Question
Find the area bounded by the circle x2 + y2 = 16 and the line √3y=x in the first quadrant, using integration.
Solution
The area of the region bounded by the circle, x2+y2=16, x=√3y, and the x-axis is the area OAB.
Solving x2+y2=16 and x=√3y, we have
(√3y)2+y2=16
⇒3y2+y2=16
⇒4y2=16
⇒y2=4
⇒y=2
(In the first quadrant, y is positive)
When y = 2, x = 2√3
So, the point of intersection of the given line and circle in the first quadrant is (2√3,2).
The graph of the given line and circle is shown below:
Required area = Area of the shaded region = Area OABO = Area OCAO + Area ACB
Area OCAO = 12×2√3×2=2√3 sq units
