`y = sqrt(x) , 0<=x<=1` Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

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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,



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`


is the volume

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