The area bounded by the loop of the curve `4y^2=x^2(4-x^2)` is given by (1) `7/3` (2) `8/3` (3) `11/3` (4) `16/3`

Options

A. `7//3` sq. units
B. `8//3` sq. units
C. `11//3` sq. units
D. `16//3` sq. units

Answer

D

Explanation

`4y^(2)=x^(2)(4-x^(2))" (1)"`
`"or "y=pm(1)/(2)sqrt(x^(2)(4-x^(2)))`
`=pm (x)/(2)sqrt((4-x^(2)))`
`rArr" "y=-(x)/(2)sqrt((4-x^(2))),`
`y=(x)/(2)sqrt((4-x^(2)))`
`y=-(x)/(2)sqrt((4-x^(2)))," "y=(x)/(2)sqrt((4-x^(2)))`

`therefore" Area "(A)=4xxint_(0)^(2)(x)/(2)sqrt((4-x^(2)))dx`
`"Let "4-x^(2)=trArr-2x dx =dt`
`rArr" "A=int_(0)^(4)sqrt(t)dt=[(t^(3//2))/(3//2)]=(2)/(3)x[sqrt(64)-0]`
`=(16)/(3)` sq. units