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)

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