If a torus is cut across its tube and straightened out, it clearly becomes a cylinder. It is also a fact that the surface area and enclosed volume of the two objects are equal. Since the two objects differ in intrinsic geometry, it is not immediately clear why these measures are the same. An analytic transformation between the two would establish the equivalences, along with whether equality holds for all intermediate states.

One convenient way to produce such a transformation is with involutes of a circle. A parametrization that maps points on the right-hand side of a circle to a horizontal line is

$x=R[sint-(t-a)cost]\phantom{\rule{5em}{0ex}}y=R[cost+(t-a)sint]$

along with its mirror reflection for points on the left-hand side:

$x=-R[sint-(t-a)cost]\phantom{\rule{5em}{0ex}}y=R[cost+(t-a)sint]$

Here is visualization of several involutes in blue for a unit circle:

The involutes all begin on the circle for $t=a$ and reach the line $y=-1$ for $t=\pi $ . One can now construct a curved line that interpolates between the two endpoints of circle and line:

The initial curved line begins at the top of the circle for $a=0$ , reaching the bottom for $a=\pi $ and returning to the top at $a=2\pi $ . This line thus has the parametrization

$\begin{array}{l}x=R[sint-(t-a)cost]\\ y=R[cost+(t-a)sint]\end{array}\phantom{\rule{4em}{0ex}}t=(\pi -a)u+a$

on the domains $0\le a\le 2\pi $ and $0\le u\le 1$ . For the corresponding line in three dimensions, simply add a constant coordinate,

$\mathbf{f}(a,u)=[x(a,u),y(a,u),0]$

or more explicitly

$\begin{array}{l}\mathbf{f}(a,u)=R\phantom{\rule{.3em}{0ex}}[sin[\pi u+(1-u\left)a\right]-(\pi -a)ucos[\pi u+(1-u\left)a\right],\\ \phantom{\rule{5.5em}{0ex}}cos[\pi u+(1-u\left)a\right]+(\pi -a)usin[\pi u+(1-u\left)a\right],\phantom{\rule{.3em}{0ex}}0\phantom{\rule{.3em}{0ex}}]\end{array}$

In order to exude a three-dimensional tube along this curved line, one needs two normals to the line. One of these can be taken proportional to
$\hat{\mathbf{z}}$ ,
but the other must be calculated from the parametrization. For a fixed value of *u*, the derivative of this parametrization with respect to the other parameter is

$\begin{array}{l}\frac{d\mathbf{f}}{da}=R\phantom{\rule{.3em}{0ex}}[cos[\pi u+(1-u\left)a\right]+(\pi -a)u(1-u)sin[\pi u+(1-u\left)a\right],\\ \phantom{\rule{4.5em}{0ex}}-sin[\pi u+(1-u\left)a\right]+(\pi -a)u(1-u)cos[\pi u+(1-u\left)a\right],\phantom{\rule{.3em}{0ex}}0\phantom{\rule{.3em}{0ex}}]\end{array}$

and the unit tangent vector along the line is

$\begin{array}{l}\mathbf{T}=\frac{1}{\sqrt{1+{\Pi}^{2}}}[cos[\pi u+(1-u\left)a\right]+\Pi sin[\pi u+(1-u\left)a\right],\\ \phantom{\rule{6.5em}{0ex}}-sin[\pi u+(1-u\left)a\right]+\Pi cos[\pi u+(1-u\left)a\right],\phantom{\rule{.3em}{0ex}}0\phantom{\rule{.3em}{0ex}}]\end{array}$

with the abbreviation $\Pi =(\pi -a)u(1-u)$ . Since the derivative of vector with constant square is always normal to the vector,

$\frac{d}{da}[\phantom{\rule{.3em}{0ex}}\mathbf{g}\xb7\mathbf{g}=c\phantom{\rule{.3em}{0ex}}]\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}\mathbf{g}\xb7\frac{d\mathbf{g}}{dp}=0$

the second normal is found from the derivative of the unit tangent vector, which when normalized is

$\begin{array}{l}\mathbf{N}=\frac{d\mathbf{T}/dp}{|d\mathbf{T}/dp|}\\ \phantom{\mathbf{N}}=\frac{1}{\sqrt{1+{\Pi}^{2}}}[-sin[\pi u+(1-u\left)a\right]+\Pi cos[\pi u+(1-u\left)a\right],\\ \phantom{\rule{6.5em}{0ex}}-cos[\pi u+(1-u\left)a\right]-\Pi sin[\pi u+(1-u\left)a\right],\phantom{\rule{.3em}{0ex}}0\phantom{\rule{.3em}{0ex}}]\end{array}$

This last calculation is considerably facilitated by the introduced abbreviation. As a check, the unit binormal is easily evaluated,

$\mathbf{B}\equiv \mathbf{T}\times \mathbf{N}=-\hat{\mathbf{z}}$

consistent with a statement above. The exuded tube can now be give parametrically by multiplying the two unit normals with circular functions and a second radial scale and adding them to the parametrization of the curved line. For later consistency the opposite of the normals will be used, so that the parametrization of the surface of the tube will be taken as

$\mathbf{f}(a,u)-rcos\phi \mathbf{N}(a,u)-rsin\phi \mathbf{B}(a,u)$

Reverting to the initial notation for brevity, and noting that $\Pi =(t-a)(1-u)$ , the explicit parametrization is

$\begin{array}{l}x=R[sint-(t-a)cost]\\ \phantom{\rule{5em}{0ex}}+\frac{rcos\phi}{\sqrt{1+(t-a{)}^{2}(1-u{)}^{2}}}[sint-(t-a\left)\right(1-u)cost]\\ y=R[cost+(t-a)sint]\\ \phantom{\rule{5em}{0ex}}+\frac{rcos\phi}{\sqrt{1+(t-a{)}^{2}(1-u{)}^{2}}}[cost+(t-a\left)\right(1-u)sint]\\ z=rsin\phi \\ t=(\pi -a)u+a\end{array}$

This can be easily visualized:

As a test of the parametrization, when $u=0$ then $t=a$ and the parametrization becomes

$\left[\right(R+rcos\phi )sina,(R+rcos\phi )cosa,rsin\phi ]$

which is that of a torus with the first two coordinates exchanged. When $u=1$ then $t=\pi $ and the parametrization becomes

$\left[R\right(\pi -a),-R-rcos\phi ,rsin\phi ]$

which is a circle displaced down the *y*-axis and moved linearly along the *x*-axis, *i.e.* a cylinder displaced downwards.

With the parametrization in hand, one can now evaluate the surface area and enclosed volume of the tube as an analytic function of the morphing parameter *u*. The concept of volume element indicates that surface area and volume can be calculated in the same manner, as an integral over the square root of the determinant of the appropriate two- or three-dimensional metric. For this purpose one must switch back to the explicit variables

$\begin{array}{l}x=R[sin[\pi u+(1-u\left)a\right]-(\pi -a)ucos[\pi u+(1-u\left)a\right]]\\ \phantom{\rule{5em}{0ex}}+\frac{rcos\phi}{\sqrt{1+{\Pi}^{2}}}[sin[\pi u+(1-u\left)a\right]-\Pi cos[\pi u+(1-u\left)a\right]]\\ y=R[cos[\pi u+(1-u\left)a\right]+(\pi -a)usin[\pi u+(1-u\left)a\right]]\\ \phantom{\rule{5em}{0ex}}+\frac{rcos\phi}{\sqrt{1+{\Pi}^{2}}}[cos[\pi u+(1-u\left)a\right]+\Pi sin[\pi u+(1-u\left)a\right]]\\ z=rsin\phi \end{array}$

where again
$\Pi =(\pi -a)u(1-u)$ .
In order to evaluate both measures of the tube, the three variables

$\begin{array}{l}d{s}^{2}=d{x}^{2}+d{y}^{2}+d{z}^{2}\\ \phantom{d{s}^{2}}=d{r}^{2}+{r}^{2}d{\phi}^{2}+(1+{\Pi}^{2})[R+\frac{(1-u)(1+u+{\Pi}^{2})}{(1+{\Pi}^{2}{)}^{3/2}}rcos\phi {]}^{2}d{a}^{2}\end{array}$

For either $u=0$ or $u=1$ one has $\Pi =0$ , so that the complication in the metric goes away for the two end cases. In the former case the metric reverts as expected to that for a torus, and likewise in the latter to that for a cylinder.

Since the coefficient of the radial parameter is one in the metric, the transition from surface area to volume is trivial. The surface area itself is surprisingly simple

$\begin{array}{l}S=\underset{0}{\overset{2\pi}{\int}}da\underset{0}{\overset{2\pi}{\int}}d\phi \phantom{\rule{.3em}{0ex}}r(1+{\Pi}^{2})[R+\frac{(1-u)(1+u+{\Pi}^{2})}{(1+{\Pi}^{2}{)}^{3/2}}rcos\phi ]\\ \phantom{S}=2\pi rR\underset{0}{\overset{2\pi}{\int}}da[1+(\pi -a{)}^{2}{u}^{2}(1-u{)}^{2}]\\ S=4{\pi}^{2}rR[1+\frac{{\pi}^{2}}{3}{u}^{2}(1-u{)}^{2}]\end{array}$

because the integration over the angular variable removes the most complicated part of the integral. The volume is then one more radial integration:

$V=\underset{0}{\overset{r}{\int}}dr\phantom{\rule{.3em}{0ex}}S=2{\pi}^{2}{r}^{2}R[1+\frac{{\pi}^{2}}{3}{u}^{2}(1-u{)}^{2}]$

Curiously while the surface area and enclosed volume are equal for the torus and cylinder as stated at the outset, this is not true of intermediate states. Here is the function in square brackets that represents the deviation from the two end cases:

Not a dramatic deviation, but one all the same. Perhaps an interesting question would be if there exists a transformation that keeps these two measures constant for all values of the interpolating parameter.

*Uploaded 2020.07.17*
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