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