Even Archimedes knew that if one takes a 2D disc $D^2$ and a strip of constant width inside it (two parallel chords and the space between them) and moves it around within the disc, then area of the strip will change, obviously. But if you construct a hemisphere 'above' the disc (such that the boundary of the disc $\mathbb{S}^1$ becomes the boundary of the hemisphere, and the disc itself is the diametrical one), the area of the projection of the strip onto the hemisphere will not change because of the curvature of the hemisphere.
Well, there appears a generalisation of the process discussed above: take a convex (for starting simplicity to make all the chords lie entirely within the figure) figure $\Omega$ on a plane $\mathbb{R}^2$, take a strip of constant width $l>0$, and construct such a regular, compact, convex surface (with the boundary orthogonal to the plane of the figure) which would preserve the area of the projection of the strip, if you move it in some (!) direction (so, the surface will, obviously, change, if you change the direction of movement). Let us call the property of preserving the area of the projection of a strip in some direction an Archimedes's property towards $\Omega$.
First of all, I tried to prove the Archimedes's result without knowing the answer (as one of my teachers offered), i.e. to counstruct the surface (the (hemi)sphere) without knowing that it must be the answer. I took a square (the answer for the square is obvious and is 'a box without two sides') and began approaching the $\mathbb{S}^1$ with inscribed regular polygons, such that the surface I wanted to build would have to be a convex one (I 'fractured' the sides of the box in a way it is shown down in the scheme). Then I wanted to prove that if the number of vertices increases, the surface gets more and more spherical, but (!) I do not know how to include the strip projection area-preserving condition. However, how do we know that the surface has to be even convex at least?
Then, there is a hypothesis of mine about such Archimedes's surfaces:
$\mathbf{Hypothesis}$. Any regular closed strictly convex surface has a section with a plane, such that one of the pieces cut off by the plane has an Archimedes's property towards the figure in the section.
So, how can one construct/build an actual Archimedes's surface for any planar convex figure?
UPD 02.03.2023: There appeared an idea to keep the problem planar, inducing some planar metric of a projection of the surface for which we are looking for. Hence, for a square, the surface is the intersection of the sphere and the cube with the square in the basis.
Probably not the kind of surface you expected, but for a square with the strip parallel to a diagonal a surface that works is a sort of double pyramid (see figure below). $BG$ and $DH$ are perpendicular to the square and their length is a half the length of a diagonal. The projection of strip $EMNF$ is the yellow surface, and you can check that its area doesn't depend on the position of the strip.