Question

Let $A = \{(x, y): |x| + |y| \geq 2, x^2 + y^2 \leq 4\}$. Then the area of the region $A$ is (in square units)

Answer 1

4π - 8

Answer 2

The region $A$ is defined by two conditions:

1. $x^2 + y^2 \leq 4$: This inequality represents the set of all points $(x,y)$ inside or on the circle centered at the origin $(0,0)$ with radius $r=2$. The area of this circular region is $Area_{circle} = \pi r^2 = \pi (2^2) = 4\pi$ square units.

2. $|x| + |y| \geq 2$: This inequality represents the set of all points $(x,y)$ outside or on the boundary of the square defined by $|x| + |y| = 2$. Let's first consider the region defined by $|x| + |y| \leq 2$. This inequality represents a square rotated by 45 degrees. Its vertices are found by setting one variable to zero:

   • If $x=0$, then $|y|=2 \Rightarrow y = \pm 2$. So, $(0,2)$ and $(0,-2)$ are vertices.

   • If $y=0$, then $|x|=2 \Rightarrow x = \pm 2$. So, $(2,0)$ and $(-2,0)$ are vertices.

   The vertices of this square are $(2,0), (0,2), (-2,0), (0,-2)$. This square has diagonals along the x and y axes, each of length $2 - (-2) = 4$. The area of a square with diagonal length $d$ is $d^2/2$. So, the area of the square region defined by $|x| + |y| \leq 2$ is $Area_{square} = \frac{4^2}{2} = \frac{16}{2} = 8$ square units.

Now, let's analyze the relationship between these two regions. The vertices of the square $(2,0), (0,2), (-2,0), (0,-2)$ all lie on the circle $x^2 + y^2 = 2^2 = 4$. For example, for $(2,0)$, $2^2 + 0^2 = 4$. This implies that the square defined by $|x| + |y| \leq 2$ is inscribed within the circle $x^2 + y^2 \leq 4$.

The region $A$ is defined by points $(x,y)$ that are:

• Inside or on the circle ($x^2 + y^2 \leq 4$)
• AND outside or on the square ($|x| + |y| \geq 2$)

Since the square is entirely contained within the circle, the area of region $A$ is the area of the circle minus the area of the square.

Area of region $A = Area_{circle} - Area_{square}$

Area of region $A = 4\pi - 8$ square units.