`y = 3-x , y=0, y=2, x=0` Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the...

For the region bounded by` y=3-x` , `y=0` , `y=2` , and` x=0` revolved  about the line` x=5` , we may apply Washer method for the integral application for the volume of a solid.

 The formula for the Washer Method  is:

`V = pi int_a^b [(f(x))^2-(g(x))^2]dx`

or

`V = pi int_a^b [(f(y))^2-(g(y))^2]dy`

where f as function of the outer radius 

         g as a function of the inner radius

To determine which form...

For the region bounded by` y=3-x` , `y=0` , `y=2` , and` x=0` revolved  about the line` x=5` , we may apply Washer method for the integral application for the volume of a solid.

 The formula for the Washer Method  is:

`V = pi int_a^b [(f(x))^2-(g(x))^2]dx`

or

`V = pi int_a^b [(f(y))^2-(g(y))^2]dy`

where f as function of the outer radius 

         g as a function of the inner radius

To determine which form we use, we consider the horizontal rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image. The given strip  has a thickness of "`dy` " which is our clue to use the formula:

`V = pi int_a^b [(f(y))^2-(g(y))^2]dy`

For each radius, we follow the `x_2-x_1` . We have `x_2=5` since it a distance between the axis of rotation and each boundary graph.

For the inner radius, we have: `g(y) = 5-(3-y) ` simplified to `g(y)=2+y`

Note: `y_(below)` for the inner radius is based from `y =3-x ` rearrange into `x= 3-y`

For the outer radius, we have: `f(y) = 5-0`  simplified to `f(y)=5` .

Then the boundary values of y  is `a=0` and `b =2` .

Then the integral will be: 

`V = pi int_0^2 [(5)^2-(2+y)^2]dy`

Expand using the FOIL method on:

`(2+y)^2 = (2+y)(2+y)= 4+4y+y^2 and 5^2=25` .

The integral becomes: 

`V = pi int_0^2 [25 -(4+4y+y^2 )]dy`

Simplify:

`V = pi int_0^2 [25 -4-4y-y^2 ]dy`

`V = pi int_0^2 [21-4y-y^2 ]dy`

Apply basic integration property:

`int (u-v-w)dy = int (u)dy-int (v)dy-int (w)dy`

`V = pi [ int_0^2 21dy - int_0^2 4ydy -int_0^2 y^2dy]`

For the integration of  `int 21 dy` , we apply basic integration property: `int c dx = cx` .

 For the integration of `int_0^2 4ydy` and `int_0^2 y^2dy` , we apply the Power rule for integration:  `int y^n dx = y^(n+1)/(n+1)` .

`V = pi [ 21y - 4*y^2/2 -y^3/3]|_0^2`

`V = pi [ 21y - 2y^2 -y^3/3]|_0^2`

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

`V = pi [ 21(2) - 2(2)^2 -(2)^3/3]-pi [ 21(0) - 2(0)^2 -(0)^3/3]`

`V = pi [ 42 - 8 -8/3]-pi [ 0-0-0]`

`V = pi [ 94/3]-pi[0]`

`V =(94pi)/3` or `98.44` (approximated value)