+845 
So far in my attempt i have found the normal area of the shaded region then realising i cannot convert it into the volume of revolution (or i don't know how to)
please explain thank you
+129852 We can use the "Ring Method" to solve this
The volume is = volume of top function rotated about the x axis from x = .5 to x = 1 less the volume of the bottom function rotated about the x axis from x = .5 to x = 1
So we have
1 1
pi ∫ (x + 1/x)^2 dx - pi ∫ ( 2x)^2 dx =
.5 .5
1 1
pi ∫ x^2 + 2 + 1 /x^2 dx - pi ∫ 4x^2 dx =
.5 .5
1
pi ∫ 2 + 1/x^2 - 3x^2 dx =
5
1 1 1
pi ( 2x ] - 1 /x ] - x^3 ] ) =
.5 .5 .5
pi * ( [ 2 - 1 ] - [ 1 - 2 ] - [ 1 - .125 ] ) =
pi ( 1 + 1 - 0.875 ) =
[ 1.125 pi ] units^3
