Equilateral triangle revolution

It's a simple application of the second Pappus's centroid theorem , which states that: the volume of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area A of F and the distance d traveled by the geometric centroid of F , in formula:

$V=d\cdot A$

So we only need the position of the geometric centre (and its distance from the rotation axis) and the area of the triangle.

The centre it's easily found at distance $\frac{r}{2\sqrt{3}}$ from the $x$ -axis, so the distance from the rotation axis is $\rho=r+\frac{r}{2\sqrt{3}}=\frac{2\sqrt{3}+1}{2\sqrt{3}}r$ , and the distance traveled by the centroid is $d=2\pi\rho$ .

Now it's the turn of the area: $A=\frac{1}{2} r h=\frac{1}{2} r\cdot \frac{\sqrt3}{2}= \frac{\sqrt3}{4} r^2$ .

So the volume is:

$V=d\cdot A=2\pi\cdot \frac{2\sqrt{3}+1}{2\sqrt{3}}r \cdot \frac{\sqrt3}{4} r^2 = \pi\cdot \frac{2\sqrt{3}+1}{2} \cdot \frac{1}{2} r^3 = \pi r^3 \left( \frac{2\sqrt{3}+1}{4} \right)$ ,

and finally we found that $c=\frac{2\sqrt{3}+1}{4}\approx1.116$