Assume for the moment that your plane curve $\gamma$ is parametrized by arc length: $\gamma:\quad s\mapsto\bigl(u(s),v(s),0\bigr)\qquad(0\leq s\leq L)\ .$ Then the body of your pipe has the following parametrization: ${\bf f}:\quad (s,t,\phi)\mapsto\left\{\eqalign{x&=u-t\dot v\cos\phi\cr y&=v+t\dot u\cos\phi\cr z&=t\sin\phi\cr}\right.\quad ,$ and putting $t:=r$ you get a parametrization of the surface of the pipe. Using the Frenet formulas $\ddot u=-\kappa\dot v$, $\ \ddot v=\kappa \dot u$, where $\kappa=\kappa(s)$ denotes the curvature of $\gamma$, we obtain $\eqalign{{\bf f}_s&=\bigl(\dot u(1-t\kappa\cos\phi),\dot v(1-t\kappa\cos\phi),0\bigr)\ ,\cr {\bf f}_t&=(-\dot v\cos\phi,\dot u\cos\phi,\sin\phi)\ ,\cr {\bf f}_\phi&=(\dot v t\sin\phi, -\dot u t\sin\phi, t\cos\phi)\ .\cr}$ From these equations one computes ${\bf f}_\phi\times{\bf f}_s=(1-t\kappa\cos\phi)\bigl(-\dot v t\cos\phi,\dot u t\cos\phi, t\sin\phi\bigr)\ ,\qquad |{\bf f}_\phi\times{\bf f}_s|=t(1-t\kappa\cos\phi)\ ,$ and $J_{\bf f}={\bf f}_t\cdot({\bf f}_\phi\times{\bf f}_s)=t(1-t\kappa\cos\phi)\ .$ The surface of the pipe now computes to $\omega=\int_0^L\int_0^{2\pi}|{\bf f}_\phi\times{\bf f}_s|_{t:=r}\ {\rm d}(s,\phi)=2\pi r L\qquad\Bigl(=2\pi r\int_a^b\sqrt{1+f'(x)^2}\ dx\Bigr)\ ,$ and its volume to $V=\int_0^L\int_0^r\int_0^{2\pi}J_{\bf f}(s,t,\phi)\ {\rm d}(s,t,\phi)=2\pi{r^2\over 2}L\qquad\Bigl(=\pi r^2\int_a^b\sqrt{1+f'(x)^2}\ dx\Bigr)\ .$
These computations show that your conjectured formulas are indeed true: The gain in volume and surface on the outside of a bend of the pipe is exactly outweighed by the loss on the inside.
In all of this we have tacitly assumed that ${\bf f}$ is injective in the considered domain. This is guaranteed as long as $\ r \kappa(s)<1$ $\ (0\leq s\leq L)$. If this condition is not fulfilled we have "overlap", i.e., the map ${\bf f}$ producing the body of the pipe is no longer injective. In this case the integral $I:=\int_0^L\int_0^r\int_0^{2\pi}J_{\bf f}(s,t,\phi)\ {\rm d}(s,t,\phi)$ is no longer equal to the actual volume of the pipe but it is equal to a "weighted" volume where each volume element ${\rm d}(x,y,z)$ is counted as many times as it is covered by the representation. Computing the actual volume will be difficult in such a case, insofar as one might have to deal with pieces of envelope surfaces turning up in the process.