How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y =1/(x^2+1)#, x=0, x=1, y=0 revolved about the y-axis?

1 Answer
Feb 21, 2017

Volume #= piln2 #

Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness #deltax# between the #x#-axis and the curve at some particular #x#-coordinate it would have an area:

#delta A ~~"width" xx "height" = ydeltax = f(x)deltax#
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If we then rotated this infinitesimally thin vertical line about #Oy# then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume #delta V# would be given by:

#delta V~~ 2pi xx "radius" xx "thickness" = 2pixdeltaA=2pixf(x)deltax#

If we add up all these infinitesimally thin cylinders then we would get the precise total volume #V# given by:

# V=int_(x=a)^(x=b)2pixf(x) dx #

So for this problem we have:

# V = int_0^1 2pix 1/(x^2+1) dx #
# \ \ \ = pi int_0^1 (2x)/(x^2+1) dx #
# \ \ \ = pi ln|x^2+1)|_0^1 #
# \ \ \ = pi (ln2-ln1)#
# \ \ \ = piln2#