Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 19

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.