How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y=6sin(5x^2)#, between #0# and #sqrt(pi/5)# revolved about #Ox#?

1 Answer
Steve M
Dec 10, 2016

# 2.4pi# (#7.54# 2 dp)

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^sqrt(pi/5) 2pix(6sin(5x^2)) dx #
# :. V = 12piint_0^sqrt(pi/5) xsin(5x^2) dx #
# :. V = 12pi [-cos(5x^2)/10]_0^sqrt(pi/5) #
# :. V = (-12pi)/10 [cos(5x^2)]_0^sqrt(pi/5) #
# :. V = (-12pi)/10 (cospi-cos0) #
# :. V = (-12pi)/10 (-1-1) #
# :. V = (24pi)/10#
# :. V = 2.4pi# (#~~7.5398...#)

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