The equation of a doughnut (and volume/surface area)
By Luqman Bajwa (P3371) on Saturday, December 2, 2000 - 12:50 am :

Being new to equations in 3D, I was wondering what the equation of a doughnut was (the ones with a hole in the middle). I think I have found the equation but I'm not too sure.

I started by considering that the doughnut to be placed such that it was perfectly rotationary symmetrical around the z axis, and that it laid flat on the xy plane/axis. I also introduced two constants:
r - the radius of the outer circular bit
R- the radius of the whole doughnut from the centre to the middle of the outer circular bit -so that the interior radius of the doughnut is (R-r), and the exterior radius is (R+r)

I then considered the relationship between x and y (which is circular):

=> x2 +y2 =A2 [where A is the radius of this circle, and a variable defined by z]

I then needed to consider the variable A in terms of z. I did this by considering the cross-section through the origin of the doughnut so i could study the problem in 2D (eg when x=0). so the radius would then be:

A=R±sqrt[r2 -z2 ]

So the equation of the doughnut would be:

x2 +y2 =(R ±sqrt[r2 -z2 ])2

I'm not sure if this is correct though. I would be grateful if someone could verify this or correct any false assumptions made.

Thanks,

Luqman

By Sean Hartnoll (Sah40) on Saturday, December 2, 2000 - 01:05 am :

Seems good to me - well done!

Sean

By Sean Hartnoll (Sah40) on Saturday, December 2, 2000 - 01:08 am :

ps. the mathematical name for a doughnut is a torus.

Perhaps you could try to use your formula to calculate areas and volumes of tori (plural of torus) - if you know about integration...

Sean

By Brad Rodgers (P1930) on Friday, December 8, 2000 - 08:13 pm :

Just wondering, how do you find the volume and area of any 3D shape generated. I would think that


V=òab(òcd y dx)dz

and that

A=òab(òcd(1+y¢2)1/2dx)dz
(less sure about the latter). I'm not sure how this would apply to a shape like a torus though. Are these right? How would you apply these equations (or possibly other ones describing area and volume) to a shape like a torus or a sphere?

Thanks,
Brad

By Kerwin Hui (Kwkh2) on Friday, December 8, 2000 - 10:48 pm :

I am not sure there is a nice general formula for the area in terms of x, y and z. I think we need to use vector calculus on that.

Of course, we can apply Pappus to torus or sphere. Pappus states that the area is the product of length of curve multiplied by the distance travelled by the centroid in order to generate the curve. Similarly for volume.

Kerwin

By Sean Hartnoll (Sah40) on Saturday, December 9, 2000 - 02:05 am :
The idea is easy, if not the practicality, all you do is integrate unity (i.e. 1) over the whole surface of interior of the surface. So where the equation of the figure comes in is not in the integrand but in the limits of the integration, and this is where you would use the equation for the shape. This is likely to be very difficult unless there is a coordinate system in which the equations for the figure are nice. You also need to know the area and volume elements (this is doing using vector calculus)

So for a sphere radius r the surface area is

A=òf = 02pòq = 0p dA

where for a sphere dA = r2sinqdqdf

Sean