`y = sqrt(x-2) , y=0 , x=4` Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving...

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)