Let [A=(x, y) R^2: y 0,2 x y √4-(x-1)^2] and [ B=(x, y) R R: 0 y 2 x, √4-(x-1)^2 .

Question

Let $$A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$$ and

$$
B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. }
$$.

Then the ratio of the area of A to the area of B is

Accepted Answer

$y^{2}+(x-1)^{2}=4$

shaded portion $=$ circular $(\mathrm{OABC})$

$$
\begin{aligned}
& -\operatorname{Ar}(\Delta \mathrm{OAB}) \\
& =\frac{\pi(4)}{4}-\frac{1}{2}(2)(1) \\
& \mathrm{A}=(\pi-1)
\end{aligned}
$$

Area $B=\operatorname{Ar}(\triangle \mathrm{AOB})+$ Area of arc of circle $(\mathrm{ABC})$

$$
=\frac{1}{2}(1)(2)+\frac{\pi(2)^{2}}{4}=\pi+1
$$

$$
\frac{\mathrm{A}}{\mathrm{B}}=\frac{\pi-1}{\pi+1}
$$

Solutions · 1