To find the area of the region defined by the set of points \( (x, y) \) such that \( x^{2} + y^{2} \le 1 \le x + y \), we need to understand the shapes of the boundaries involved.
- The inequality \( x^{2} + y^{2} \le 1 \) describes a circle centered at the origin with radius 1.
- The inequality \( x + y \ge 1 \) describes a line. This line intersects the circle to form a segment on the upper boundary of the circle.
First, let's find the points of intersection between the circle and the line:
- From the equation \( x + y = 1 \), we can express \( y \) as \( y = 1 - x \).
- Substitute \( y = 1 - x \) into the circle's equation:
\(x^2 + (1 - x)^2 = 1\)
Simplifying, we get:
\(x^2 + 1 - 2x + x^2 = 1\)
\(2x^2 - 2x = 0\)
\(2x(x - 1) = 0\) - Thus, the solutions are \( x = 0 \) or \( x = 1 \).
- For \( x = 0 \), \( y = 1 \) and for \( x = 1 \), \( y = 0 \). The intersection points are \( (0, 1) \) and \( (1, 0) \).
The region bounded by the circle (\( x^2 + y^2 = 1 \)) from \( (0, 1) \) to \( (1, 0) \), and above the line \( x + y = 1 \), forms a segment of the circle.
To calculate the area of the region, we find the area of the circular segment and subtract the area of the triangle formed by the intersection points.
- Area of the circular segment:
The angle subtended by the segment at the center of the circle is \( \frac{\pi}{2} \), since it corresponds to a quarter of the circle. The area of a quarter-circle is: \(\frac{\pi \cdot 1^2}{4} = \frac{\pi}{4}\) - Area of the triangle:
The triangle with vertices at \( (0, 1), (1, 0), (0, 0) \) has a base and height of 1 unit each. Its area is: \(\frac{1 \cdot 1}{2} = \frac{1}{2}\) - Area of the region:
Subtracting the area of the triangle from the area of the quarter-circle: \(\frac{\pi}{4} - \frac{1}{2}\)
Therefore, the area of the region is \(\left(\frac{\pi}{4} - \frac{1}{2}\right)\) square units.