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Prove that the area in the first quadrant enclosed by the axis, the line x = √3y and the circle x2 + y2 = 4 is π/3.

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To find an area in the first quadrant enclosed by the x – axis,

x = √3y

x2 + y2 = 4

Or \((\sqrt{3y})^2 + y^2 = 4\)

Or 4y2 = 4

Or y = \(\pm 1\)

And x = \(\pm \sqrt{3}\)

Equation (i) represents a line passing through (0, 0), (– √3, – 1), (√3,1).

Equation (ii) represents a circle centre (0,0) and passing through (±2, 0), (0, ±2).

Points of intersection of line and circle are (– √3, – 1) and (√3,1).

These are shown in the graph below: -

Required enclosed area = Region OABO

= Region OCBO + Region ABCA

Hence proved that the area in the first quadrant enclosed by the axis, the line x = \(\sqrt{3}y\) and the circle x2 + y2 = 4 is \(\frac{\pi}{3}.\)

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