Comparing V and SA of a torus vs sphere

[Quinn Pochekailo]

Homework Statement

A torus has a major radius and a minor radius. When R>r by a magnitude of at least 4x, it comes to be a slim ring looking shape. When R>r by a magnitude of 1/2, the shape looks to be a donut. When R=r, the torus shape looks more like a sphere except with a small gap in the center of the shape.

I want to be able to show that volume and surface of a torus is greater than that of a sphere at the nanoscale level.

The purpose of this problem is to show that a torus at the nanoscale level can carry more particles and has a greater surface area, making this shape a more effective method than a spherical liposome.

Homework Equations

Torus:
V= 2*pi^2*R*r^2
SA = 4*pi^2*R*r

Sphere:
V= (4/3) * pi * r^2
SA = 4 * pi * r^2

The Attempt at a Solution

I started first by comparing the V and SA using R=5 and r=2.5 for simplicity and because I want the donut looking shape compared to the small ring (by using R= (1/2)r )

Final numbers for torus:
V= 616.225
SA = 492.98

Final numbers for sphere:
V = 26.17
SA = 78.5

I then calculated the magnitude of difference in V and SA, which turned out to be 23x and 6.3x respectively.

I expected to see the same results of magnitude at the nanoscale level.

Because I have to keep in mind the major radius of the torus, I made R+r = 1e-9. For the sphere it was just r = 1e-9.

Stating that R = 1/2r, I plugged that in my equation and made
(1/2r) + r = 1e-9.
r + 2r = 2e-9
3r = 2e-9
r = 6.7 e -10

R = .5 (6.7e-10) <---- R = 1/2r
R = 3.35e-10

Figuring out my values, I then used those numbers in my V and SA formulas

Torus:
V= 2.97 e -29
SA = 8.9 e-18

Sphere:
V= 4.2e-18
SA = 1.256e-17

As you can see, the calculations from the first results are very different and at the nanoscale this shows that the volume of a torus is actually less than that of a sphere. This makes sense because I am multiplying by a very small number (R) unlike that in the sphere formula.

Did I do my work wrong somewhere? Why is that at the nanoscale level the torus is less effective than the sphere?

 

[SammyS]

For one thing that formula for the volume of a torus is only good for R ≥ r .

 

[Quinn Pochekailo]

For one thing that formula for the volume of a torus is only good for R ≥ r .

How would the formula change if R<r ? I would assume it looks closer to the formula for the volume of a sphere.

 

[haruspex]

show that volume and surface of a torus is greater than that of a sphere

That doesn't really mean anything. There has to be some constraint against simply making the sphere arbitrarily larger or smaller.

If you mean that a torus has a greater area than a sphere of the same volume, that is true and quite easy to show.

If you fix on a torus with R=r then you could ask whether it has a greater volume and area than a sphere of the same r (or R)

Torus:
V= 2*pi^2*R*r^2
SA = 4*pi^2*R*r

Sphere:
V= (4/3) * pi * r^2
SA = 4 * pi * r^2

If you really mean r to be the same in both, then trivially you can make the torus larger in both respects by making R sufficiently large.

Judging from this:

I made R+r = 1e-9. For the sphere it was just r = 1e-9

you want R+r of the torus equal to the radius of the sphere. With that, the sphere will obviously have the greater volume.

In short, you need to define the radius of the sphere that's to be compared with a given torus.

 

[haruspex]

How would the formula change if R<r ? I would assume it looks closer to the formula for the volume of a sphere.

Draw a figure of a cross-section through such a torus. You no longer have a complete 2πr circumference as two circles overlap. You need to do a bit of integration to find the area (could be nasty - haven't checked).