The area enclosed by the curve `y=sqrt(4-x^2),ygeqsqrt(2)sin((xpi)/(2sqrt(2)))` , and the `x-a xi s` is divided by the `y-a xi s` in the ratio. `(pi^2-8)/(pi^2+8)` (b) `(pi^2-4)/(pi^2+4)` `(pi-4)/(pi-4)` (d) `(2pi^2)/(2pi+pi^2-8)`

Options

A. `(pi^(2)-8)/(pi^(2)+8)`
B. `(pi^(2)-4)/(pi^(2)+4)`
C. `(pi-4)/(pi-4)`
D. `(2pi^(2))/(2pi+pi^(2)-8)`

Answer

D

Explanation

`y=sqrt(4-x^(2)),y=sqrt(2)sin ((xpi)/(2sqrt(2)))" intersect at "x=sqrt(2)`

Area of the left of y-axis is `pi`
Area to the right of y-axis
`=overset(sqrt(2))underset(0)int(sqrt(4-x^(2))-sqrt(2)sin""(xpi)/(2sqrt(2)))dx`
`=((xsqrt(4-x^(2)))/(2)+(4)/(2)sin^(-1)""(x)/(2))_(0)^(sqrt(2))+((4)/(pi)cos""(xpi)/(2sqrt(2)))_(0)^(sqrt(2))`
`=(1+2xx(pi)/(4))+(4)/(pi)(0-1)`
`=1+(pi)/(2)-(4)/(pi)`
`=(2pi-pi^(2)-8)/(2pi)` sq. units
`therefore" Ratio"=(2pi^(2))/(2pi+pi^(2)-8)`.