# 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. 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`

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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]-]`

=`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`

Approved by eNotes Editorial Team