# 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?

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I'm not 100% sure. But I think it is correct. –  Erik Jun 26 '12 at 11:15
Your volume seems to agree with a differential form of [Pappus' centroid theorem][1] . The first theorem would give $$A=2\pi r \int_{x_1}^{x_2}\sqrt{1+(f'(x)^2}\;dx.$$ But I don't know if the theorem relies on completing the solid of revolution [1]: en.wikipedia.org/wiki/Pappus%27s_centroid_theorem –  Ross Millikan Jun 26 '12 at 13:26
@RossMillikan: As a sense, I feel the formulas are so but I know that parallel curves may not be a function. (Please see my previous question:math.stackexchange.com/questions/146675/parallel-functions) How can it affect formulas if parallel curve is not a function? –  Mathlover Jun 26 '12 at 13:55
I don't think this matters unless your r is greater than the radius of curvature of your function. In that case you will get overlap as shown in the figures on your previous question. –  Ross Millikan Jun 26 '12 at 14:02
if so, you think it is not the general formulas what we wrote above for volume and surface for all cases. They are limited with overlap status. If so, How can we define general formula? –  Mathlover Jun 26 '12 at 14:05

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 $$V=\int_0^L\int_0^r\int_0^{2\pi}J_{\bf f}(s,t,\phi)\ {\rm d}(s,t,\phi)=2\pi{r^2\over 2}L\qquad\Bigl(=\pi r^2\int_a^b\sqrt{1+f'(x)^2}\ dx\Bigr)\ .$$
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.