Posted by **shah** on Sunday, May 27, 2012 at 6:40am.

The area enclosed between the x-axis, the curve y=x(2-x) and the ordinates x=1 and x=2 is rotated through 2π radians about x-axis. (

a)Calculate the volume of the solid revolution formed.

(b)Calculate the rotating area.

- calculus -
**Reiny**, Sunday, May 27, 2012 at 8:14am
I will find the area first

( I read that as the area to be rotated, not the surface area of the rotated solid)

b) Area = ∫(2x - x^2) dx from 1 to 2

= [x^2 - (1/3)x^3] from 1 to 2

= (4 - (1/3)(8) ) - (1 - 1/3)

= 2/3

a) we need y^2

y^2 = x^2(4 - 4x + x^2) = 4x^2 - 4x^3 + x^4

volume = π∫y^2 dx from 1 to 2

= π∫(4x^2 - 4x^3 + x^4) dx from 1 to 2

= π[(4/3)x^3 - x^4 + (1/5)x^5] from 1 to 2

= π( (4/3)(8) - 16 + (1/5)(32) ) - (4/3 - 1 + 1/5) )

= π( 16/15 - 8/15)

= 8π/15

check my arithmetic

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