Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 1

For the region bounded by y=-x+1 revolve about the x-axis, we can also apply the Shell Method using a horizontal rectangular strip parallel to the axis of revolution (x-axis).We may follow the formula for Shell Method as:
V = int_a^b 2pi * radius*height*thickness
where:
radius (r)= distance of the rectangular strip to the axis of revolution
height (h) = length of the rectangular strip
thickness = width of the rectangular strip as dx or dy .
As shown on the attached file, the rectangular strip has:
r=y
h =f(y) or h = x_2-x_1
h = (1-y) -0 = 1-y
Note: y =-x+1 can be rearrange into x=1-y .
Thickness = dy
Boundary values of y: a=0 to b =1 .
Plug-in the values on to the formula V = int_a^b 2pi * radius*height*thickness, we get:
V = int_0^1 2pi* y*(1-y)*dy
Apply basic integration property: intc*f(x) dx = c int f(x) dx .
V = 2pi int_0^1 y*(1-y)*dy
Simplify: V = 2pi int_0^1 (y-y^2)dy

Apply basic integration property:int (u-v)dy = int (u)dy-int (v)dy to be able to integrate them separately using Power rule for integration: int y^n dy = y^(n+1)/(n+1) .
V = 2pi *[ int_0^1 (y) dy -int_0^1 (y^2)dy]
V = 2pi *[y^2/2 -y^3/3]|_0^1
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = 2pi *[(1)^2/2 -(1)^3/3] -2pi *[(0)^2/2 -(0)^3/3]
V = 2pi *[1/2 -1/3] -2pi *[0-0]
V = 2pi *[1/6] -2pi *[0]
V = (2pi )/6 -0
V = (2pi )/6 or pi/3
We will get the same result whether we use Disk Method or Shell Method for the given bounded region revolve about the x-axis on this problem.