How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y = 15e−x^2#, y = 0, x = 0, x = 1 revolved about the y-axis?

1 Answer
Dec 8, 2016

# V=2pi [15/2e-1/4] #

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(15e-x^2) dx #
# :. V=2pi int_0^1 (15ex-x^3) dx #
# :. V=2pi [15ex^2/2-x^4/4]_0^1 #
# :. V=2pi [15/2e-1/4] #