Expert: Paul Klarreich - 9/24/2007

Question
QUESTION: Dear Mr. Paul Klarreich
I am an Assistant Engineer, and gone through your reply to Mr.Aswin Kumar Dated 3/3/2006, regarding the volume of Torus.

Generally the cross section of Torus is taken as full circle and the volume formulas are derived. If the cross section is segment of a circle, How to calculate the volume? The segment may have any included angle and its
Location may be any where from 0 to 360deg. Consider the 0 deg is on X axes And torus rotates about Y axes.

For e.g. Torus center radius (R) be 25mm and section radius (r) be 10mm
And the arc segment is 45deg starting from 0deg to 45deg. How to calculate its volume?
I have struck in this problem for some months. Please help to solve.



ANSWER: Questioner:   Pachamuthu
Category:  Calculus
Private:  No
 
Subject:  Volume of Torus which has circular segment cross section
Question:  Dear Mr. Paul Klarreich
I am an Assistant Engineer, and gone through your reply to Mr.Aswin Kumar Dated 3/3/2006, regarding the volume of Torus.

Generally the cross section of Torus is taken as full circle and the volume formulas are derived. If the cross section is segment of a circle, How to calculate the volume? The segment may have any included angle and its
Location may be any where from 0 to 360deg. Consider the 0 deg is on X axes And torus rotates about Y axes.

For e.g. Torus center radius (R) be 25mm and section radius (r) be 10mm
And the arc segment is 45deg starting from 0deg to 45deg. How to calculate its volume?
I have struck in this problem for some months. Please help to solve.
.....................................
Hi, Pachamuthu,

I am not sure exactly which 'segment' you are referring to:

A. Do you mean that you are going to cut slices of the 'donut' where the central angle is at the origin?  In that case you are actually talking about a 'sector', not a segment.

If that is the case, you simply use the fact that you are taking a fraction of the volume, where the fraction is just  theta/360  or  theta/2pi.


B. Do you mean that you will take a slice of the cross-section circle, then rotate THAT about the z-axis?  

In that case, you use the formula:

Sector area = (theta/360) * area of circle.

Then you simply rotate that about the z-axis; multiply by the path traveled, which is  2pi * torus radius.

(torus radius is your 25 mm)

Please note the vocabulary.  If you have a circle and

1. You draw a simgle line through it, the piece cut off is a segment.

2. You draw two radii, the portion between them is a sector.


---------- FOLLOW-UP ----------

QUESTION: Dear Mr.Paul Klarreich,
Thanks for the reply. My case is option “B”. The sector area = theta/360 * area of cross section circle and that sector is rotated about “z” axes. Let the torus center be ‘R”. The center of torus be (0,0,0), and the center for the section circle is (R,0,0). Let the section circle radius be “r”. If the section is full circle the torus outer radius is R+r and the inner radius is R-r. Our section is a sector with “theta” deg included angle.

Option1: If the sector is located at 0deg side, the torus will be between “R” and “R+r”. Option2: If the sector is located at 180deg side, the torus will be between “R” and “R-r”.
Please note that theta is measured wrt (R,0,0).

The circumference length and volume of option1 will be more than option2 for the same value of “theta”. so, the volume varies with the location of sector. My doubt is, how to calculate volume for different location?

Please excuse me for poor English.  

Answer
Questioner:   Pachamuthu
Category:  Calculus
Private:  No
 
B. Do you mean that you will take a slice of the cross-section circle, then rotate THAT about the z-axis?  

In that case, you use the formula:

Sector area = (theta/360) * area of circle.

Then you simply rotate that about the z-axis; multiply by the path traveled, which is  2pi * torus radius.

(torus radius is your 25 mm)

Please note the vocabulary.  If you have a circle and

1. You draw a simgle line through it, the piece cut off is a segment.

2. You draw two radii, the portion between them is a sector.


---------- FOLLOW-UP ----------

Thanks for the reply. My case is option “B”. The sector area = theta/360 * area of cross section circle and that sector is rotated about “z” axes. Let the torus center be ‘R”. The center of torus be (0,0,0), and the center for the section circle is (R,0,0). Let the section circle radius be “r”. If the section is full circle the torus outer radius is R+r and the inner radius is R-r. Our section is a sector with “theta” deg included angle.

Option1: If the sector is located at 0deg side, the torus will be between “R” and “R+r”. Option2: If the sector is located at 180deg side, the torus will be between “R” and “R-r”.
Please note that theta is measured wrt (R,0,0).

The circumference length and volume of option1 will be more than option2 for the same value of “theta”. so, the volume varies with the location of sector. My doubt is, how to calculate volume for different location?

Please excuse me for poor English.
===========================================
Hi, Pachamuthu,

OK, then.  If it is option B, meaning you cut a slice (sector) of the circle, then revolve that about the z-axis, then the scheme is this:

1. Compute the area of the sector.  Easy.  Area = (1/2)theta * r^2  (in radians) or  theta/360 * pi r^2  using degrees.

2. Determine the path traveled by the CENTROID of the sector area.

That means you have to do a centroid calculation -- you have to find the x-coordinate, which gives you the length of the rotation, 2pi * x-coordinate.

         Sector moment
Centroid(x) = -------------
         Sector area

..........................................

Suppose the central angle of your sector covers angles from @1 to @2.  This takes care of all the cases.  

If @1 = 0 and @2 = 2pi, it's the whole circle.
If @1 = -pi/2 and @2 = pi/2, it's the half on the outside.
If @1 = pi/2 and @2 = 3pi/2, it's the half on the inside.


Set up this integral:

Each area element is a small 'sectorette' (OK, I made that up.) which goes from t to t+dt and  from  r to  r+dr.  [I'll use 't' for the variable angle so I don't have to keep typing theta's.]

This area element, is a 'pseudo-rectangle.  

Two of its sides go along the radius and have length  dr.

Two of its sides go 'around' -- they are bits of arc and have length  r dt.

The area element is  dA =  r dr dt

The area integral is:

{{
|| r dr dt =
}}

{t2
| r^2/2[0 to r0] dt =  
}t1

{t2
| r0^2/2 dt =  
}t1

r0^2/2 * [t2 - t1] = the area of a sector.

NOW THE MOMENT HAS ARRIVED.  I mean, we have to compute the x-moment, which requires that we multiply the integrand by x = r cos t

The moment integral is:

{{
|| r cos t * r dr dt =
}}

{{
|| r^2 cos t dr dt =
}}

{{
|| r^2 dr cos t dt =
}}

{
| r^3/3 [0 to r0]  cos t dt =
}

{
| r0^3/3  cos t dt =
}

r0^3/3  sin t [t2 - t1] =

r0^3/3 (sin t2 - sin t1)

Now the value of  x-bar is:

r0^3/3 (sin t2 - sin t1)
------------------------
 r0^2/2 [t2 - t1]


2 r0 (sin t2 - sin t1)
------------------------
  3 [t2 - t1]


I think that's it, basically.