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?

1 Answer

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.