x + y = 4 , y = x , y = 0 Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.

We can use a rectangular strips to represent the region bounded by x+y=4, y=0 , and y=x revolved about the x-axis. As shown on the attached graph, we consider two sets of rectangular strip perpendicular to the x-axis (axis of revolution) to be able to use the Disk Method.  This is the case since the upper bound of the rectangular strip differs before and after x=2 .
 In this method, we follow the formula: V = int_a^b A(x) dx since we are using a vertical orientation of each rectangular strip with a thickness =dy.
Note: A = pir^2 where r= length of the rectangular strip.
 We may apply r = y_(above) - y_(below) .
For the region within the boundary values of x: [ 0,2] , we follow r = x-0=x
For the region within the boundary values of x: [ 2,4] , we follow r = 4-x-0=4-x
Note: x+y=4 can be rearranged as y=4-x.
Then the integral set-up will be:
V = int_0^2 pi*(x)^2dx+int_2^4 pi*(4-x)^2dx
For the first integral: int_0^2 pi*(x)^2dx , we may apply Power rule of integration: int x^n dx = x^(n+1)/(n+1) .
int_0^2 pi*(x)^2dx= pi* x^((2+1))/((2+1))|_0^2
                        =(pix^3)/3|_0^2
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
(pix^3)/3|_0^2 =(pi(2)^3)/3-(pi(0)^3)/3
          =(8pi)/3- 0
          =(8pi)/3
For the indefinite integral of int_2^4 pi*(4-x)^2dx , we may u-substitution by letting u =4-x then du =-dx or (-1)du =dx .
The integral becomes :
int pi*(4-x)^2dx =int pi*u^2*(-1) du
Apply basic integration property: intc*f(x) dx = c int f(x) dx.
int pi*u^2*(-1) du = -pi int u^2 du
Apply power rule for integration: int x^n dy= x^(n+1)/(n+1).
-pi int u^2 du =-pi* u^((2+1))/((2+1))
                      = (-piu^3)/3
Plug-in u=4-x on (-piu^3)/3  we get:
int_2^4 pi*(4-x)^2dx =(-pi(4-x)^3)/3|_2^4 or ((x-4)^3pi)/3|_2^4
Apply the definite integral formula: int _a^b f(x) dx = F(b) - F(a) .
((x-4)^3pi)/3|_2^4 =((4-4)^3pi)/3-((2-4)^3pi)/3
                  = 0 - (-8pi)/3
                = (8pi)/3
Combing the two definite integrals, we get:
 
V = int_0^2 pi*(x)^2dx+int_2^4 pi*(4-x)^2dx
V = (8pi)/3+(8pi)/3
V =(16pi)/3 or 16.76 (approximated value).