How to calculate the volume of the solid generated by revolving an arbitrary closed shape around an axis? I got a arbitrary and complex closed shape in a plane which I will revolve around an axis (The distance between the center of mass of the shape and the axis is known). Since this shape is arbitrary is not obvious how to define it using a function(s), I only know its area. Is it possible to calculate the volume of revolution?
Note. There are quite a lot of existing questions about computing the volume of revolution solids but they use a function to define the shape of the solid.
 A: Call $r_k$ the distance from the axis of an elemental area $\Delta A_k$.
Then the elemental volume it generates in a revolution would be approximately
$$
\Delta V_{\,k}  = 2\pi \,r_{\,k} \, \Delta A_{\,k} 
$$
Now, suppose that you have a figure composed by $n$ elemental areas, so
$$
\left\{ \matrix{
  A = \sum\limits_{k = 1}^n {\Delta A_{\,k} }  \hfill \cr 
  V \approx 2\pi \sum\limits_{k = 1}^n {\,r_{\,k} \,\Delta A_{\,k} }  \hfill \cr}  \right.
$$
The distance of the barycenter (centroid) from the axis is by definition
$$
r_{\,g}  = {1 \over {\sum\limits_{k = 1}^n {\Delta A_{\,k} } }}\sum\limits_{k = 1}^n {\,r_{\,k} \,\Delta A_{\,k} }  = {1 \over A}\sum\limits_{k = 1}^n {\,r_{\,k} \,\Delta A_{\,k} } 
$$
so that 
$$
V \approx 2\pi \,r_{\,g} \,A
$$
Can you proceed from here same as for Riemann sum ?
Note that if the axis crosses the area, and you take a revolution of $2 \pi$, then the above formula
will provide the volume between the more external surface and the more internal, which will define a "cavity".
Otherwise, when the axis crosses the area, you shall take the centroid of the "left" and "right" part separately
and turn by $\pi$ radians only, .. etc.
See the Pappus Theorem
