The area bounded by the curve `y=3/|x| and y+|2-x|=2` is

Options

A. `(pi)/(2)`
B. `pi`
C. `2pi`
D. `3pi`

Answer

D

Explanation

`y^(2)(2-x)=x^(2)`
`therefore" "y^(2)=(x^(3))/(2-x)`
`therefore" "y=pm(x^(3//2))/(sqrt(2-x))," clearly "x in [0,2]`
`"Consider "y=(x^(3//2))/(sqrt(2-x))`
When x=0, y=0
Also, when x increases from `'0'" to "'2',y" increases from '0'" to " 'oo'`
Hence graph of given relation is as shown in the following figure :

`"Required Area is "A=2overset(2)underset(0)int(x^(3//2))/(sqrt(2-x))dx`
`"Put "x=2 sin^(2)theta`
`therefore" "A=2overset(pi//2)underset(0)int(2sqrt(2)sin^(3) theta)/(sqrt(2-2 sin^(2) theta))4 sin theta cos theta d""theta`
`=16overset(pi//2)underset(0)intsin^(4)theta d""theta`
`=16overset(pi//2)underset(0)int((1-cos 2theta)^(2))/(4)d""theta`
`=4overset(pi//2)underset(0)int(1-2 cos 2theta+cos^(2)2theta)d""theta`
`=4overset(pi//2)underset(0)int(1-2 cos 2theta +(1+cos 4theta)/(2))d""theta`
`=4(theta+sin 2theta+(theta+(sin 4 theta)/(4))/(2))_(0)^(pi//2)`
`=4((pi)/(2)+0+(pi)/(4))`
`=3pi`