Question

18.The area of the circle x,y2-16 exterior to the parabola y-6x is44(A)(47-V3)(B)3(4IN3)(C)3(87-V3)(D)3(8A+3)

Answer 1

The given equation of circle is x 2 + y 2 =16 and equation of parabola is y 2 =6x. The area of the bounded region by the given curves is represented by the shaded region.

The required area becomes,

Area=2[ Area( OADO )+Area( ADBA ) ] =2[ ∫ 1 2 6x dx + ∫ 2 4 16− x 2 dx ] =2[ 6 { x 3 2 3 2 } 0 2 ]+2 [ x 2 16− x 2 + 16 2 sin −1 x 4 ] 2 4 = 4 6 3 ( 2 2 )+2[ 4π− 12 −8× π 6 ]

Simplify further,

Area= 16 3 3 +8π−4 3 − 8 3 π = 4 3 [ 4 3 +6π−3 3 −2π ] = 4 3 [ 3 +4π ]

The area of circle is π r 2 . Since the given equation of circle shows that the radius is 4, so substitute the value r=4 in the area of circle.

Area of circle=π ( 4 ) 2 =16π

So the required area becomes,

Area=16π− 4 3 [ 4π+ 3 ] = 4 3 [ 4( 3π )−4π− 3 ] = 4 3 [ 8π− 3 ] sq. units

Therefore, option (C) is correct.