Find the area of the circle `x^2+y^2=16 which is exterior to the parabola y^2=6x` by using integration.

Answer

`(4)/(3)(8pi-sqrt(3))` sq. units

Explanation

The given curves are
`x^(2)+y^(2)=16" …(i)"`
`y^(2)=6x" ….(ii)"`
`"Solving "x^(2)+6x=16`
`"or "(x-2)(x+8)=0`
`therefore" "x=2` (as x=-8 is not possible)
Thus, the point of intersection are `A(2,2sqrt(3)) and A'(2,-2sqrt(3))`.
The graphs of the curves is as shown in the following figure.

From the figure, required area is
`="Area of semicircle"+2overset(2)underset(0)int(sqrt((16-x^(2)))-sqrt(6x))dx`
`=8pi+2[(x)/(2)sqrt(16-x^(2))+(16)/(2)sin^(-1)""(x)/(4)-sqrt(6)(2x^(3//2))/(3)]_(0)^(2)`
`=8pi+2[2sqrt(2)+(4pi)/(3)-sqrt(6)(2cdot2^(3//2))/(3)]`
`=(4)/(3)(8pi-sqrt(3))` sq. units