Given `y=sqrt(x)` ,and to find the volume of the solid of rotation about x-axis,

So, this can be solved using the shell method. And is as follows,

`y=sqrt(x)`

=>`x=y^2`

and the `0<=y<=1` these are the limits of y

now in the shell method the volume is given as ,
`V= 2*pi int _c ^d p(y)h(y) dy`

here p(y) is the average radius about x axis =y

and h(y) is the function of height`=y^2` and c=0 and d=1 as the range of y

so, the volume is:

`V=2*pi int _c ^d p(y)h(y) dy`

=`2*pi int _0 ^1 (y)(y^2) dy`

=`2*pi int _0 ^1 (y^3) dy`
=`2*pi [y^4/4] _0 ^1`
`=2*pi[[1/4]-[0]]`

=`pi/2`

is the volume