Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 26

For the region bounded by y=2/(x+1) ,y=0 , x=0 and x=6 and revolved about the x-axis, we may apply Disk method. For the Disk method, we consider a perpendicular rectangular strip with the axis of revolution.
As shown on the attached image, the thickness of the rectangular strip is "dx" with a vertical orientation perpendicular to the x-axis (axis of revolution).
We follow the formula for the Disk method:V = int_a^b A(x) dx where disk base area is A= pi r^2 with r =y=f(x) .
For the r, we consider the length of the strip= y_(above) - y_(below).
Then r =f(x)= 2/(x+1)-0
f(x) = 2/(x+1)
The boundary values of x will be a=0 to b=6 .
Plug-in the f(x) and the boundary values to integral formula, we get:
V = int_0^6 pi (2/(x+1))^2 dx
V = int_0^6 (4pi)/(x+1)^2 dx
To solve for the indefinite integral, we may apply u-substitution by using u = x+1 then du =dx .
The integral becomes:
V = int (4pi)/(u^2) du
Apply basic integration property: int c f(x) dx = c int f(x) dx .
V =4pi int 1/u^2 du
Apply Law of exponent: 1/x^n = x^(-n) and Power rule of integration: int x^n dx = x^(n+1)/(n+1)
V =4pi int u^(-2) du
V= 4pi u^((-2+1))/((-2+1))
V=4pi* u^(-1)/(-1)
V=(-4pi)/u

Plug-in u=x+1 on V=(-4pi)/u , we get:
V=(-4pi)/(x+1) with boundary values a=0 to b =6 .
Apply definite integration formula: int_a^b f(y) dy= F(b)-F(a) .
V = (-4pi)/(6+1) -(-4pi)/(0+1)
V =(-4pi)/7 +4pi)
V =(24pi)/7 or 10.77 (approximated value)