Posted by fairuz on Sunday, May 27, 2012 at 9:27am.
The curve intersects x=2 at (2,1/5)
The y-intercept is at (0,1)
Using discs,
v = ∫π(2^2)(1/5) dy [0,1/5] + ∫π((1-y/2y)^2 (y-1/5) dy [1/5,1]
4π/5 + π/4 (y^2-1-2ylny)/4y [1/5,1]
= π/2 (4-ln5)
Using shells,
v = ∫2πx(1/(1+2x)) dx [0,2]
= π/2 ((2x+1)-ln(2x+1)) [0,2]
= π/2 (4-ln5)
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