calculus
posted by shah .
The area enclosed between the xaxis, the curve y=x(2x) and the ordinates x=1 and x=2 is rotated through 2π radians about xaxis. (
a)Calculate the volume of the solid revolution formed.
(b)Calculate the rotating area.

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