Find the volume of the solid obtained by rotating the region under the graph of the function f(x) = (2)/(x+1) about the x-axis over the interval [0,3]

Answer 1

using discs (washers),

v = ∫[0,3] π(R^2-r^2) dx
where r = 1/2 and R = y = 2/(x+1)
v = π∫[0,3] (4/(x+1)^2 - 1/4) dx
= π (-4/(x+1) - x/4) [0,3]
= π ((-4/4 - 3/4) - (-4/1))
= 9π/4

Using shells,
v = ∫[1/2,2] 2πrh dy
where r=y and h = x = (2/y - 1)
v = 2π∫[1/2,2] y(2/y - 1) dy
= 2π (2y - 1/2 y^2)[1/2,2]
= 9π/4