0$$, this represents the upper semi-circle.

Let's find the intersection points of the line and the semi-circle by substituting $$y = 2x$$:

$$(x-1)^2 + (2x)^2 = 4$$

$$x^2 - 2x + 1 + 4x^2 = 4$$

$$5x^2 - 2x - 3 = 0$$

$$(5x + 3)(x - 1) = 0$$

Since we are looking for the intersection in the first quadrant ($$y \ge 0 \implies x \ge 0$$), the intersecting point is at $$x = 1$$. At this point, $$y = 2(1) = 2$$.

Step 1: Area of Region A

Region $$A$$is defined by$$y \geq 0$$and$$2x \leq y \leq \sqrt{4-(x-1)^2}$$.

This region is bounded below by the $$x$$-axis (when $$x < 0$$) and the line $$y = 2x$$(when$$x \ge 0$$), and above by the semi-circle.

For $$x \in [-1, 0]$$, $$2x$$is negative, so the effective lower bound is$$y = 0$$.

For $$x \in [0, 1]$$, the lower bound is $$y = 2x$$.

The area of $$A$$can be found by taking the area under the circle from$$x = -1$$to$$x = 1$$and subtracting the area of the triangle formed by$$y = 2x$$from$$x = 0$$to$$x = 1$$.

The interval from $$x = -1$$to$$x = 1$$ spans from the left edge of the circle to its center. This forms exactly a quarter of the circle.

$$\text{Area of quarter circle} = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2)^2 = \pi$$

The area of the right triangle under $$y=2x$$(base =$$1$$, height = $$2$$) is:

$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1$$

Therefore, the Area of $$A = \pi - 1$$.

Step 2: Area of Region B

Region $$B$$is defined by$$0 \leq y \leq \min\left\{2x, \sqrt{4-(x-1)^2}\right\}$$.

This region lies entirely in the first quadrant and is bounded above by the lower of the two graphs.

For $$x \in [0, 1]$$, the line $$y = 2x$$ is below the circle.

For $$x \in [1, 3]$$, the circle is below the line.

The area of $$B$$is the sum of the area under the line from$$x = 0$$to$$x = 1$$and the area under the circle from$$x = 1$$to$$x = 3$$.

As established above, the area under the line (the triangle) is $$1$$.

The interval from $$x = 1$$to$$x = 3$$ spans from the circle's center to its right edge. This forms another quarter of the circle.

$$\text{Area of quarter circle} = \pi$$

Therefore, the Area of $$B = \pi + 1$$.

Final Answer

The ratio of the area of $$A$$to the area of$$B$$ is:

$$\frac{\pi - 1}{\pi + 1}$$