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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...

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`

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