A solid is generated by revolving the region bounded by `y = x^2/2` and `y = 2` about the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one fourth of...

A solid is generated by revolving the region bounded by `y = x^2/2` and `y = 2` about the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one fourth of the volume is removed. Find the diameter of the hole.

Expert Answers
kseddy123 eNotes educator| Certified Educator

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 radius`x_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`