How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=x#, #0<=x<=1# rotated about the x-axis?

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Answer 1 · 2016-05-30T15:05:04

Volume #V=pi/3" "#cubic units

Using the cylindrical shell method. The differential is

#dV=2pi*r*h*dr#

#dV=2*pi*y*(1-x)*dy#

but #x=y#, therefore

#dV=2*pi*y*(1-y)*dy#

our limits for x are #0 rarr 1#
our limits for y are #0 rarr 1#

We solve the volume by integrating both sides with limits #y=0# to #y=1#

#int dV=2*pi*int y*(1-y)*dy#

#V=2*pi*int_0^1 (y-y^2)*dy#

#V=2*pi* [y^2/2-y^3/3]_0^1#

#V=2*pi* [1^2/2-1^3/3-(0^2/2-0^3/3)]#

#V=2*pi* [1/2-1/3-0]#

#V=2*pi* 1/6#

#V=pi/3" "#cubic units

God bless....I hope the explanation is useful.