The area of the circle `x^2+y^2=16` exterior to the parabola `y^2=6x` is

Options

A. `(4)/(3)(4pi-sqrt(3))`
B. `(4)/(3)(4pi+sqrt(3))`
C. `(4)/(3)(8pi-sqrt(3))`
D. `(4)/(3)(8pi+sqrt(3))`

Answer

C

Explanation

For the point of intersection of the circle `x^(2)+y^(2)=16` and parabola `y^(2)=6x`

`x^(2)+6x=16`
`impliesx^(2)+6x-16=0`
`implies(x+8)(x-2)=0`
`implies x= -8, 2`
` :. ` Required area
`=` area of circle `-2xx`(area of shaded region)
`=pi(4)^(2)-2[int_(0)^(2)sqrt(6x)+int_(2)^(4)sqrt(16-x^(2))dx]`
`=16pi-2sqrt(6)[x^(3//2)/(3//2)]_(0)^(2)-2[(x)/(2)sqrt(16-x^(2))+(16)/(2)"sin"^(1)(x)/(4)]_(2)^(4)`
`=16pi-(4sqrt(6))/(3)(2^(3//2)-0)-2[0+8"sin"^(-1)(1)-sqrt(16-4)-8"sin"^(-1)(1)/(2)]`
`=16pi-(4sqrt(6)xx2sqrt(2))/(3)-2[8xx(pi)/(2)-sqrt(12)-8xx(pi)/(2)]`
`=16pi-(16sqrt(3))/(3)-(16pi)/(3)+4sqrt(3)`
`=(32pi)/(3)-(4sqrt(3))/(3)`
`=(4)/(3)(8pi-sqrt(3))` sq. units.