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
Explanation:
If you imagine an almost infinitesimally thin vertical line of thickness
#delta A ~~"width" xx "height" = ydeltax = f(x)deltax#
If we then rotated this infinitesimally thin vertical line about
#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=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...# )