Sorry if the title is vague but I can't think of anything better. What follows is the (translated) text of a question from an old examination.
Let $\gamma:[0,L]\to \mathbb{R}^3$ be a $C^3$ regular curve with $\|\dot{\gamma}\|\equiv 1$. For all $t \in [0,L]$, let $E_{t}$ be a compact set such that $E_t$ lies in the affine plane by $\gamma(t)$, orthogonal to $\dot{\gamma}(t)$. Suppose also that the barycenter of $E_t$ is $\gamma(t)$. Finally, suppose $E_t\cap E_{t'}=\emptyset$ for all $t\neq t'$. Then if $E:=\bigcup_{t\in[0,L]}E_t$, prove that (assuming suitable regularity conditions for the family of sets $\left\{E_t\right\}$) $$ |E|=\int_{0}^{L}|E_t|dt$$
where with $|\cdot|$ I mean the measure of the set (either volume or area, whatever makes sense).
I know that if $\gamma$ is a line segment, then this is just a consequence of Fubini's theorem. But if $\gamma$ is an arbitrary curve? My idea would be to use a regular parametrization of each compact $E_t$ with a map $P_t:\Omega \to \mathbb{R}^3$, where $\Omega\subset \mathbb{R}^2$ is a fixed compact for all $t$; it doesn't hurt to assume that $(0,0)\in \Omega$ and $P_t(0,0)=\gamma(t)$ too. Then (since the $E_t$ are mutually disjoint, and $\gamma$ is regular) I can construct a regular parametrization for $E$ given by $$P:[0,L] \times \Omega \to \mathbb{R}^3,\quad P(t,x,y):=P_t(x,y)$$ Then $$\int_{0}^{L}|E_t|dt = \int_{0}^{L}\left[\int_{\Omega}\left\|\frac{\partial P}{\partial x}\wedge \frac{\partial P}{\partial y}\right\|dxdy \right]dt$$ $$ |E|=\int_{[0,L]}\left[\int_{\Omega}|\det D_P(t,x,y)|dxdy\right]dt=\int_{0}^{L}\left[\int_{\Omega}\left|\left\langle\frac{\partial P}{\partial t},\frac{\partial P}{\partial x}\wedge \frac{\partial P}{\partial y}\right \rangle\right|dxdy\right]dt$$ It only remains to show that $$\left|\left \langle \frac{\partial P}{\partial t},\frac{\partial P}{\partial x}\wedge \frac{\partial P}{\partial y}\right \rangle\right| =\left\|\frac{\partial P}{\partial x}\wedge \frac{\partial P}{\partial y}\right\|$$ Which is true, provided that $\frac{\partial P}{\partial t} $ is parallel to $\dot{\gamma}(t)$ (which is orthogonal to both $\frac{\partial P}{\partial x}$ and $\frac{\partial P}{\partial y}$) and that $\left\|\frac{\partial P}{\partial t}\right\|=1$. But this seems to be true only for the points in the form $(\gamma(t),0,0)=P(t,0,0)$. I suppose I could add these as regularity conditions, but I think I don't really have a precise idea of what I'm doing. Is such an assumption actually reasonable? Or, is there a proof which can work under weaker assumptions?