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  • Find the area of the region in the first quadrant enclosed by x-axis, line x = sqrt 3 y and the circle x^2 + y^2 = 4.

Q : 6        Find the area of the region in the first quadrant enclosed by \small x-axis, line  \small x=\sqrt{3}y and the circle \small x^2+y^2=4
 

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The area of the region bounded by \small x=\sqrt{3}y and  \small x^2+y^2=4 is ABC shown:

The point B of the intersection of the line and the circle in the first quadrant is (\sqrt3,1).

Area ABC = Area ABM + Area BMC where, M is point in x-axis perpendicular drawn from the line.

Now,area of ABM = \frac{1}{2}\times AM\times BM = \frac{1}{2}\times \sqrt{3}\times 1 =\frac{\sqrt3}{2}     ............(1)

and Area of BMC = \int^2_{\sqrt{3}} ydx

= \int^2_{\sqrt3} \sqrt{4-x^2} dx

= \left [ \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{\sqrt3}

= \left [ 2\times\frac{\pi}{2}-\frac{\sqrt3}{2}\sqrt{4-3}-2\sin^{-1}\left ( \frac{\sqrt3}{2} \right ) \right ]

= \left [ \pi - \frac{\sqrt3\pi}{2}-2\frac{\pi}{3} \right ]

= \left [ \pi-\frac{\sqrt3}{2}-\frac{2\pi}{3} \right ]

= \left [ \frac{\pi}{3}-\frac{\sqrt3}{2} \right ]                                              ..................................(2)

then adding the area (1) and (2), we have then

The Area under ABC = \frac{\sqrt3}{2} +\frac{\pi}{3}-\frac{\sqrt3}{2} = \frac{\pi}{3}\ units.

 

 

 

 

 

Posted by

Divya Prakash Singh

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