# Find the volume of the solid obtained by revolving about the line x=4 the region bounded by `y=x-x^2, y=0`

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Student Comments

aruv | Student

Given curve

`y =x-x^2`

`y=x-x^2=-(x^2-x+1/4-1/4)=`

`y-1/4=-(x-1/2)^2`

`(x-1/2)^2=1/4-y`

`x=(1/2)+sqrt(1/4-y)`

y varies from 0 t0 1/4

Thus

`V=int_0^(1/4) pi(1/2+sqrt(1/4-y))^2dy`

`=int_0^(1/4)pi(1/4+1/4-y+sqrt(1/4-y))dy`

`=pi(y/2-y^2/2-2/3(1/4-y)^(3/2))_0^(1/4)`

`=pi(1/8-1/32-0-(0-0-2/3(1/4)^(3/2)))`

`=pi(3/32+1/12)`

`=pi(17/96)`

`=(17pi)/96` cubic unit