y = x^3 , x = 0 , y = 8 Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.

We can use a shell method when a bounded region represented by rectangular strip is parallel to the axis of revolution. It forms of infinite number of thin 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
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(x) or h=x_2 - x_1
The x_1 will be based from the boundary line x=0.
The x_2 will be base on the equation y =x^3 rearranged into x= root(3)(y)
h = root(3)(y)-0
h=root(3)(y)
For boundary values, we have y_1=0 to y_2=8 (based from the boundary line).
Plug-in the values on
 V = int_a^b 2pi *radius*height*thickness, , we get:
V =int_0^8 2pi y*root(3)y*dy
Apply basic integration property: intc*f(x) dx = c int f(x) dx.
V = 2pi int_0^8 y* root(3)(y)dy
Apply Law of Exponent: root(n)(y^m)=y^(m/n) then root(n)(y)= y^(1/3)and y^n*y^m = y^(n+m)
V = 2pi int_0^8 y y^(1/3)dy
V = 2pi int_0^8  y^(1/3+1)dy
V = 2pi int_0^8  y^(4/3)dy
Apply power rule for integration: int y^n dy= y^(n+1)/(n+1).
V = 2pi y^(4/3+1)/(4/3+1) |_0^8
V = 2pi y^(7/3)/(7/3) |_0^8
V = 2pi y^(7/3)*(3/7) |_0^8
V = (6pi y^(7/3))/7 |_0^8
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a).
V = (6pi (8)^(7/3))/7 -(6pi (0)^(4/3))/7
V =(768pi)/7-0
V =(768pi)/7  or 344.68 (approximated value).