I'm just curious, this is an answer I came up in statistics, meaning that the radius of said sphere is $-\root3\of\pi\sigma$ (yes, negative), where $\sigma$ is the standard deviation. I feel like this is a good connection between statistics and geometry, and I wanted some light on what exactly means to divide a volume of sphere by $\pi$, or the area of the unit circle. The formula now is $-\dfrac 43\sigma^3=K$. I know that $V(\sigma)=\dfrac43\pi\sigma^3=-\pi K$ is proportional to $K$ with proportionality of $-\pi$, but that tells me nothing. Well, not nothing, there is a bijection between arbitrary spheres and all $\sigma$, but it seems obvious. I also know that $K=\dfrac{-V(\sigma)}{\pi}$, but this is just from definition. Maybe it will light for some of you. Thanks.
By the way, this is where $-\dfrac43\sigma^3$ come from:
$$\int_{\mu-\sigma}^{\mu+\sigma}(x-\mu)^2-\sigma^2dx=-\dfrac43\sigma^3=\dfrac{V(\sigma)}{-\pi}=\dfrac{\dfrac43(\sigma\root3\of\pi)^3}{-\pi}$$