What is the volume of the solid generated when S is revolved about the line #y=3# where S is the region enclosed by the graphs of #y=2x# and #y=2x^2# and x is between [0,1]?

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Answer 1 · 2015-02-23T00:15:15

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First let us determine where the two functions intersect

#2x^2=2x #

#2x^2-2x=0 #

#x(2x-2)=0 #

#x=0 # and #2x-2=0# so #x=1 # also

So the interval over which we will integrate is #0<=x<=1 #

I am going to use the method of washers to find the volume

Outer radius is #3-2x^2#

Inner radius is #3-2x #

Integral for volume is

#piint_0^1(3-x^2)^2-(3-2x)^2dx #

#piint_0^1[9-12x^2+4x^4]-[9-12x+4x^2]dx #

#piint_0^1 9-12x^2+4x^4-9+12x-4x^2dx #

#piint_0^1 4x^4-16x^2+12xdx #

#pi[4/5x^5-16/3x^3+6x^2] #

#pi[4/5-16/3+6-0] #

#pi[-68/15+6]=(22pi)/15 #