Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 15

For a region bounded by y=x (slant line), x=0 (along the vertical axis), and y = 3 (horizontal line) revolved about the line y=4 , we may apply the Washer Method. It considers multiple disc with a hole. Basically, it can be just two disc method in which we take the difference of the volume of the bigger and smaller disc.
It follows the formula: A = pi(R_(outer))^2-(r_(i.n.n.e.r))^2]
Let R_(outer) as a function of f and r_(i.n.n.e.r) as function of g .
Then 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 .
We use a rectangular strip representation that is perpendicular to the axis of rotation as shown on the attached image.
For the inner radius, we have:
g(x)=4-3=1
For the outer radius, we have: f(x)=4-x
The boundary values of x will be a=0 to b=3 .
The integral to approximate the volume of the solid is:
V=pi int_0^3 [(4-x)^2-1^2] dx
Expand using FOIL method on (4-x)^2 = (4-x)*(4-x) = 16-8x+x^2 and 1^2=1 .
The integral becomes:
V=pi int_0^3 [16-8x+x^2 -1] dx
Simplify:V=pi int_0^3 [15-8x+x^2] dx
Apply basic integration property: int (u+-v+-w) dx= int (u) dx+-int (v) dx+-int (w) dx
V=pi [ int_0^3 15 dx - int_0^3 8xd +int_0^3 x^2 dx]
Apply basic integration property: int c dx = cx and Power rule for integration: int x^n dx= x^(n+1)/(n+1).
V=pi [15x- 8*x^2/2 + x^3/3]|_0^3
V=pi [15x- 4x^2 + x^3/3]|_0^3
Apply definite integration formula: int_a^b f(x) dx = F(b)-F(a).
V=pi [15*(3)- 4(3)^2 +(3)^3/3]-pi [15(0)- 4(0)^2 + (0)^3/3]
V=pi [45- 36+9] -pi [0-0+0]
V = 18pi or 56.55 (approximated value)