Prove that the area in the first quadrant enclosed by the axis, the line `x=sqrt(3)y` and the circle `x^2+y^2=4\ i s\ pi//3.`

Explanation

Obviously` x^2 + y^2 = 4` is a
circle having centre at` (0, 0)` and radius` 2` units.
For graph of line x `=sqrt3y`
For intersecting point of given circle and line
putting `x=sqrt3y "in" x^2+y^2=4`
we get
`(sqrt3y)^2+y^2=4`
`3y^2+y^2=4`
`4y^2=4`
`y=pm1`,`y=pm1`
Intersecting points are`(sqrt3,1),(-sqrt3,-1)`
Shaded region is required region
Now required area`= int_0^sqrt3 x/sqrt3dx+int_sqrt3^2 sqrt(4-x^2)`
`=1/sqrt3[x^2/2]_0^sqrt3+[(xsqrt(4-x^2))/2+4/2sin^-1(x/2)]_sqrt3^2`
`=1/(2sqrt3)(3-0)+[2sin^-1(1)-(sqrt3/2+2sin^-1(sqrt3/2)]`
`=sqrt3/2+[2(pi/2)-sqrt3/2-(2pi)/3]`
`=sqrt3/2+pi-sqrt3/2-(2pi)/3`
`=pi-(2pi)/3=pi/3`sq.unit