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 strips, 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*

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 =(sqrt(x-2)) - 0 = sqrt(x-2)

*thickness* `= dx`

For boundary values of x, we have `a=2` to `b=4` .

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

`V = int_2^4 2pi *x* sqrt(x-2) * dx`

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

`V = 2pi int_2^4 x* sqrt(x-2)dx`

To determine the indefinite integral, we may apply u-substitution by letting `u = x-2` then `du = dx` . The `u=x-2` can be rearranged as `x=u+2` .

The integral becomes:

`V = 2pi int (u+2)* sqrt(u)dx`

Apply Law of Exponent:` sqrt(u) = u^(1/2)` and `u^n*u^m = u^(n+m)` .

`V = 2pi int (u+2)* u^((1/2))dx`

`V = 2pi int u^((3/2)) +2u^((1/2)) dx`

Apply basic integration property:i`nt (u+v)dx = int (u)dx+int (v)dx` .

`V = 2pi [int u^((3/2))dx +int 2u^((1/2)) dx]`

Apply Power rule for integration: `int x^n dx= x^(n+1)/(n+1)` .

`V = 2pi [ u^(3/2+1)/(3/2+1) +2u^(1/2+1)/(1/2+1)]`

`V = 2pi [ u^(5/2)/(5/2) +2*u^(3/2)/(3/2)]`

`V = 2pi [ u^(5/2)*(2/5) +2u^(3/2)*(2/3)]`

`V = 2pi [ (2u^(5/2))/5 +(4u^(3/2))/3]`

Plug -in` u =x-2` , we get:

`V = 2pi [ (2(x-2)^((5/2)))/5 +(4(x-2)^((3/2)))/3] ` with boundary values:` a=2` to `b=4` .

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

`V = 2pi [ (2(4x-2)^((5/2)))/5 +(4(4-2)^((3/2)))/3] -2pi [ (2(4-2)^((5/2)))/5 +(4(4-2)^((3/2)))/3]`

`V=2pi [ (8sqrt(2))/5 +(8sqrt(2))/3] - 2pi [0 +0]`

`V=2pi [ (8sqrt(2))/5 +(8sqrt(2))/3] - 2pi [0 +0]`

`V =2pi[(64sqrt(2))/15]- 2pi[0]`

`V =(128sqrt(2)pi)/15-0`

`V =(128sqrt(2)pi)/15` or `37.92` (approximated value)

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