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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?

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  • $\begingroup$ When you construct the two objects from identical material, does this material fill the inside of the objects or are you only making a thin layer of that material over the surface of each object? $\endgroup$
    – David K
    Nov 11 '20 at 19:43
  • $\begingroup$ @DavidK I was imagining that the volume, outer surface area, and material are consistent between both objects, so the thickness of the material is physically limited to the entire volume of the [smaller] object. But I see what you're saying; by having two solid objects of equal density the math would be much simpler. So, that would be a fine way to approach it as well. $\endgroup$
    – Jamesfo
    Nov 11 '20 at 20:29
  • $\begingroup$ Equal volume and equal density will give you equal mass. That's why I asked. $\endgroup$
    – David K
    Nov 11 '20 at 21:19
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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).

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$$V_{cube}=a^3=\frac{4{\pi}r^3}{3}=V{sphere}$$

$$\frac{a^3}{r^3}=\frac{4{\pi}}{3}$$

$$\frac{a}{r}=\sqrt[3]{\frac{4{\pi}}{3}}$$

$$\frac{{Area_{cube}}}{{Area_{sphere}}}=\frac{6a^2}{4{\pi}r^2}=\sqrt[3]{\frac{216\times 16{\pi}^2}{9 \times64{\pi}^3}}=\sqrt[3]{\frac{6}{\pi}}>1$$

$$Area_{cube} > Area_{sphere}$$

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