y = -x + 1 Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis.

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.

For the region bounded by y=-x+1 and revolved about the x-axis, we may apply Disk method. For the Disk method, we consider a perpendicular rectangular strip with the axis of revolution. Note: The bounded region is only from the first quadrant.
As shown on the attached image, the thickness of the rectangular strip is "dx " with a vertical orientation perpendicular to the x-axis (axis  of revolution).
We follow the formula for the Disk method:V = int_a^b A(x) dx  where disk's base area is A= pi r^2 with r =y=f(x) .
 Note: r = length of the rectangular strip.
We may apply r = y_(above)-y_(below) .
 Then r =(-x+1)-0 = -x+1
The boundary values of x is a=0 to b=1 .
Plug-in the f(x) and the boundary values to integral formula, we get: 
V = int_0^1 pi (-x+1)^2 dx
Apply basic integration property: intc*f(y) dy = c int f(y) dy .
V =pi int_0^1 (-x+1)^2 dx
To find the indefinite integral, let u = -x+1 then du = -dx or (-1) du =dx
The integral becomes: V =pi int (u)^2*(-1) du
Apply Power rule for integration:  int x^n dx = x^(n+1)/(n+1) .
V =pi * u^(2+1)/(2+1)*(-1)
V =-(piu^3)/3
Plug-in u=-x+1 on V=-(piu^3)/3 , we get:
V=-(pi(-x+1)^3)/3 or (pi(x-1)^3)/3  with boundary values a=0 to b=1
Apply the definite integral formula: int _a^b f(x) dx = F(b) - F(a) .
V =(pi(1-1)^3)/3 -(pi(0-1)^3)/3
V =0 - (-pi)/3
V = pi/3 or 1.05 (approximated value)