We can use a shell method when a bounded region represented by a rectangular strip is parallel to the axis of revolution. It forms an 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_2` will be based on the equation `y =4x^2 ` rearranged into `x= sqrt(y/4)` or `x =sqrt(y)/2`

`h =sqrt(y)/2-0`

`h=sqrt(y)/2`

For boundary values, we have `y_1=0` to `y_2=4` (based from the boundary line).

Plug-in the values:

`V = int_a^b` **2pi*radius*height*thickness**, we
get:

`V = int_0^4 2pi*y*sqrt(y)/2*dy`

`V = int_0^4 2pi*y*y^(1/2)/2*dy`

`V = int_0^4 piy^(3/2)dy`

Apply basic integration property: `intc*f(x) dx = c int f(x) dx.`

`V = pi int_0^4 y^(3/2)dy`

Apply power rule for integration: `int y^n dy= y^(n+1)/(n+1).`

`V = pi *(y^(3/2+1))/((3/2+1))|_0^4`

`V = pi *(y^(5/2))/((5/2))|_0^4`

`V = pi *y^(5/2)*2/5|_0^4`

`V = (2pi y^(5/2))/5|_0^4`

Apply definite integration formula:` int_a^b f(y) dy= F(b)-F(a)` .

`V = (2pi (4)^(5/2))/5-(2pi (0)^(5/2))/5`

`V = (64pi)/5 -0`

`V = (64pi)/5` or `40.21` (approximated value)

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