Question

Find the area bounded by the circle x² + y² = 16 and the line `sqrt3 y = x` in the first quadrant, using integration.

Answer 1

The area bounded by the circle x² + y² = 16 , x = `sqrt3 y = x` , and the x-axis is the area OAB.

Solving x² + y² = 16 , x = `sqrt3 y = x` we have

`(sqrt3y)^2 + y^2 = 16`

⇒3y² + y² = 16

⇒4y² = 16

⇒y² = 4 

⇒ y = 2 (In the first quadrant, y is positive)

When y = 2, x = `2sqrt3`

So, the point of intersection of the given line and circle in the first quadrant is `(2sqrt3, 2)`

The graph of the given line and cirlce is shown below:

Required area =  Area of the shaded region = Area OABO = Area OCAO + Area ACB

Area OCAO = `1/2 xx 2sqrt3 xx 2 = 2sqrt3` sq units

Area ABC = `int_(2sqrt3)^4 ydx`

= `int_(2sqrt3)^4 sqrt(16 - x^2) dx`

`= [x/2 sqrt(16 - x^2) + 16/2 sin^(-1) x/4]_(2sqrt3)^4`

`=[(0 + 8sin^(-1) 1) - ((2sqrt3)/3 xx 2 + 8 xx sin^(-1) sqrt3/2)]`

`= 8 xx pi/2 - 2sqrt3 - 8 xx pi/3`

= `((4pi)/3 - 2sqrt3)` sq unit

∴ Required area = `((4pi)/3 - 2sqrt3) + 2sqrt3 = (4pi)/3` sq units