Question

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

Answer 1

The given circle is x2 + y2 = 4 ...(1)

Its centre is (0,0) and radius is 2

Given line is x = √3y ...(2)

solving (1) & (2) to get point of intersection

(√3y)2 + y2 = 4

4y2 = 4

y2 = 1

y = 1, -1 but y ≥ 0 for first quadrant.

for y = 1, x = √3.1 = √3

So, point of intersection is (√3,1)

Now, the require area = area of shaded region

= area(OAMA) + area(MANM)

= (1/√3)((3/2) - 0) + (1/2)(0 - √3) + 2[sin-1 - sin-1 (√3/2)]

= (√3/2) - (√3/2) + 2((π/2) - (π/3)) = π/3 sq. unit