Is it true that two 3D bodies of different shapes cannot have the same volume to area ratio unless both have exactly the same volume and area?

Question

I found this elegant physics question in the Q/A section of the research gate. My own judgment is that this is true except, and only except, for complete spherical shapes. . In other words, it applies well to both hemispheres and to all shapes other than solid spheres. I tried to test the accuracy of this question to no avail. The exception of the sphere versus the cube puzzles me. The reason may be the physical component of the question: the cube and the sphere both have a degree of freedom! I don't have a strict explanation. Why?

Answer 1

If I got you question right, I think that your assertion is false. Take an object 1 with Volume/Area ratio equal to $\frac{V_1}{A_1} = R_1$ and a second object with $\frac{V_1}{A_1} = R_2$. Two remarks before going on:

• in general $R_1 \ne R_2$;

• the Volume/Area ratio of a family of scaled objects is proportional to a characteristic dimension $\ell_1$ of the object of the family,

  $\frac{V_i}{A_i} = R_i(\ell_i) = K_i \ell_i$,

  where the constant $K_i$ depends on the shape of the objects.

- as the dimension of the object goes to zero $\ell \rightarrow 0$, the Volume/Area ration goes to zero as well,

$\lim_{\ell_i \rightarrow 0} \frac{V_i}{A_i}(\ell_i) = \lim_{\ell_i \rightarrow 0} R(\ell_i) = K_i \lim_{\ell_i \rightarrow 0} \ell_i = 0$.

Comparing the volume to area ratio of the two families of objects to find the ratio of the characteristic length we get the following equation

$\dfrac{V_1}{A_1} = \dfrac{V_2}{A_2}$$ \qquad \rightarrow \qquad$$K_1 \ell_1 = K_2 \ell_2$$ \qquad \rightarrow \qquad$$\dfrac{\ell_2}{\ell_1} = \dfrac{K_1}{K_2}$.

Answer 2

This is obviously false as @QiaochuYuan says. Take any two shapes from this Wikipedia page under the heading Mathematical examples. They have the SA/V ratio parametrized by the shape's characteristic length.

For instance:

SA/V for a tetrahedron is $\rho_{\text{tetrahedron}} = {6\sqrt {6} \over {a}}$ where $a$ is the edge length.

SA/V for a octahedron is $\rho_{\text{octahedron}} = {3\sqrt {6} \over {b}}$ where $b$ is the edge length.

These may be scaled so that ${\rho_{\text{tetrahedron}} \over \rho_{\text{octahedron}}} = 1$.

The shapes have different volumes, surface areas, but their SA/V ratios are equal.