Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 5

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 hollow pipes or representative cylinders.
In this method, we follow the formula: V = int_a^b 2pirhdy
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 x-axis (axis of rotation). We can let:
r=y
h= f(y) or h =x_2-x_1
h=y^(1/5)-y^(1/2)
Note: y = x^5 is expressed as x = y^(1/5) and y = x^2 is expressed as x = y^(1/2) .
thickness = dy
Boundary values of y: a=0 to b=1 .

Plug-in the values on V = int_a^b 2pirhdy
V = int_0^1 2pi*y*(y^(1/5)-y^(1/2)) dy
Apply basic integration property: intc*f(x) dx = c int f(x) dx
V = 2pi int_0^1 y( y^(1/5)-y^(1/2)) dy

Apply Law of Exponent: y^n*x^m = y^((n+m)).
V = 2pi int_0^1 (y^((1/5+1))-y^((1/2+1))) dy
V = 2pi int_0^1 (y^(6/5)-y^(3/2)) dy
Apply basic integration property:int (u-v)dx = int (u)dx-int (v)dx .
V = 2pi [int_0^1 (y^(6/5))dy -int_0^1 (y^(3/2)) dy]
Apply Power rule for integration: int y^n dy= y^(n+1)/(n+1).
V = 2pi [y^((6/5+1))/((6/5+1)) -y^((3/2+1))/((3/2+1))]|_0^1
V = 2pi [y^((11/5))/((11/5)) -y^((5/2))/((5/2))]|_0^1
V = 2pi [y^(11/5)*(5/11) -y^(5/2)*(2/5)]|_0^1
V = 2pi [(5y^(11/5))/11 -(2y^(5/2))/5]|_0^1
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = 2pi [(5(1)^(11/5))/11 -(2(1)^(5/2))/5]-2pi [(5(0)^(11/5))/11 -(2(0)^(5/2))/5]
V = 2pi [5/11 -2/5]-2pi [0 -0]
V= 2pi [3/55]-2pi [0]
V= (6pi)/55-0
V= (6pi)/55 or 0.343 (approximated value)