Question

The area of the region $\left\{\left(x,y\right):x^2+y^2\le 1\le x+y\right\}$ is
(a) $\frac{\mathrm{\pi }}{5}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{2}-\frac{1}{2}$

(d) $\frac{{\mathrm{\pi }}^2}{2}$

Answer 1

Non of the given option is correct.



To find the points of intersection of the line and the circle substitute y = 1 − x in x² + y² = 1,we get A(0, 1) and B(1, 0).

Therefore, the required area of the shaded region,

$$\begin{gathered} A={\int }_0^1\left(y_1-y_2\right)\mathit{d}x\left(\mathrm{Where},y_1=\sqrt{1-x^2}\mathrm{and}{y}_2=1-x\right) \\ ={\int }_0^1\left[\left(\sqrt{1-x^2}\right)-\left(1-x\right)\right]\mathit{d}x \\ =={\int }_0^1\left(\sqrt{1-x^2}-1+x\right)\mathit{d}x \\ ={\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(x\right)-x+\frac{x^2}{2}\right]}_0^1 \\ =\left[\frac{1}{2}\sqrt{1-1^2}+\frac{1}{2}{\sin }^{-1}\left(1\right)-\left(1\right)+\frac{{\left(1\right)}^2}{2}\right]-\left[\frac{\left(0\right)}{2}\sqrt{1-{\left(0\right)}^2}+\frac{1}{2}{\sin }^{-1}\left(0\right)-\left(0\right)+\frac{{\left(0\right)}^2}{2}\right] \\ =\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right)\mathrm{square}\mathrm{units} \end{gathered}$$