Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 59

To find the volume of the torus generated by revolving the region bounded by the graph of circle about the y-axis, we may apply Washer method. In this method, rectangular strip representation that is perpendicular to the axis of rotation. We follow the formula for Washer method: V =pi int_a^b[(f(y))^2-(g(y))^2]dy using a horizontal rectangular strip representation with thickness of dy .
The given equation: (x-3)^2 +y^2=1 is in a form of (x-R)^2+y^2=r^2 .
We set up the function of each radius based on the following formula:
inner radius: f(y)= R -sqrt(r^2-y^2)
Then the function for the graph from x=3 to x=4 will be: f(y)= 3-sqrt(1 -y^2)
outer radius: g(y)= R+sqrt(r^2-y)
Then the function for the graph from x=2 to x=3 will be: g(y)= 3+sqrt(1-y^2)
From the attached image, the boundary values of y are: a= -1 and b =1 .
Plug-in the values on the formula, we set up:
V =pi int_(-1)^(1) [(3+sqrt(1-y^2))^2-(3-sqrt(1 -y^2))^2]dy
=pi int_(-1)^(1) [(3+6sqrt(1-y^2) +1-y^2)-(3-6sqrt(1 -y^2) +1 -y^2)]dy
=pi int_(-1)^(1) [ 3+6sqrt(1-y^2) +1-y^2 -3+6sqrt(1 -y^2) -1 +y^2]dy
=pi int_(-1)^(1) [12sqrt(1-y^2)]dy
Apply the basic integration property: int c f(x) dx - c int f(x) dx .
V =12pi int_(-1)^(1) [sqrt(1-y^2)]dy
From integration table, we may apply the integral formula for function with roots:
int sqrt(a^2-u^2)du= (u*sqrt(a^2-u^2))/2+a^2/2 arcsin(u/a)
Then,
V =12pi int_(-1)^(1) [sqrt(1-y^2)]dy
= 12pi * [(y*sqrt(1-y^2))/2+1/2 arcsin(y/1)] |_(-1)^(1)
= 12pi * [(ysqrt(1-y^2))/2+ arcsin(y)/2] |_(-1)^(1)
=[6piysqrt(1-y^2)+ 6piarcsin(y)] |_(-1)^(1)
Apply definite integral formula: .
V =[6piysqrt(1-y^2)+ 6piarcsin(y)] |_(-1)^(1)
=[6pi(1)sqrt(1-1^2)+ 6piarcsin(1)]-[6pi(-1)sqrt(1-(-1)^2)+ 6piarcsin(-1)]
=[6pisqrt(1-1)+ 6piarcsin(1)]-[-6pisqrt(1-1)+ 6piarcsin(-1)]
=[6pisqrt(0)+ 6piarcsin(1)]-[-6pisqrt(0)+ 6piarcsin(-1)]
=[6pi*0+ 6pi*(pi/2)]-[-6pi*0+ 6pi*(-pi/2))]
=[0+ 3pi^2]-[0+ (-3pi^2)]
= 3pi^2 -(-3pi^2)
=3pi^2 +3pi^2
=6pi^2 or 59.22 (approximated value)