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

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Question

(A) The region is defined by two inequalities: $1$. $x^2 + y^2 \leqslant 1$ (the interior of a circle centered at the origin with radius $1$). $2$. $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The region is defined by two inequalities: $1$. $x^2 + y^2 \leqslant 1$ (the interior of a circle centered at the origin with radius $1$). $2$. $x + y \geqslant 1$ (the region above or on the line $x + y = 1$). The intersection of these two regions is a circular segment. The line $x + y = 1$ passes through $(1, 0)$ and $(0, 1)$. The distance from the origin $(0, 0)$ to the line $x + y - 1 = 0$ is $d = \frac{|0 + 0 - 1|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$. In a circle of radius $r = 1$,the area of a circular segment cut by a chord at distance $d$ from the center is given by $A = r^2 \cos^{-1}(\frac{d}{r}) - d\sqrt{r^2 - d^2}$. Substituting $r = 1$ and $d = \frac{1}{\sqrt{2}}$: $A = (1)^2 \cos^{-1}(\frac{1/\sqrt{2}}{1}) - \frac{1}{\sqrt{2}} \sqrt{1 - (\frac{1}{\sqrt{2}})^2}$ $A = \cos^{-1}(\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{2}}$ $A = \frac{\pi}{4} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{4} - \frac{1}{2}$.

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

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