If the area bounded by the curves $x^2+y^2\le 4, x+y\ge 2,$ and $y\ge 1$ is $\frac{2\pi }{K}-\frac{\sqrt{K}}{2}-\frac{1}{2}$ sq. units, then the value of $K$ is equal to

Solution

The required area is $A=\left({\int }_1^2\sqrt{4-y^2}dy\right)-\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{}∆APQ$
$\Rightarrow A={\left[\frac{y\sqrt{4-y^2}}{2}+\frac{4}{2}{\sin }^{-1}\left(\frac{y}{2}\right)\right]}_1^2-\frac{1}{2}\times 1\times 1$
$\Rightarrow A=0+2{\sin }^{-1}\left(1\right)-\frac{\sqrt{3}}{2}-2{\sin }^{-1}\left(\frac{1}{2}\right)-\frac{1}{2}$
$=\pi -\frac{\sqrt{3}}{2}-\frac{\pi }{3}-\frac{1}{2}$
$=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}-\frac{1}{2}$ sq. units

Hence, the correct answer is 3.