Question

Integral, help!

Answer 1

\[\int_0^{2\pi} \sqrt{(R+r \cos t)^2 +r^2} dt \]

R and r are constants and R>r

Answer 2

Out of curiosity, how did you come across this one? It kinda resembles an arc length integral.

Answer 3

it really had the same impression about it

Answer 4

does

Answer 5

Ahhh of course! It is an arc length integral. If you travel on the surface of a donut, this is a specific path, let me draw something up since it's a little weird.

Answer 6

The donut has two radii, R and r, R is from the center of the hole to the center of the donut itself, and then r is the radius from the center of the donut to its surface like this: So if we keep these constant then we can put an angle theta to R and an angle phi to r which allows us to shift them around.

We can travel all the way around the top of the donut with these two parameters, and if we vary theta from 0 to 2pi we get a path that walks along the donut like the frosting on top kind of lays. If we vary phi from 0 to 2pi we get the other path that takes us through the center of the donut hole.

If we go through both of these paths at the same time, angles at the same angular velocity, we get my path.

I can show you how I actually derived it but it involves using the surface metric tensor, so I don't know how important that is or if I can reformulate this without using it or not I have to think.

Answer 7

I think this set of equations should describe the surface coordinates:
\(x(\theta,\phi) = (R+r \cos \phi)\cos \theta\)
\(y(\theta,\phi) = (R+r \cos \phi)\sin \theta\)
\(z(\theta,\phi) = R \sin \theta\)

The path I'm taking is \(\theta = \phi=t\) from 0 to \(2\pi\)

So that's where this came from haha =P

Answer 8

Right, I'm sort of familiar with the parameterization of a torus, but not at all with tensor calculus. As for the integral, I suspect the closed form (if there is one) is in terms of one of those elliptical or hypergeometric fucntions, both of which are beyond my pay grade :)

Answer 9

but again that does quite seem to have a good form

Answer 10

does not*

Answer 11

Haha yeah it's just a random fun question I came up with to play around, I'm trying to see if I can just somehow work out a geometric argument instead for the length of this curve.

I'm not entirely sure if it's allowed to just cut the torus apart, it seems like this would make it so easy: