`x^(2)+y^(2)=4` is the equation of a circle whose centre is (0, 0) and radius is 2. On solving the equation of circle to the straight line `x=sqrt(3)y` we obtained the points of intersection `A(sqrt(3), 1)` and `D(-sqrt(3), -1)`.
Now required area
`=` area of OACO + area of CABC
`=int_(0)^(sqrt(3))(x)/sqrt(3)dx+int_(sqrt(3))^(2)sqrt(4-x^(2))dx`
`=[(x^(2))/(2sqrt(3))]_(0)^(sqrt(3))+[(x)/(2)sqrt(4-x^(2))+(4)/(2)"sin"^(-1)(x)/(2)]_(sqrt(3))^(2)`
`=(sqrt(3))/(2) +[(0+2"sin"^(-1)1)-((sqrt(3))/(2)+2"sin"^(-1)(sqrt(3))/(2))]`
`=(sqrt(3))/(2)+2*(pi)/(2)-(sqrt(3))/(2)-2*(pi)/(3)`
`=(pi)/(3)` sq. units.
