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I am trying to determine the surface area of a torus that has been twisted. I know the non-twisted case can be solved by either using Pappus's Theorem or a method along the lines of this post.

To be more precise, suppose we take a circle of radius $1$ sitting in the $xz$-plane with center at $(2,0,0)$ and revolve it about the $z$-axis so that the center of the revolving circle always rests in the $xy$-plane.

Right before attaching this rotated object back up upon itself to form a torus, we rotate the circle by some amount $\theta$ and then attach the starting circle to the rotating circle.

From the outside, this torus looks just like a normal torus, but it certainly is not because if we consider where the point $(3,0,0)$ goes in the process of the revolution, it is clear that some nonzero rotation will increase the length of the path sweeped out by the point.

My question is, how (if?) does the area change? Moreso, how do you calculate this new area? (I think this may depend upon the "fervor" in which which one twists the torus. For example, if we hope to acheive a twist by $\pi/2$, we can do it in the last $25\%$ of the revolution or the last $50\%$ of the revolution (actually, I have no idea if the area would change in this case!).

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It is still the same torus, but you introduce different coordinate systems on it. To see that, consider building such coordinate systems.

Consider a circle $C_2$ of radius $r_2$, situated in $(x,y)$ plane, parametrized by $$(x,y,z)=(r_2 \cos(\theta), r_2 \sin(\theta), 0)$$ Now for each point of this circle $C_1$ we build a circle of radius $r_1$ about it, so that the normal to circle $C_1$ aligns with the tangent to $C_2$ at this point:

enter image description here

Now, when we parametrize points on this circle, we have a choice of where to choose the origin. One possibility is to choose $\phi = 0$ to correspond to the point on $(x,y)$ plane, but this is not the only choice.

Consider a choice, where $\phi = \alpha \theta$ corresponds to the point on $(x,y)$ plane, for some constant $\alpha$. This gives the following parametrization of the points on the torus: $$\begin{eqnarray} x(\theta, \phi) &=& \left(r_2 + r_1 \cos(\phi - \alpha \theta) \right) \cos(\theta) \\ y(\theta, \phi) &=& \left(r_2 + r_1 \cos(\phi - \alpha \theta) \right) \sin(\theta) \\ z(\theta, \phi) &=& -r_1 \sin\left( \phi - \alpha \theta\right) \end{eqnarray} $$ Here is how tori look for different values of $\alpha$:

enter image description here

We now proceed to compute the surface area of such a torus, keeping $\alpha$ arbitrary. The Euclidean interval $\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$, after some algebra: $$ \begin{eqnarray} \mathrm{d}s^2 &=& \left( \alpha^2 r_1^2 + \left(r_2 + r_1 \cos( \phi - \alpha \theta) \right)^2 \right) \mathrm{d} \theta^2 - 2 \alpha r_1^2 \mathrm{d} \theta \mathrm{d} \phi + r_1^2 \mathrm{d} \phi^2 \\ &=& h_{\theta\theta} \mathrm{d} \theta^2 + 2 h_{\theta \phi} \mathrm{d} \theta \mathrm{d} \phi + h_{\phi\phi} \mathrm{d} \phi^2 \end{eqnarray} $$ This allows to find the measure on the torus, induced by its embedding into $\mathbb{R}^3$: $$ \mathrm{d} S = \sqrt{ h_{\theta\theta} h_{\phi\phi} - h_{\theta\phi}^2} \mathrm{d}\theta \mathrm{d} \phi = r_1 \left(r_2 + r_1 \cos(\phi - \alpha\theta) \right) \mathrm{d} \theta \mathrm{d} \phi $$ Therefore, the surface area of the torus is: $$ S = \int_{0}^{2 \pi} \int_{0}^{2\pi} r_1 \left(r_2 + r_1 \cos(\phi - \alpha\theta) \right) \mathrm{d} \theta \mathrm{d} \phi = 2 \pi r_1 r_2 \int_0^{2\pi} \mathrm{d} \theta = 4 \pi^2 r_1 r_2 $$ It is explicitly independent on $\alpha$, as expected.

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Just a nitpick: Diffeomorphisms can indeed change volume - e.g., the plane is diffeomorphic to an open disc. –  Neal Sep 14 '12 at 4:33
    
@Neal Thanks, I have removed that statement. –  Sasha Sep 14 '12 at 4:35
    
That is the exact explanation I was hoping for. It is counterintuitive to me that twisting the torus can increase the length of a fixed band, but twisting a region does not change its area. I guess this is similar to the idea that a shear transformation on a rectangle preserves the area. Thank-you for an excellent explanation. –  NoClue Sep 14 '12 at 16:41
    
@NoClue Was glad to help. I had fun answering it too. –  Sasha Sep 14 '12 at 16:58
    
@NoClue If you regard a flat torus as a rectangle with opposite edges identified in an orientation-preserving manner, then a shear transformation on the rectangle is the same as the twist you asked about. So your analogy is very analogous :) –  Neal Sep 16 '12 at 15:56
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