How to get the trace surface of an n-gon rotated around a central point with constant orientation (integral) First post here. Tried to find an answer myself for 2 days but have failed, maybe someone here can help me, or give me a pointer at least.
Let $H\subset\Re^n$ be a hypersphere, for this first example that is a circle with $r=1$ in vectorspace $\Re^2$.
The second object is a n-dimensional convex set $C\subset\Re^n$, like an polygon in vectorspace $\Re^2$, for example a triangle/square/hexagon.
If we translate $C$ across all points of $H$, what is the n-dimensional size of all points traced by this action, so what is the area in $\Re^2$, the volume in $\Re^3$...
Note that the transforms performed are strictly translations. $C$ is never rotated and always maintains the same orientation.
Also the radius of $C$ and $H$ are always set in a way that the resulting object does not intersect with itself.
I'd wager we need some sort of integral to do this, but I can't figure out which and how.
I know for a truly ring shaped object, or anything with circular symmetry I could simply use a transformation to polar coordinates in $\Re^2$.
Is something similar to this possible for a roughly ring shaped object traced by a polygon?
Specifically I'm thinking of an hexagon with $r=0.5$ traced around a ring with $r=1$ for the first example in $\Re^2$.
Then after that is the question how to do that in $\Re^n$. I'd appreciate any answer.
Visual representation
Edit: Added a picture for the 2D part of the question. The black line is the circle around which we move a shape, here shown with a hexagon/triangle and some copies and an arrow to demonstrate the movement. The orientation of the hexagon can't change, and the traced object does not intersect with itself.
Edit2: I substantially edited my question to be clearer.
 A: This problem belongs to integral geometry, a field somewhat remote from the mathematical mainstream. But be aware that this field has provided the mathematical foundations of modern medical imaging.
Let $C\subset{\mathbb R}^2$ be a compact convex set with nonempty interior and boundary curve $\partial C$. If $C$ is moved along a circle of sufficiently large radius $R$ in the described way an annular region $A\subset{\mathbb R}^2$ is traced out. It turns out that the area of $A$ is given by the following magical formula:
$${\rm area}(A)=2R\cdot L(\partial C)\ .\tag{1}$$
A quick sanity check: If $C$ is a circular disc of radius $r$ then $A$ is a circular annulus of outer radius $R+r$ and inner radius $R-r$. One therefore has
$${\rm area}(A)=\pi\bigl((R+r)^2-(R-r)^2\bigr)=4\pi R r=2R\cdot 2\pi r\ .\quad\checkmark$$
Idea of proof. Assume that $\partial C$ is smooth. During the motion of $C$ at each instant there is a front  $F\subset\partial C$ which occupies new territory, whereby this front is continuuosly changing. Consider an infinitesimal parallel translation $d{\bf z}$. Then the newly traced territory is a narrow band  of area $|d{\bf z}|\cdot w$, whereby $w$ is the width of the momentaneous front (hence of $C$) measured orthogonally to $d{\bf z}$. This already indicates that ${\rm area}(A)=2\pi R\cdot \overline{w}$, whereby $\overline{w}$ denotes the mean width of $C$ over all possible directions. It is a fact that $$\overline{w}={1\over\pi}L(\partial C)\ ,\tag{2}$$
so that $(1)$ follows.
But we can avoid the use of $(2)$: Consider a line element $ds$ on $\partial C$. This line element is lying on the front during exactly half of the geometrical process. (Here we use that we move along a circle.) The new territory traced out by this line element is  $\pi R\cdot|ds|$ times the mean value of $\sin$ over the interval $[0,\pi]$, which is ${2\over\pi}$. It follows that our line element traces out $2R\cdot|ds|$ new area, and since this is true for all line elements on $\partial C$ we again arrive at $(1)$. 
