Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 20

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)