Make a curve smaller to find the volume of a shell So I've got to calculate the volume of steel necessary to make the carcass of the starship Enterprise. To to so I divided the ship into a few curves and used geogebra to make approximate regressions to find the curves. The first way I did it was calculating the surface of revolution of the many curves and then multiplicating by a certain height (5cm) to find a volume. But I was told by my teacher that I needed to make my work richer, to do so I had to calculate the volume of the carcass by another way. So the idea is to calculate the integral of the original curve and then find one curve that is "smaller" that goes "inside" the original one and calculate it's integral. The next step would be to subtract the integral from the smaller curve from the integral of the original curve. 
The problem is that I don't know how to find this "smaller" curve. I want to start with an exponential curve, and for each point on that curve I want to find the orthogonal vector and map that point a distance $a$ along the orthogonal vector. Is there a way for me to take this original curve and find this curve that has all the tangents parallel to the original one always keeping a defined constant distance? 

 A: We begin with $(x_1,y_1)$ on your exponential curve. For now, let's assume we are working with $e^x$ (the generalization should be clear, but I leave the work to the OP unless more help is needed.) Thus, we can write any point on our first curve as $(x_1,e^{x_1})$ where $y_2=e^{x_1}$. I will also assume that the constant distance along orthogonal line is $1$.
We clearly have that the orthogonal line is $y=e^{x_1}-e^{-x_1}(x-x_1)$. The most important thing here is the slope of $-e^{-x_1}$. What this slope means is that the difference in the $y$ coordinates of our point $(x_1, y_1)$ on the exponential curve and our point $(x_2, y_2)$, which we will call $\Delta y$, is equal to the change in the $x$-coordinates ($\Delta x$) scaled by $e^{-x_1}$. Symbolically, this means that
$$\Delta y = e^{-x_1} \Delta x$$
If we draw a triangle with hypotenuse $1$ (the length along orthogonal line), and legs $\Delta x$ and $\Delta y$ and substitute the above equality and the Pythagorean Theorem, we find that
$$(\Delta x)^2e^{-2x_1}+(\Delta x)^2 = 1 \implies \Delta x=\frac{1}{\sqrt{1+e^{-2x_1}}} \implies \Delta y=\frac{e^{-x_1}}{\sqrt{1+e^{-2x_1}}}$$
All that is left is to note the direction we want to transform $(x_1,y_1)$. The new curve will be above and to the left, meaning that we want to decrease $x_1$ by $\Delta x$ and we want to increase $y_1$ by $\Delta y$. Writing this out, we find that
$$x_2 = x_1-\frac{1}{\sqrt{1+e^{-2x_1}}}$$
$$y_2 = y_1+\frac{1}{\sqrt{1+e^{-2x_1}}} = e^{x_1}+\frac{1}{\sqrt{1+e^{-2x_1}}}$$
Note that, if we run along all of the values $x_1$ could take on, we draw out our new curve. Thus, we can replace $x_1$ with $t$ and get a parametric curve
$$\left(t-\frac{1}{\sqrt{1+e^{-2t}}},e^t+\frac{e^{-t}}{\sqrt{1+e^{-2t}}}\right)$$
We must let $t\in(-\infty,\infty)$ for this to draw out all of the needed points. Here is a picture from a Desmos graph I made. It shows the orthogonal line as well as the points where the line intersects both curves (the purple curve is the parametric function)
  
Here is a link to the Desmos graph I made where you can play around with this curve (which I would duplicate to your own Desmos profile if you want to keep it, as it may at some point be deleted by either myself or Desmos, though I tend to note delete my more important graphs) 
Unfortunately, this still doesn't solve your integration problem. In order to integrate you are going to use calculus on these parametric equations. In order to do this, I will quote this Wikipedia Article:

When a curve is defined by its parametric form $(x(t),y(t))$ in some
  interval $[a,b]$, the volumes of the solids generated by revolving the
  curve around the $x$-axis or the $y$-axis are given by
  $$V_x = \int_a^b \pi y^2 \frac{dx}{dt} dt$$
  $$V_y = \int_a^b \pi x^2 \frac{dy}{dt} dt$$

Subtract this from the volume you get revolving the outside curve along the interval you want, and you are golden. If you are afraid you made an error, just scale an exponential curve to be roughly the same shape as the parametric one you need and do the integration. If your answers are close, then you can sleep soundly!  
All that is left is for you to generalize the methods I used. Instead of $y_1 = e^{x_1}$ you will have that $y_1 = f(x_1)$, and instead of setting the hypotenuse of the triangle to be $1$ you will set it to be $c$ (where $c$ is your constant). The only trouble I can see is in scaling extra constants in $f(x_1)$, which we didn't have here. Let me know if you have trouble generalizing.
