The area of the region {(x, y): x^{2}+y^{2} l | vedclass

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Question

(C) The given region is defined by the inequalities $x^{2}+y^{2} \leq 1$ and $x+y \geq 1$.The inequality $x^{2}+y^{2} \leq 1$ represents the interior

Vedclass · Accepted Answer

$\frac{\pi}{4}-\frac{1}{2}$

Vedclass · Answer

(C) The given region is defined by the inequalities $x^{2}+y^{2} \leq 1$ and $x+y \geq 1$.The inequality $x^{2}+y^{2} \leq 1$ represents the interior of a circle with center $(0, 0)$ and radius $1$.The inequality $x+y \geq 1$ represents the region on or above the line $x+y=1$.The intersection points of the circle $x^{2}+y^{2}=1$ and the line $x+y=1$ are $(1, 0)$ and $(0, 1)$.The area of the region is the area of the circular segment bounded by the arc of the circle and the chord connecting $(1, 0)$ and $(0, 1)$.This area is equal to the area of the circular sector (corresponding to the first quadrant arc) minus the area of the right-angled triangle formed by the origin $(0, 0)$, $(1, 0)$, and $(0, 1)$.Area of the circular sector in the first quadrant $= \frac{1}{4} 	imes \pi 	imes (1)^{2} = \frac{\pi}{4}$.Area of the right-angled triangle $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$.Therefore, the required area $= \frac{\pi}{4} - \frac{1}{2}$.

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

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