Area enclosed between the curve `y^2(2a-x)=x^3` and line `x=2a` above X-axis is (a) `pia^2` sq units (b) `(3pia^2)/2` sq units (c) `2pia^2` sq units (d) `3pia^2` sq units

Options

A. `pia^(2)` sq unit
B. `(3pia^(2))/(2)` sq unit
C. `2pia^(2)` sq unit
D. `3pia^(2)` sq unit

Answer

B

Explanation

Given equation of curve is

which is symmetrical about X-axis and passes through origin.
Also, `(x^(3))/(2a-x)lt 0` ffor `xgt2a`
or `x lt0`
So, curve does not lie in xgt2 and xlt0, therefore curve lies wholly on
`0lexle2a`
`therefore`Requried area`=int_(0)^(2a)(x^(3//2))/(sqrt(2a-x))dx`
Put `x=2asin^(2)theta`
`impliesdx=2a2sinthetacosthetad theta`
`therefore` Required area`=int_(0)^(pi//2)8a^(2)sin^(4)thetad theta`
`=9a^(2)[(3)/(4)*(1)/(2)*(pi)/(2)]=(3pia^(2))/(2)` sq unit