Question

Find the area of the region in the first quadrant enclosed by x-axis,line x=√3y and the circle x²+y²=4

Answer 1

The area of the region bounded by the circle,x2+y2=4,x=√3y,and the x-axis is the

area OAB.

The point of intersection of the line and the circle in the first quadrant is(√3,1).

Area OAB=Area ΔOCA+Area ACB

Area of OAC=1/2×OC×AC=1/2×√3×1=√3/2...(1)

Area of ABC=∫2√3ydx

=∫2√3√4-x2dx

=[x/2√4-x2+4/2sin-1x/2]2√3

=[2×π/2-√3/2√4-3-2sin-1(√3/2)]

=[π-√3π/2-2(/3)]

=[π-√3/2-2π/3]

=[π/3-√3/2]...(2)

Therefore,area enclosed by x-axis,the line x=√3y,and the circle x2+y2=4 in the first

quadrant=√3π/2+/3-3√/2π=/3units.