The area of the region $\{(x, y): x^2 + y^2 \leq 1 \leq x + y\}$ is

The condition $x^2 + y^2 \leq 1 \leq x + y$ means two things simultaneously:

$x^2 + y^2 \leq 1$ → region inside (or on) the unit circle centered at origin

$x + y \geq 1$ → region above (or on) the line $x + y = 1$

The required area is the overlapping region — the part of the circle that lies above the line.

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Substituting $y = 1 - x$ into $x^2 + y^2 = 1$:

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

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

$2x^2 - 2x = 0$

$2x(x - 1) = 0$

$x = 0 \text{ or } x = 1$

The intersection points are $A(0, 1)$ and $B(1, 0)$.

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The required region is a circular segment (the slice between the chord $AB$ and the arc $AB$).

Area of Segment $=$ Area of Sector $-$ Area of Triangle

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The points on the unit circle are:

$B(1, 0)$ → angle $= 0$

$A(0, 1)$ → angle $= \dfrac{\pi}{2}$

The angle subtended at the centre $O(0,0)$:

$\theta = \dfrac{\pi}{2}$

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Area of Sector $OAB$:

$= \dfrac{1}{2} r^2 \theta$

$= \dfrac{1}{2}(1)^2\left(\dfrac{\pi}{2}\right)$

$= \dfrac{\pi}{4}$

Area of Triangle $OAB$ with vertices $O(0,0)$, $A(0,1)$, $B(1,0)$:

$= \dfrac{1}{2} \times 1 \times 1$

$= \dfrac{1}{2}$

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Required Area $= \dfrac{\pi}{4} - \dfrac{1}{2}$

$= \dfrac{\pi - 2}{4}$

$= \dfrac{1}{2}\left(\dfrac{\pi}{2} - 1\right)$ sq. units