Prove that the area in the first quadrant enclosed by the axis, the line x = √3y and the circle x2 + y2 = 4 is π/3.
To find an area in the first quadrant enclosed by the x – axis,
x = √3y
x2 + y2 = 4
Or
Or
Or
And
Equation (i) represents a line passing through (0,0), ( – √3, – 1), (√3,1).
Equation (ii) represents a circle centre (0,0) and passing through (±2,0), (0,±2).
Points of intersection of line and circle are ( – √3, – 1) and (√3,1).
These are shown in the graph below: -
Required enclosed area = Region OABO
= Region OCBO + Region ABCA
Hence proved that the area in the first quadrant enclosed by the axis, the line and the circle
is π/3.