`(x-3)^2+y^2 =1 ` Find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y-axis.

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marizi eNotes educator| Certified Educator

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)

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