y =x , y = 0 , y=4, x = 5 Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 5

For the region bounded by y=x  ,y=0  , y=4  and x=5  and revolved about the line x=5 , we may also apply the Shell method. we are to use two sets of vertical rectangular strips parallel to the line x=5 (axis of revolution). In this case, we need two sets of rectangular strip since the upper bound of the rectangular strip before and after x=4 differs. 
We follow the formula: 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, both rectangular strip has:
r=5-x
h= y_(above) - y_(below)
thickness =dx
For the  rectangular strip representing the bounded region  from x=0 to x=4, we may let:
h = x -0 = x
For the  rectangular strip representing the bounded region  from x=4 to x=5 , we may let:
h =4 -0 = 4
Plug-in the values correspondingly, we get:
V = int_0^4 2pi*(5-x)(x) dx +2piint_4^5 (5-x)(4) dx
or
V =2pi int_0^4 (5-x)(x) dx +2piint_4^5 (5-x)(4) dx
 For the first integral, we solve it as:
2pi int_0^4 2pi*(5x-x^2) dx
= 2pi * [ 5x^2/2 -x^3/3]|_0^4
= 2pi * [ (5(4)^2/2 -(4)^3/3) - (5(0)^2/2 -(0)^3/3)]
= 2pi * [ (40 - 64/3) -(0- 0)]
= 2pi * [ 56/3]
= (112pi)/3
 For the second integral, we solve it as:
2pi int_4^5 2pi*(20-4x) dx
= 2pi * [ 20x -4x^2/2]|_4^5
= 2pi * [ 20x -2x^2]|_4^5
= 2pi * [ (20(5) -2(5)^2) - (20(4) -2(4)^2)]
= 2pi * [ (100 - 50) -(80-32)]
= 2pi * [ 50 -48]
= 2pi*[2]
=4pi
Combing the two results, we get:
V=(112pi)/3+4pi
V=(124pi)/3 or 129.85 ( approximated value).
 We will get the same result whether we use Disk Method or Shell Method for the given bounded region on this problem.

For the region bounded by y=x ,y=0 , y=0 and x=5 and revolved about the line x=5 , we may apply Disk method. For the Disk method, we consider a perpendicular rectangular strip with the axis of revolution.
As shown on the attached image, the thickness of the rectangular strip is "dy" with a horizontal orientation perpendicular to the line x=5 (axis  of revolution).
We follow the formula for the Disk method:V = int_a^b A(y) dy  where disk's base area is A= pi r^2 with r =x=f(y) .
Note: r = length of the rectangular strip. We may apply r = x_2-x_1.
Then r = f(y)=5-y
Boundary values of y: a=0 to b=4 .
Plug-in the values in the formula V = int_a^b A(y)dy , we get:
Then the integral will be: 
V =int_0^4pi (5-y)^2dy
Apply basic integration property: intc*f(y) dy = c int f(y) dy.
V =pi int_0^4 (5-y)^2dy
To find the indefinite integral, let u = 5-y then du = -dy or (-1) du =dy
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=5-y on V=-(piu^3)/3 , we get:
V=-(pi(5-y)^3)/3 or (pi(y-5)^3)/3  with boundary values: a=0 to b=4
 
Apply the definite integral formula: int _a^b f(x) dx = F(b) - F(a) .
V =(pi(4-5)^3)/3 -(pi(0-5)^3)/3
V = (-pi)/3 -(-125pi)/3
V =(-pi)/3 +(125pi)/3
V =(124pi)/3 or 129.85 (approximated value)