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Given a circle $C$ in the xz-plane which does not intersect the $z$-axis, we can build a smooth 2-torus with surface area $(2\pi a)(2\pi b)$ where $a$ is the radius of the circle $C$ and $b$ is the distance from the $z$-axis to the center of $C$.

Now, a circle has rotational symmetry and the surface area formula in this case becomes a product of arclengths. I am wondering under which conditions a surface area of a surface of revolution remains a product of arclengths. For instance, suppose that rather than $C$ being a circle, we instead take $C$ to be an ellipse, which we can rotate in the $xz$-plane before revolving about the $z$-axis. Will the surface area of the resulting surface $S$ of revolution be invariant under rotations in the $xz$-plane which fix the center of $C$ before revolving $C$ around the $z$-axis [provided that $C$ does not intersect the $z$-axis in the $xz-$plane]? What if instead of $C$ being a smooth (quadratic) curve, we take $C$ to be a convex polygon in the $xz$-plane which does not intersect the $z$-axis. Is the surface area of the resulting surface $S$ of revolution invariant under rotating $C$ in the $xz$-plane while fixing the center point, before revolving around the $z$-axis?

Essentially, I am wondering under which conditions the surface area of a surface $S$ of revolution is given by a product of arclengths. Here, for instance, when we revolve $C$ [contained in the $xz$-plane] around the $z$-axis, the center point of $C$ traverses a circle around the $z$-axis as we revolve. What if rather than using a circle to create a surface of revolution, we “revolve” $C$ around the $z$-axis using an ellipse [which projects to an ellipse in the $xy$-plane]: will the surface area still be a product of arclengths?

It feels like I'm circling (pardon the pun) around a theorem relating curvature to the surface area of $S$. We have, for instance, Gauss–Bonnet: $$ 2\pi \chi = \int_S \mathcal{K}dS $$ and I'm wondering whether the vanishing of either side of Gauss–Bonnet can tell us that the surface area is a product of arclengths. In the case that $C$ is a convex polytope (polygon) in the $xz$-plane, the curvature of $C$ is concentrated at the vertices, and I'm wondering whether rotating $C$ in the $xz$-plane before we revolve around the $z$-axis can create non-zero curvature on $S$ which therefore tells us that the surface area is no longer a product of arclengths.

Ultimately, I am asking: does there exist a simple closed curve $C$ in the $xz$ plane which does not intersect the $z$ axis for which we can create a surface of revolution $S$ by revolving $C$ around the $z$-axis such that the surface area of $S$ is not the arclength of $C$ multiplied by $2\pi b$ where $b$ is the distance from the $z$-axis to the “center” of $C$. I am especially interested in the case that $C$ is not smooth [just piecewise linear, for instance].

Edit/Update: Let's fix a definition of $b$ as the distance from the $z$-axis to the centroid of the convex hull of $C$ in the $xz$-plane.

Let's fix a definition of “center” as centroid of the convex hull of $C$ in the $xz$-plane.

Another question here is what happens to surface area of $S$ if we rotate $C$ as we revolve [does surface area only “see” integer number of rotations per revolution, for instance? Does it see anything at all?]. But I suppose that is a separate question.

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    $\begingroup$ If you're going to allow $b$ to be as vague as "the distance from the z-axis to the center of $C$ in some sense", then of course there will be a suitable value of $b$. You're just defining it by asking it to reproduce the surface area of $S$. $\endgroup$
    – Michael Engelhardt
    Commented Dec 12, 2022 at 2:05
  • $\begingroup$ I have in mind a definition of center here as something more like "barycenter," or the centroid of the convex hull of $C$ $\endgroup$
    – locally trivial
    Commented Dec 12, 2022 at 2:24
  • $\begingroup$ I suppose the theorem I'm reaching for is along the lines of ``the n-volume of a product of n 1-manifolds is given by the product of their 1-volumes," in the smooth case. And then I'm wondering under what conditions this still holds in the non-smooth case, for a product of piecewise-smooth curves [or manifolds with corners]. More generally, given manifolds M_1, \dots, M_r of dimension d_1, \dots, d_r with D=∑di then relating the D volume of the product to the product of the d_i-volumes, and then how-far into the non-smooth case does this still hold? [Is there a theorem related to this?] $\endgroup$
    – locally trivial
    Commented Dec 12, 2022 at 2:50
  • $\begingroup$ You can make $b$ arbitrarily close to the minimal distance of $C$ from the z-axis (by having $C$ have many wrinkles there), or arbitrarily close to the maximal distance of $C$ from the z-axis (by having $C$ have many wrinkles there), or anywhere in between. $\endgroup$
    – Michael Engelhardt
    Commented Dec 12, 2022 at 6:52
  • $\begingroup$ I am confused: If $C$ is a circle then b is not the minimal distance of C to the z-axis; it's the distance from the z axis to the center of the circle. The surface area for the torus being (2pi a)(2pi b) only works for this specific value of b. What do you mean by ``having many wrinkles there?" Do you mean that if we concentrate curvature of C near this point for b (in the xz-plane) [i.e., nearest the z-axis, or furthest away from the z-axis, respectively] then we can have these given values for b? What are your assumptions on C? [just continuous, for instance?] $\endgroup$
    – locally trivial
    Commented Dec 12, 2022 at 17:36

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No, the surface area of the surface of revolution $S$ is in general not given by the arc length of $C$ multiplied by $2\pi b$, with $b$ the distance of the centroid of the convex hull of $C$ from the axis. As a simple counterexample, take $C$ to be the union of a semicircle of radius $R$ and a straight segment connecting the endpoints of the semicircle. Place the straight segment parallel to the axis, at a distance $a$ from the latter, and have the semicircle face outward, away from the axis. Then: $$ b=a+\frac{4}{3\pi } R $$ The arc length is $$ L=(2+\pi )R $$ The surface area of $S$ is $$ \Sigma = 2\pi (2+\pi ) aR + 4\pi R^2 $$ and therefore $\Sigma \neq 2\pi b L$.

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  • $\begingroup$ The correct alternative definition of $b$ which will make $\Sigma =2\pi b L$ (not just for this example, but in general) is to take $b$ to be the distance of the centroid of $C$ itself from the axis, not the one of its convex hull. $\endgroup$
    – Michael Engelhardt
    Commented Dec 13, 2022 at 6:17
  • $\begingroup$ So if we define ``center" as centroid of $C$, and $b$ to be the distance from the $z-$axis to the center of $C$, then you are saying that surface area will be invariant under rotating $C$ in the $xz$-plane before revolving $C$ around the $z$-axis, provided that such rotation fix the centroid of $C$. There are no smoothness assumptions on $C$ [for instance, continuity suffices]. I suspect that if we rotate as we revolve, then surface area can change, but that is a separate question. $\endgroup$
    – locally trivial
    Commented Dec 13, 2022 at 16:21
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    $\begingroup$ @Locallytrivial - I fully agree. $\endgroup$
    – Michael Engelhardt
    Commented Dec 13, 2022 at 16:27

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