Question

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?

Answer 1

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

Explanation:

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.