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