If the line integral can be interpreted as the area under the curve $C$ on a scalar field, then the surface integral would be the volume under a surface on a scalar field right?
If so, whats the difference between a surface and volume integrals?
[Math] Whats the difference between surface integral and volume integral
calculus
Related Solutions
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.
Given the multitude of different comments and given the first answer saying that your formula "hold in general", I would like to emphasize: It is NOT true that $V'(r)=S(r)$ for the Volume $V$ and Surface $S$ of general convex bodies scaled by $r$.
What is true, is that locally
$S\approx V/h$,
where locally means that we are considering a small piece of surface, small enough that it is approximately a piece of a plane, and denote by $S$ its area and by $V$ the volume of the body that you get from extending the surface along its normal direction by length $h$ (normal = orthogonal to the surface)
When you want to consider the global case, that is you want to "sum up" the local formulae, the sphere is a special case because when you enlarge the sphere then all small pieces of surface as described above are by the same normal length $h$. For other convex bodies the summing up does not work.
The problem with general convex bodies is that when you enlarge the body some of the small pieces of surface are extended more than others. For example, if you try to apply your formula to rectangles with side lengths $2r$ and $r$ (and center in the origin), you get the încorrect formula $S(r)=V'(r)=(2r^2)'=4r$. The difference to the correct formula $S(r)=6r$ comes from the fact that dividing by $h$, as you do, is "unjust" to the long sides. These long sides get extended by only $h/2$ in their normal direction when $r$ is enlarged to $r+h$.
Best Answer
Yes, you are right. The two dimensional integral $\int\! f(x,y)\,dx\,dy$ can be thought of as finding the volume to the body that is described by the function $f(x,y)$ and $z=0$ (and the integration intervals). Any one or two dimensional integral can be visualized like that: $x$ and $y$ axis for the arguments of the integrand and $z$ axis for the value of the integrand at position $x,y$. This fails for three dimensional integrals because we would need a fourth dimension to visualize the value of the integrand.
Where is the difference? Well the integrated structure has different dimensions for surface and volume integrals. The Riemannian sum corresponding to a surface integral devides the surface into small squares (or other shape) and sums the value for those squares, while the volume integrals acts on a body and devides it into small cubes (or other 3-dimensional shape) and sums the values for those cubes.
When I first worked with three dimensional integrals I found it useful to "misuse" the average of a function to imagine what the result of the integral will be (be warned that this is very subjective of course). Imagine a integral of the function $f(x,y,z)$ over some three dimensional shape $S$ with volume $V$. The average of the function $f$ over this shape is given as $$ \left<f\right>_S = \frac{1}{V} \int_S\!f(x,y,z)\,dx\,dy\,dz $$ therefore the integral itself is given as the product of the average value of the function $f$ times the volume of the integrated shape $$ \int_S\!f(x,y,z)\,dx\,dy\,dz = \left<f\right>_S \cdot V . $$ Maybe this helps to get a feeling for those integrals.