Area lying in the first quadrant and bounded by the circle `x^(2)+y^(2)=4` the line `x=sqrt(3)y` and x-axis , is

Options

A. `pi` sq units
B. `(pi)/(2)` sq units
C. `(pi)/(3)` sq units
D. None of these

Answer

C

Explanation

Intersection points of the curves are (0, 0) and `((sqrt(3,1))`
Therefore,
Required area `=int_(0)^(1)(x_(2)-x_(1))dy=int_(0)^(1)(sqrt(4-y^(2))-sqrt(3)y)dy`
`=[(1)/(2)ysqrt(4-y^(2))+(1)/(2)(4)sin^(-1).(y)/(2)-(sqrt(3)y^(2))/(2)]_(0)^(1)`

`=(sqrt(3))/(2)+2sin^(-1)((1)/(2))-(sqrt(3))/(2)-2sin^(-1)0`
`=(pi)/(3)` sq units
Alternate Method
Area `=(theta)/(360^(@))xxpir^(2)=(30)/(360)xxpi(2)^(2)=(pi)/(3)` sq units