I’ve developed a 3D rendering program for bodies of non-spherical revolution, by which I mean that as the curve is rotated about the vertical axis it is modulated by an arbitrary closed curve. Moreover, the curve itself may change with the azimuth. An example body is shown in the animation below. The still figure is the planiform curve. Every radial profile is unique, even before the modulation.
I wish to calculate the volume and surface area of such a shape. I know the equations of the vertical ribs and so can calculate for example, the area of the radial cross-sections. What I attempted was to deconstruct Pappus’s theorems to the differential level and reconstruct them for non-circular paths. I arrived at the following $$\begin{matrix} dV=C\left( \theta \right)A\left( \theta \right)d\theta & V=\int_{0}^{2\pi }{\hat{C}\left( \theta \right)\hat{A}\left( \theta \right){{g}^{2}}\left( \theta \right)d\theta } \\ dS=c\left( \theta \right)L\left( \theta \right)d\theta & S=\int_{0}^{2\pi }{\hat{c}\left( \theta \right)\hat{L}\left( \theta \right)\sqrt{{{g}^{2}}\left( \theta \right)+{{{\dot{g}}}^{2}}\left( \theta \right)}\,\,d\theta } \\ \end{matrix}$$
where $g(\theta)$ is the normalized (or dimensionless, if you prefer) closed curve, $A(\theta)$ is area of the radial section, $C(\theta)$ is its centroid, $L(\theta)$ is its length, and $c(\theta)$ is its centroid. The circumflex indicates the raw vertical profile, i.e., before the modulation. Note that these quantities themselves have to be calculated by regular one-dimensional integrals. (Also, all centroids are relative the vertical axis.)
I can derive the volume equation from the standard definition of volume in cylindrical coordinates. The surface area integral above is an ansatz solution which I need help with. I cannot find a general surface area equation in cylindrical coordinates. The ones I find are specifically for circular cylinders. I arrived at the ansatz by looking at the equation for the perimeter of a curve and superposing a vertical component on it.
I’ve tested the above integrals against solids for which I know the volume and/or surface area and they seem to be correct (so far). However, I would like to somehow verify the ansatz solution.
Here’s an interesting tidbit for those who are interested. Consider a simple case where there’s a single vertical profile, e..g., $\hat C(\theta) \hat A(\theta) = \hat C\hat A$, a constant. Then
$$V=\hat{C}\,\hat{A}\,\int_{0}^{2\pi }{{{g}^{2}}\left( \theta \right)d\theta }$$
But this integral is just twice the planiform area $A_p$. Thus, $V=2\hat C\hat A A_p$. Compare this to the regular body of revolution, which by Pappus’s $2^{nd}$ theorem is $V_{rot}=2\pi \hat C\hat A$.
This means that the volume for any closed curve is just $V=V_{rot} \frac{A_p}{\pi}$. I think that’s interesting.
