We have some arbitrary shape in $n$ dimensional space. That shape has some $n$-dimensional measure, $A$. We make a copy of the shape and move it orthogonal to the original in the $n+1$-st dimension a distance of $h$. The $n+1$ dimensional measure of this new cylindrical shape is $A.h$. Now we translate the second $n$ dimensional shape, keeping it in its $n$ dimensional plane to an arbitrary place all the while maintaining the connections with the original shape. The $n+1$ dimensional measure is still $A.h$. How can I prove this?
This fact was mentioned to me for $n=2$ in high school. But they left out the proof. This is shown for the special case of a 3-dimensional cylinder below.
In terms of my attempt, I was able to show it for $n=1$ when we can only have a line. In two dimensional space, moving the "clone" of the line leads to a parallelogram. In the figure below, line AB is cloned to DC and we get a rectangle. Then DC is moved to FE. The parallelogram ABEF has the same area as ABCD because it gains the area BEFX and loses the area AXCD. But the areas of BEFX and AXCD are the same. This is because BEFX is the triangle BEC minus the triangle XFC. Similarly, AXCD is the triangle AFD subtracted by the triangle CXF. But the triangles BEC and AFD are congruent, thus having the same area.
I can also do this for a parallelopipid in 3 dimensional space in a rather painful manner. But I haven't been able to prove this for an arbitrary area in 2-d space (like even a circle) let alone $n$ dimensional space.
Mathematically, this is a consequence of Fubini's theorem. Suppose $C$ is the original shape in $\mathbb{R}^{n + 1}$ and $C'$ is the new shape. Then writing $x \in \mathbb{R}^{n + 1}$ as $(x', x_{n + 1})$, by Fubini's theorem, \begin{align} m_{n + 1}(C') &= \int_{\mathbb{R}^{n + 1}}1_{C'}(x', x_{n + 1})\,dx \\ &= \int_{\mathbb{R}}\int_{\mathbb{R}^n} 1_{C'}(x', x_{n + 1})\, dx' \, dx_{n + 1} \\ &= \int_{\mathbb{R}}m_n(C'_{x_{n+1}})\,dx_{n + 1}. \end{align} Here $C'_s$ is the cross section of $C'$ at $x_{n + 1} = s$, i.e. $C'_{s} = \{x' \in \mathbb{R}^n : (x', s) \in C'\}$. In your construction of $C'$, for each $s$, $C'_s$ is a translate of $C_s$, so $C_s$ and $C'_s$ have the same measure. By the above equation, $C$ and $C'$ have the same measure.