# 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.

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:

• 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

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

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

• A Conical [Frustum] 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: $$R_1 = \frac{A}{2}$$ $$R_2 = \frac{B}{2}$$
• Did you try yourself? – Qwerty Jul 4 '16 at 15:34
• Could you be more specific about the cones? – Moti Jul 4 '16 at 16:34
• Yes, and my results didn't add upp. I made the asumption that the cones where directly constructed by the diagonals of the cylinder A. Which might have been the fault, either that or something else went wrong. I realizer now that there would be a point in adding my approach in th question, thanks to you @Qwerty I will make a little update. – Victor Ekekrantz Jul 4 '16 at 16:45
• Yes, I could. I will add the specification in the question to make it easier to read. @Moti – Victor Ekekrantz Jul 4 '16 at 16:47
• Suggest you solve first the challenge of intersection of cylinder and cone. – Moti Jul 5 '16 at 0:42

## 1 Answer

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 $$V=V_a-V_b-2V_f=\frac{\pi A^3}{4}-\frac{\pi B^3}{4}-\frac{A-B}{3}\left(\frac{\pi A^2}{4}+\frac{\pi B^2}{4}+\frac{\pi}{4}A B\right)$$

• Very good declaration of the solution, thank you @Wouter :) – Victor Ekekrantz Jul 4 '16 at 19:13
• You can try also the second theorem of Pappus: mathworld.wolfram.com/PappussCentroidTheorem.html – cgiovanardi Jul 4 '16 at 22:56
• This is not right - you need to subtract the INTERSECTION OF THE CONE AND CYLINDER. – Moti Jul 5 '16 at 0:41
• It's a conical frustum - the tiny cones, the intersection of the cones with cylinder $B$, are already subtracted. – Wouter Jul 5 '16 at 4:59
• According to the picture this is not the case. The cone and the cylinder are intersecting and the "cone" base is shaped by the cylinder - think of it as a combustion turbine that need to move in a cylinder. Thus the volume? – Moti Jul 8 '16 at 7:02