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.