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

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

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

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Given curve is

`y=x-x^2`

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

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

axis of revolution is x=4

thus

`V=int_0^(1/4)pi((4-1/2-sqrt(1/4-y))^2-9)dy`

`=int_0^(1/4)((9/2-sqrt(1/4-y))^2-9)dy`

`=piint_0^(1/4)(81/4+1/4-y-9sqrt(1/4-y)-9)dy`

`=piint_0^(1/4)(46/4-y-9sqrt(1/4-y))dy`

`=pi((46y)/4-y^2/2+9(2/3)(1/4-y)^(3/2))_0^(1/4)`

`=pi((46/16-1/32)-6(1/8))`

`=pi(23/8-1/32-6/8)`

`=(67pi)/32`  cubic unit

In my answer I forget and get axis of revolution x=0.

So that answer is not correct for this question.