# Surface area of a region of a torus

What is the simplest way to calculate the surface area of a region of a torus?

This is a cross section through a torus, the dashed line is the central axis. I am interested in the external surface area of the region marked with the solid black line.

A few notes about the diagram: The angle alpha is between π/2 and π radians, and the vertical line of the sector is parallel to the axis of the torus (dashed line on the diagram).

Given the simple geometry, is there a more simple/direct approach than a long process of integration to determine this area? Is there a known expression I can use here?

Any help would be welcome.

SteamKing
Staff Emeritus
Homework Helper
What is the simplest way to calculate the surface area of a region of a torus?

This is a cross section through a torus, the dashed line is the central axis. I am interested in the external surface area of the region marked with the solid black line.

A few notes about the diagram: The angle alpha is between π/2 and π radians, and the vertical line of the sector is parallel to the axis of the torus (dashed line on the diagram).

Given the simple geometry, is there a more simple/direct approach than a long process of integration to determine this area? Is there a known expression I can use here?

Any help would be welcome.
You don't need to do any complicated integrations to find the area of a portion of the SA of the torus unless you need the practice.

The torus can be thought of as a cylinder which has been turned on itself until the top and bottom surfaces have coincided:

http://en.wikipedia.org/wiki/Torus

Since you know the central angle of the black line and the radius r of the toroidal segment, you can easily calculate the length of the arc formed by this line. Since you also know the radius of the torus R, then the SA of the toroidal segment will be 2π * (R + r) * r * α, where r * α is the arc length of the black line.

Thank you SteamKing for the very quick reply. I did originally think of something similar when I first looked at this problem and there is something here that I do not understand.

If I understand what you have said, we are multiplying the circumference of the outermost external edge of the torus [2π(R+r)] by the length of the arc. The issue I have with this approach is that if you look at the torus from above (along its axis), then a point at the very edge will cover a larger distance than any other point on the arc when you move it around the ring: i.e. not all of the arc moves through the same distance (points further out move further). So I would expect the actual surface area to be smaller than the area calculated for a straight cylinder.

A simple way to test this is to move the segment - for example if you were to rotate the segments position (keeping the same angle α) so that it faced further in towards the center of the torus it would inscribe a smaller area when you rotate it around the ring. So the actual area would be smaller. This tells me there should be a term in the solution that contains information about the orientation of the arc within the ring itself.

I'm sorry if I have misunderstood.

If no-one else has any ideas then would a possible solution be: if I find some way to calculate the average distance of every point on the arc from the axis of the torus, then replace (R+r) term in your solution with this distance? - would that work?

Many thanks

SteamKing
Staff Emeritus
Homework Helper
A simple way to test this is to move the segment - for example if you were to rotate the segments position (keeping the same angle α) so that it faced further in towards the center of the torus it would inscribe a smaller area when you rotate it around the ring. So the actual area would be smaller. This tells me there should be a term in the solution that contains information about the orientation of the arc within the ring itself.

If you read the wiki article on the torus, you'll see that calculations involving the volume and surface area can be done using Pappus' centroid theorem.

I think for the general orientation of the arc, the arc length will still be r * α, but the circumferential length of the SA of the torus will be determined by the circumference of the centroid of the arc, which would be on the line which bisects the angle α.

P. 14 of the following article shows the calculation of the centroid or a circular arc:

http://www.eng.auburn.edu/~marghitu/MECH2110/staticsC3.pdf [Broken]

which centroid is 2r sin (α/2) / α, measured from the center of the tube along the angle bisector. To get this centroidal distance measured in the coordinate system of the torus, we must multiply

(2r sin (α/2) / α) by sin (π/2 - α/2)

for the torus shown in your diagram. The SA of the toroidal segment would then be:

SA = (arc length) * 2π * (centroidal distance)

SA = (r * α) * 2π * [(2r sin (α/2) / α) * sin (π/2 - α/2) + R], where all angles are measured in radians.

You should be able to check this formula by splitting up the torus into a couple of pieces, calculate the SA of each piece, and then see if the sum of the SAs equals the area of the torus, as determined by the formula from the torus wiki.

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Thank you again SteamKing, I can't believe how much time you've given to my problem.

I wonder if I might be able to ask a few more questions.

1) In the line below....
To get this centroidal distance measured in the coordinate system of the torus, we must multiply

(2r sin (α/2) / α) by sin (π/2 - α/2)
.... should the term we are multiplying by, sin (π/2 - α/2) not be cos(π/2 - α/2)?

[I realize that we could also simplify by applying trig identity cos(π/2 - α/2) = sin(α/2) which does turn it back into a sine function too!]

If it helps there's an updated version of the diagram here to jog the memory: https://www.dropbox.com/s/c4dkwbxprehl8vo/surface-area-torus2.png?dl=0

2) Should the term (2r sin (α/2) / α) have a 2 at the front of it? If we are integrating between the limits -α/2 and α/2, then eventually we end up with α/2+α/2 = α (I may have made a mistake? This may just be a confusion from the coincidence of α in my example being twice the value of α in the .pdf link.

3) This may be a silly question, but I don't completely understand where the ∫Rdθ comes from in equation 3.19 of the .pdf link you sent. If I follow the reference back to pages 5-6 in the .pdf I can understand where equation 3.5 comes from, but I am wondering: does the ∫Rdθ in Equation 3.19 represent the mass term in the equation at the top of page 6 because the arc is of uniform 'mass'? In other words does it scale the other integral to a unit value? I'm sorry if this sounds confusing.
I can follow the rest of the derivation very well.

Thank you for all of your help.

SteamKing
Staff Emeritus
Homework Helper
Thank you again SteamKing, I can't believe how much time you've given to my problem.

I wonder if I might be able to ask a few more questions.

1) In the line below....

.... should the term we are multiplying by, sin (π/2 - α/2) not be cos(π/2 - α/2)?

[I realize that we could also simplify by applying trig identity cos(π/2 - α/2) = sin(α/2) which does turn it back into a sine function too!]

If it helps there's an updated version of the diagram here to jog the memory: https://www.dropbox.com/s/c4dkwbxprehl8vo/surface-area-torus2.png?dl=0

Yes, you are right about the cosine.

2) Should the term (2r sin (α/2) / α) have a 2 at the front of it? If we are integrating between the limits -α/2 and α/2, then eventually we end up with α/2+α/2 = α (I may have made a mistake? This may just be a confusion from the coincidence of α in my example being twice the value of α in the .pdf link.

According to the formula for the centroid of the arc, rc = r sin (β) / β, where β is half the angle covered by the arc.

In your diagram of the torus, β = α / 2, so
rc = r sin (α/2) / (α / 2) = 2r sin (α/2) / α after simplifying.

3) This may be a silly question, but I don't completely understand where the ∫Rdθ comes from in equation 3.19 of the .pdf link you sent. If I follow the reference back to pages 5-6 in the .pdf I can understand where equation 3.5 comes from, but I am wondering: does the ∫Rdθ in Equation 3.19 represent the mass term in the equation at the top of page 6 because the arc is of uniform 'mass'? In other words does it scale the other integral to a unit value? I'm sorry if this sounds confusing.
I can follow the rest of the derivation very well.

Thank you for all of your help.

If you look at Fig. 3.13 in the link, you'll see that the arc is split up into smaller segments which each have length ds. The centroid of each segment is xc units from the origin.
Since we are dealing with a circular arc of radius R, then ds = R dθ, from the definition of the radian (dθ is the angle subtended by the segment ds). We can also calculate the location of the centroid of ds in terms of the radius of the circle and the angle made by the middle of this arc with the x-axis:

xc = R cos (θ)

so the first moment of the arc segment about the y-axis is:

dM = xc*ds = R cos (θ) * R dθ = R2cos(θ)dθ

The total moment is the integral of dM w.r.t. θ.