`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...

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.