geometryintegrationsolid of revolution
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.
Draw a picture. We have a parabola whose axis of symmetry is the line $y=2$, and which opens leftward. The apex of the parabola is at $(4,2)$.
The line $y=1$ divides the part of the parabola to the right of the $y$-axis into two parts, the fat part above the line $y=1$, and a much thinner part below the line $y=1$. It is not clear from the wording which part is being spun. We will assume it is the fat part. Minor modification will take care of things if it is the thinner part.
It is most convenient to use the Method of Cylindrical Shells. Look at a thin horizontal strip, going from height $y$ to height "$y+dy$."
Spin this strip about the $x$-axis. We get a cylindrical shell of thickness "$dy$." The shell has radius $y$ and height $x=4y-y^2$. So the shell has volume approximately equal to $2\pi y(4y-y^2)\,dy$. "Add up" (integrate) from $y=1$ to $y=4$. We get volume $$\int_{y=1}^4 2\pi y(4y-y^2)\,dy.$$
Remark: If we want to integrate with respect to $x$, things get a little more complicated. You did it along those lines, so we write down suitable expressions. The volume from $x=0$ to $x=3$ is equal to $$\int_{x=0}^3 \pi[(2+\sqrt{4-x})^2-1^2]\,dx.$$ To this we must add the volume from $x=3$ to $x=4$, which is $$\int_{x=3}^4 \pi[(2+\sqrt{4-x})^2-(2-\sqrt{4-x})^2]\,dx.$$
Referring to the diagram below the exercise is to find the volume of the solid of revolution when revolving the blue region about the $x$-axis.
If one varies $x$ between the values of $-\sqrt{2}$ and $\sqrt{2}$ and uses the annulus method then
\begin{eqnarray} V&=&\int_{-\sqrt{2}}^{\sqrt{2}}\pi(R^2-r^2)\,dx\\ &=&2\pi\int_0^{\sqrt{2}}R^2-r^2\,dx\tag{1} \end{eqnarray}
where $R=\sqrt{4-x^2}$ and $r=x$. This gives a value
\begin{equation} V=\frac{16\sqrt{2}\pi}{3} \end{equation}
As a check we can also find the volume of the solid of revolution formed by revolving the red region about the $x$-axis. The blue plus red volumes should sum to the volume $V=\dfrac{32\pi}{3}$.
The "red volume" can be found by the cylindrical shell method. We will use the sector on the right and double the result so that
\begin{equation} V=2\int_0^\sqrt{2}2\pi rh\,dy \end{equation}
where $r=y$ and $h=\sqrt{4-y^2}-y$. So
\begin{equation} V=4\pi\int_0^\sqrt{2}y\sqrt{4-y^2}-y^2\,dy \end{equation}
which has a value of
\begin{equation} V=\frac{32\pi}{3}-\frac{16\sqrt{2}\pi}{3} \end{equation}
And we see that the "blue volume" and "red volume" sum correctly to the volume of the sphere.
Addendum: Note that if one integrates equation $(1)$ over the range $[-2,2]$ that will give the volume of the solid of revolution formed by revolving both the green and the blue regions (with the green volume subtracted from the blue volume), which is not the volume asked for in the question.
Best Answer
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