Answer :

To find the area enclosed by

y = √3x ...(i)

x^{2} + y^{2} = 16 ...(ii)

On solving the equation (i) and (ii),

Or x^{2} + = 16

Or 4x^{2} = 16

Or x^{2} = 4

Or x =

∴ Y = ±

Equation (i) represents a parabola with vertex (0,0) and axis as x - axis.

Equation (ii) represent axis a circle with centre (4,0) and meets axes at (0,0) and (4,0).

They intersect at A (2,) and C ( – 2, – ).

These are shown in the graph below: -

Area of the region OAB = Area OAC + Area ACB

The area of the region in the first quadrant enclosed by x - axis, the line y = √3x and the circle x^{2} + y^{2} = 16 is

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