Given two objects of equal volume, one cubic and one spherical, which will have the greater outer surface area, and how much greater will it be versus the other object? I know this is a math forum, so a mathematical explanation is great but I'm also hoping for layman's terms and a simple ratio.
As a bonus, assuming the objects are constructed from identical material, if we know the mass of one can we know the mass of the other by using the same ratio? Why or why not?
On
$$V_{cube}=a^3=\frac{4{\pi}r^3}{3}=V{sphere}$$
$$\frac{a^3}{r^3}=\frac{4{\pi}}{3}$$
$$\frac{a}{r}=\sqrt[3]{\frac{4{\pi}}{3}}$$
$$\frac{{Area_{cube}}}{{Area_{sphere}}}=\frac{6a^2}{4{\pi}r^2}=\sqrt[3]{\frac{216\times 16{\pi}^2}{9 \times64{\pi}^3}}=\sqrt[3]{\frac{6}{\pi}}>1$$
$$Area_{cube} > Area_{sphere}$$
The volume of a sphere with radius $r$ is $V_{sphere} = \frac{4}{3}\pi r^3$. The volume of a cube with side $s$ is $V_{cube} = s^3$. If you set these equal, you see that: $s^3 = \frac{4}{3}\pi r^3$ or $s = (\frac{4}{3}\pi)^{1/3} r$.
The surface area of a sphere is $S_{sphere} = 4\pi r^2$, and the surface area of a cube is $S_{cube} = 6s^2$. Using the expression for $s$ from above, $S_{cube} = 6(\frac{4}{3}\pi)^{2/3}r^2$.
Thus, $\frac{S_{cube}}{S_{sphere}}= \frac{6(\frac{4}{3}\pi)^{2/3}}{4\pi} \approx 1.2,$ so the surface area of the cube is greater.
This is not at all surprising since a sphere has the smallest surface area of all surfaces that enclose a given volume (see: https://en.wikipedia.org/wiki/Sphere). As an interesting aside, since animals give off heat in proportion to their surface area but generate heat in proportion to their volume, animals living in colder climates tend to be rounder than their relatives in warmer climates (see: https://www.amazon.com/Engineering-Animals-How-Life-Works/dp/0674048547).