Calculus of a Single Variable, Chapter 7, 7.3, Section 7.3, Problem 47

Given ,
y=(x^2)/2 and y=2
A solid is generated by revolving the region bounded by these curves about y-axis.
So ,let us find the volume of the solid generated. It is as follows:
Using the shell method we can find the volume of the solid
V= 2*pi int_a^b p(x)h(x)dx
but we have to find the range of x
so ,
let us find the intersection points
=>
2= x^2/2
=> x^2=4
=>x^2-4=0
so,x=+-2 ,
as only the ranges is between 0 to a positive number,
So, 0<=x<=2
So, volume
= 2*pi int_0^2 (x)(2-(x^2/2))dx
=2*pi int_0^2(2x-(x^3)/2)dx
=2*pi [2x^2/2 -x^4/8]_0^2
=2*pi[[4-2]-[0]]
=4pi
Now given that a hole, centered along the axis of revolution, is drilled through this solid so that one fourth of the volume
so the volume of the hole = 1/4* 4pi = pi
now we have to find the diameter .
as we know that the radius of the hole can be calculated by getting the range of the x with respect to the hole .
now we can find the volume of the hole
it is given as
V= 2*pi int_a^b p(x)h(x) dx
but V_hole = pi

=> pi = 2*pi int_a^b p(x)h(x) dx
let a =0 and b = x_0 so we have to find the radiusx_0 and then the diameter.
=>pi = 2*pi int_0^(x_0) (x)(2-x^2/2) dx
=>pi= 2*pi int_0^(x_0) (2x-x^3/2) dx
=>pi = 2*pi [2x^2/2 - x^4/8]_0^(x_0)
=> 1=2[(x_0)^2 - (x_0)^4/8]
=> 1 = 2(x_0)^2 -(x_0)^4/4
let u= (x_0)^2
=>1=2u - u^2/4
=>4= 8u - u^2
=> u^2-8u +4=0
so u = (-(-8)+-sqrt(64-16))/2
=(8+-sqrt(48))/2
= (8+-4sqrt(3))/2
= 2(2+-sqrt(3))
= 4+-2sqrt(3)
so, u= (x_0)^2
x_0=sqrt(4+-2sqrt(3) )
as x_0<2 if it is beyond the hole is not possibe in the solid
so ,
x_0=sqrt(4-2sqrt(3) ) =0.7320 is the radius of the hole
now the diameter of the hole is 2*(x_0) = 2*(sqrt(4-2sqrt(3) ))=1.464