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.



Conditions for surface area of surface of revolution to be product of arclengths
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 $\Sigma \neq 2\pi b L$.


