Smaller area enclosed by the circle $x^2+y^2=4$ and  the line 

$x+y=2$

The smaller area enclosed by the circle $x^2+y^2=4$ and the line, $x+y=2$ is represented by the shaded area 

$\text{ ACBA as : }$

It can be observed that

Area of  $ACBA=\text{ Area of }OACBO$

-Area of $\mathrm{△}AOB$

$A={\int }_0^2 \sqrt{4-x^2}dx-{\int }_0^2 (2-x)dx$

$\begin{array}{l}={\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}\right]}_0^2\\ =\left[2\times \frac{\pi }{2}\right]-[4-2]\\ =\pi -2\text{ sq. units }\end{array}$