# Projecting from volume to surface area of a sphere

Does it make sense to, and is there a formula for the sphere whose surface consists of all the points inside another sphere?

I realize that I'm trying to compare cubed units with squared units here but I think the assumption I'm making is that these spheres are more like clumps of sand, so another way to ask this question is: given a ball of sand, how big is the spherical shell I could construct out of grains of sand?

Hopefully this isn't nonsense.. I have a slight fear that it could be.

• There could be a formula, but it will depend upon the thickness of the shell.
– N74
Jan 30, 2017 at 22:37
• ... defined by inequalities $r^2 \leq x^2+y^2+z^2 \leq R^2$... Jan 30, 2017 at 22:49
• it this helps, in my thought experiment that this question is related to, I imagine a solid sphere of radius r whose volume is filled with tiny balls and then I imagine all those tiny balls moving to the surface of a sphere. I'd like to know the radius of that sphere as it related to r. FYI, this is for a sci-fi story I'm trying to write :) Jan 30, 2017 at 23:44

The region between two spheres of radius $R_{1} < R_{2}$ has volume $\frac{4}{3}\pi (R_{2}^{3} - R_{1}^{3})$. If it were filled with sand grains of side length $r \ll R_{1}$, and if these grains were rearranged into a spherical shell of radius $R$ and thickness $r$, the shell would have volume (roughly, to within a factor of $2$ or so due to gaps between grains and other "local" irregularities) $4\pi R^{2}r \approx \frac{4}{3}\pi (R_{2}^{3} - R_{1}^{3})$, or radius $$R \approx \sqrt{\frac{R_{2}^{3} - R_{1}^{3}}{3r}}.$$ For instance, a ball $1$ meter in radius ($R_{1} = 0$, $R_{2} = 1$) filled with $1$ millimeter sand grains ($r = 0.001$) would make a spherical shell about $$\sqrt{\frac{1}{0.003}} \approx 18.26\ \text{meters}$$ in radius.