Correct Answer - A
Given equation of curve is `y=sqrt(16-x^(2))` and the wquation of line is X-axis i.e., y = 0
`:. sqrt(16-x^(2))=0 …(i)`
`rArr 16-x^(2)=0`
`rArr x^(2)=16`
`rArr x=pm4`
So, the intersection points are (4, 0) and (-4, 0)
`:. " Area of curve , " A=int_(-4)^(4)(16-x^(2))^(1//2)dx`
`=[x/2sqrt(4^(2)-x^(2))+4^(2)/2sin^(-1).x/4]_(-4)^(4)`
`=[4/2sqrt(4^(2)-4^(2))+8sin^(-1).4/4]-[-4/2sqrt(4^(2)-(-4)^(2))+8sin^(-1)(-4/4)]`
`[20+8. pi/2-0+8. pi/2]=8pi "sq units"`
