The volume and surface area of pipe? A line segment turns around a curve with right angle from point A to point B.
I would like to find  the closed region volume and surface area that figured out in the picture. 

Could you please give me hint how to define the volume and surface area with integrals ?
Is it correct that volume formula is as shown below?
$$V=\pi r^2\int _{x_1}^{x_2} dS=\pi r^2\int _{x_1}^{x_2} \sqrt{1+(f'(x))^2}dx$$
I do not know how to define the surface of the shape?
UPDATE:
important note: the offset curves that are the parallels of a function may not be functions: My related questions about parallel functions:
Parallel functions.
What is the limit distance to the base function if offset curve is a function too?
Thanks a lot for answers and advice. 
 A: 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.
