Let $A, B :[a,b]\subset\mathbb{R}\to\mathbb{R}^2$ be two functions of class $C^{1}([a,b])$ such that two segments (or intervals) $[A(t\_1),B(t\_1)]$ and $[A(t\_2), B(t\_2)]$ never intersect for $t\_1, t\_2\in [a,b], \ t\_1\neq t\_2$. Prove that the area of the region described by the segment $[AB]$, $R=\{[A(t),B(t)] \ |\ t\in[a,b]\}$, can be computed by the formula:


$$\text{Area}(R)=\int\_{a}^b \displaystyle\Big |\det\begin{pmatrix}x(t) & y(t) \\ x'(t) & y'(t) \end{pmatrix}\Big |\ \text{d}t$$


where $B(t)-A(t)=(x(t),y(t)), \ t\in[a,b]$.


P.S. For $U,V\in\mathbb{R}^2$, by the segment $[U,V]$ I mean the set $\{\lambda U+(1-\lambda)V,\ \lambda\in [0,1]\}$.


**Question:** Can there be found a similar formula in the space? (when $A, B :[a,b]\subset\mathbb{R}\to\mathbb{R}^3)$


I found this propetry in a book of *Giuseppe Peano, Applicazioni geometriche del calcolo infinitesimale, 1887, page 238*, but I see that his proof is not complete, and I found some really hard problems in trying to fullfil it. You can find the book here: <https://archive.org/details/applicazionigeo01peangoog>


Note that this property is a nice generalization of *Mamikon Theorem* (see the book *New Horizons in Geometry,
by Tom M. Apostol* or simply search on the internet), found long time ago by *Peano*. This is why I'm asking for a generalization in $\mathbb{R}^3$, because *Mamikon theorem* works for space curves too. For details, see that article, pages 27-28: <http://dolecki.perso.math.cnrs.fr/Peano=vulgarization_1307.pdf>



Old Peano theorem (demonstration is missing details)
$\newcommand{\bR}{\mathbb{R}}$
I think that there is something wrong with the formula you wrote. Suppose that


$$ A(t)=(t,0),\;\;B(t)=(t,1),\;\;\;t\in [0,1]. $$


In this case the segment $[A(t), B(t)]$ is the vertical segment from $(t,0)$ to $(t,1)$. The region $R$ is thus the unit square $[0,1]\times [0,1] \subset\bR^2$. We have $x=0$, $y=1$ so the integrand you wrote is $0$.


Here is one way to compute the area of the region $R$ in the general case. We introduce new coordinates $(t,s)$ on $R$ as follows. Let


$$A(t)= ( p(t), q(t) )\;\;\; t\in [a,b]. $$


Every point $(u,v)$ in $R$ can be uniquely described by the parameters $t,s$


$$ u= p(t)+s x(t),\;\;v=q(t)+s y(t), t\in [a,b], s\in [0,1]. $$


Geometrically to locate a point in $R$ we first need to understand on which of the segments $[A(t), B(t)]$ it lies. This is the role of the parameter $t$. The parameter $s$ fixes the position of the point on this segment.


We have


$$ du\wedge dv= \bigl(\dot p dt+s \dot{x} dt+xds\bigr)\wedge \bigl(\dot{q}dt+s\dot{y} dt +y ds\bigr) $$


$$=\bigl(\; (\dot{p}+s\dot{x})y-(\dot{q}+s\dot{y})x\;\bigr) dt\wedge ds $$


$$ =\underbrace{\bigl(\dot{p}y-\dot{q}x +s(\dot{x}y-\dot{y}x)\;\bigr)}\_{\rho(t,s)} \;dt\wedge ds. $$


We deduce that 


$$ {\rm Area}\;(R) =\pm\int\_{[a,b]\times [0,1]}\rho(t,s) dt dt= \pm\int\_a^b\bigl(\;(\dot{p}y-\dot{q}x ) +\frac{1}{2}(\dot{x}y-\dot{y}x)\;\bigr). $$


Note that in the special case described earlier we have $p=t$, $t=0$, $x=0$, $y=1$ $a=0$, $b=1$ and the above formula yields the correct answer.


Perhaps the generalization you are looking for is the coarea formula. It states that for a Riemann manifold $(M, g)$ and $f: M\to\bR$ we have


$$ {\rm vol}\;(M,g) =\int\_{\bR} \left(\int\_{f^{-1}(t)} \frac{1}{|\nabla^g f(p)|} dV\_{f^{-1}(t)}\right) dt. $$


Above $f^{-1}(t)$ is a smooth hypersurface for almost all $t$ and $dV\_{f^{-1}(t)}$ is the volume element on this hypersurface defined by the induced metric. In the case at hand, use the function $f:R\to\bR$ described in the coordinates by $f(t,s)=t$.


**Edit 1.** In $n$ dimensions $A(t)=(a^1,\dotsc, a^n(t))$, $X(t)=B(t)-A(t)=(x^1(t),\dotsc, x^n(t))$. Parametrize the surface $R$ as above


$$ u^k=a^k(t)+sx^k(t),\;\;k=1,\dotsc, n,\;\;t\in [a,b],\;\; s\in [0,1]. $$


We have


$$du^k =(\dot{a}^k+s\dot{x}^k) dt +x^kds,\;\; k=1,\dotsc, n, $$


$$ (du^k)^2= (\dot{a}^k+s\dot{x}^k)^2 dt^2 +2(\dot{a}^k+s\dot{x}^k)x^k dtds+(x^k)^2 ds^2,\;\;k=1,\dotsc, n. $$


The induced metric on $R$ is then


$$\sum\_k (du^k)^2= \underbrace{\left(\sum\_k(\dot{a}^k+s\dot{x}^k)^2 \right)}\_{E}\,dt^2+ 2\;\underbrace{\left(\sum\_k (\dot{a}^k+s\dot{x}^k)x^k\right)}\_{F}\;dtds+\underbrace{\left(\sum\_k (x^k)^2\right)}\_{G}\; ds^2. $$


The area element on $R$ is then


$$dS=\sqrt{EG-F^2} dt ds. $$


