How should this volume be calculated? It's the volume within a cylinder minus two cones and another cylinder. I would need some help getting my thinking straight here. I have made a summation of what's needed to know to solve this very problem that's been troubeling me for a while.
Since it's a geometric problem it's not exactly easy to describe, I will add info as soon as it's sought after. PS; There's a picture linked down below in the description.
Description: This problem involves calculating tha volume of the space between the outer cylinder A and the inner cylinder B plus two cone like shapes who's constructed between the circumference of A and the circumference of B.
I really recommend looking at the very draft picture linked down below.


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*Graphical description of the figure
The greystriped area would describe the searched volume if it where to me rotaded around the vertical axex.
For the two cylinders drawn in the figure it follows for both that they have the same height as diamiter.
Diamiter A = Height A
Diamiter B = Height B
Useful formulas:


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*Volume of a cylinder: $V=\pi r^2 h$

*Volume of a cone: $V=\frac{\pi r^2 h}{3}$


Result form: The Volume is to be presented using the following variables
$$V\ is\ Volume$$
$$A\ is\ the\ mesure\ of\ cylinder\ A$$
$$B\ is\ the\ mesure\ of\ cylinder\ B$$

Updates


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*The B cylinder and the A cylinder share the same midpoint.

*The cones are to be described more precisely Conical Frustums.


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*A Conical [Frustum][3] could be described as an segment of a cone. In the case of this problem the bottom conical frustum it's formed between the bottom circle in cylinder A and the bottom circle of cylinder B. Here is a descriptive picture, these variables follows the picture: $$R_1 = \frac{A}{2}$$ $$R_2 = \frac{B}{2}$$



Indeed, the volume should be the volume of $A$, minus the volume of $B$, minus twice the volume of a conical frustum. The height of the conical frustum is $h=(A-B)/2$.
Thus
$$V_A=\pi(A/2)^2A=\frac{\pi A^3}{4}$$
$$V_B=\frac{\pi B^3}{4}$$
$$V_f=\frac{h}{3}\left(\frac{\pi A^2}{4}+\frac{\pi B^2}{4}+\sqrt{\frac{\pi A^2}{4} \frac{\pi B^2}{4}}\right)$$
(see volume of conical frustum)
$$V_f=\frac{h}{3}\left(\frac{\pi A^2}{4}+\frac{\pi B^2}{4}+\frac{\pi}{4}A B\right)$$
Putting it all together
$$V=V_a-V_b-2V_f=\frac{\pi A^3}{4}-\frac{\pi B^3}{4}-\frac{A-B}{3}\left(\frac{\pi A^2}{4}+\frac{\pi B^2}{4}+\frac{\pi}{4}A B\right)$$
