# A "significant" solid with volume $\frac{\sqrt{3}}{3}\pi r^3$?

In a painting:

there is this formula of volume:

$$V = \frac{\sqrt{3}}{3}\pi r^3$$

It seems to me this is the formula of the volume of some polyhedron inscribed or circumscribed to a sphere of radius $$r$$, but I am not expert of the field.

I had the task of finding the meaning of the different formulas, that were taken from a web site whose address has been lost, so I need to find the regular solid that corresponds to such formula, if it exists.

So I am asking help from somebody more expert then me in the field of geometry of solids, if there is some "significant" solid with the volume given by that formula.

Note: this question is twin of another question.

• I think this question needs to be reformulated in order not to be closed. The solid could be anything. For example, any cone with a circular base of radius $r$ and height $\sqrt{3} r$ (e.g. upright cone with base angle $\pi/3$) will do. Commented Feb 1, 2020 at 13:05
• The appearance of $\pi$ suggests that this is not the formula for the volume of a (flat-sided) polyhedron, whether inscribed or circumscribed about a sphere. Rather, there must be a circular component. @LeeDavidChungLin's suggestion —a cone of radius $r$ and height $\sqrt{3}r$— seems to be the "most natural" one to me. (The desired volume is $\frac13\cdot \pi r^2\cdot \sqrt{3} r$; ie, $\frac13\cdot(\text{area of base})\cdot(\text{height})$.) I'll note that such a cone arises from revolving an equilateral triangle of side $2r$ about an axis of symmetry.
– Blue
Commented Feb 1, 2020 at 13:39
• Why are you not stating the webpage? What is there to hide? Commented Feb 6, 2020 at 3:21
• You asked this question just last week, not months ago. Why should you be interested in a formula on a webpage that you forgot? Commented Feb 6, 2020 at 6:02
• Next time, include such information in your questions. I've done it for you here. Please edit your other question to include the same information. Commented Feb 9, 2020 at 6:33

The appearance of $$\pi$$ suggests that this is not the formula for the volume of a (flat-sided, straight-edged) polyhedron, whether inscribed or circumscribed about a sphere. Rather, there must be a circular component.
@LeeDavidChungLin's commented suggestion ---a cone of radius $$r$$ and height $$\sqrt{3}\,r$$--- seems to be the "most natural" one: the target volume decomposes as $$\frac13\cdot \pi r^2 \cdot \sqrt{3}\,r \;=\; \frac13\cdot(\text{area of base})\cdot(\text{height})$$
I'll note that such a cone arises from revolving an equilateral triangle of side $$2r$$ about an axis of symmetry.