# `y=x^3/2 , y=0 , x=3` 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 y-axis.

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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= x^3/2-0 =x^3/2`

thickness `= dx`

Boundary values of `x` from `a=0` to `b =3` .

Plug-in the values on `V = int_a^b 2pi` ** radius*height*thickness*, we get:

`V = int_0^3 2pi*x*(x^3/2)*dx`

Simplify: `V = int_0^3 pi*x*(x^3)dx`

Apply Law of Exponent: `x^n*x^m = x^((n+m)).`

`V = int_0^3 pi*x*(x^3)dx`

`V = int_0^3 pi(x^4)dx`

Apply basic integration property: `intc*f(x) dx = c int f(x) dx` and Power rule for integration formula : `int x^n dx =x^(n+1)/(n+1)` .

`V = int_0^3 pi(x^4)dx`

`= pi* int_0^3 (x^4)dx`

` = pi*x^((4+1))/((4+1))|_0^3`

` =(pix^5)/5|_0^3`

Apply the definite integral formula: `int _a^b f(x) dx = F(b) - F(a)` .

`V =(pi(3)^5)/5 -(pi(0)^5)/5`

`V = (243pi)/5 -0`

`V =(243pi)/5 ` or `152.68` (approximated value)