Calculus of a Single Variable, Chapter 7, 7.3, Section 7.3, Problem 10

To be able to use the shell method, a rectangular strip from the bounded plane region should be parallel to the axis of revolution. By revolving multiple rectangular strip, it forms infinite numbers of these hollow pipes or representative cylinders.
In this method, we follow the formula: V = int_a^b (length * height * thickness)
or 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 .
For the bounded region, as shown on the attached image, the rectangular strip is parallel to y-axis (axis of rotation). We can let:
r=x
h=f(x) or h=y_(above)-y_(below)
h=8-x^(3/2)
thickness = dx
Boundary values of x from a=0 to b =4 .
Plug-in the values on V = int_a^b 2pi * radius*height*thickness, we get:
V = int_0^4 2pi*x*(8-x^(3/2))*dx.
V = int_0^4 2pi(8x-x^(5/2))dx.
Apply basic integration property: int c*f(x) dx = c int f(x) dx .
V = 2pi int_0^4 (8x-x^(5/2))dx .
Apply basic integration property:int (u-v)dx = int (u)dx-int (v)dx .
V = 2pi [ int_0^4 (8x) dx-int_0^4(x^(5/2))dx]
Apply Power rule for integration: int x^n dx= x^(n+1)/(n+1) .
V = 2pi [8*x^(1+1)/(1+1) -x^((5/2+1))/((5/2+1))]|_0^4
V = 2pi [8*x^2/2 -x^((7/2))/((7/2))]|_0^4
V = 2pi [(8x^2)/2 -x^((7/2))*((2/7))]|_0^4
V = 2pi [4x^2 -(2x^(7/2))/7]|_0^4
Apply the definite integral formula: int _a^b f(x) dx = F(b) - F(a) .
V = 2pi [4(4)^2 -(2(4)^(7/2))/7]-2pi [4(0)^2 -(2(0)^(7/2))/7]
V = 2pi [64- 256/7]-2pi [0-0]
V =2pi [192/7] - 2pi [0]
V = (384pi)/7 or 172.34 (approximated value).