General proof for square–cube law Can someone present a general (and easy) proof for square-cube law?
For similar objects 1 and 2,
$$
\frac{A_1}{A_2}=k^2 \ \mathrm{and} \ \frac{V_1}{V_2}=k^3,
$$
where k is the scale of objects 1 and 2.
 A: For a cube we have that

*

*$A_1=l_1^2$

*$V_1=l_1^3$
and

*

*$A_2=l_2^2$

*$V_2=l_2^3$
then by $k=\frac{l_1}{l_2}$
$$\frac{A_1}{A_2}=\left(\frac{l_1}{l_2}\right)^2=k^2 \ \mathrm{and} \ \frac{V_1}{V_2}=\left(\frac{l_1}{l_2}\right)^3=k^3$$
For a complex boby we can think to divide it in many small cubes and apply the same reasoning to obtain the same result as a limit.
A: Preliminaries:
One way to define similarity of two three-dimensional shapes $X$ and $Y$ would be to say that when you put $X$'s center of mass on the origin of the three-dimensional grid and scale the length of each coordinate axis by some number positive number $k$, you obtain $Y$, but possibly rotated. The scaling can be represented as the linear transformation
$$S = \begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{bmatrix}$$
while the rotation can also be represented as some orthogonal transformation $U$. In other words, there exists some rotation transformation $U$ and some scaling transformation $S$ such that $U(S(X)) = Y$.
Proof of the "Cube" Part:
Recall that
$$V(X) = \int_{X} dV = \iiint_X dx dy dz$$
and performing the change of variables
$$\vec{r} = \begin{bmatrix} x \\  y \\ z\end{bmatrix} \to U \cdot S \cdot \begin{bmatrix} x \\  y \\ z\end{bmatrix} = \begin{bmatrix} x' \\  y' \\ z' \end{bmatrix} = \vec{r}~'$$
the integral from above becomes
$$V(X) = \iiint_X dx dy dz = \iiint_{U(S(X))} \left|\frac{\partial \vec{r}}{\partial U(S(\vec{r}))}\right| dx' dy' dz' = \iiint_{Y} \left|U S \right|^{-1} dx' dy' dz'$$
But $|US| = |U|\cdot |S| = |S| = k^3$ as the determinant of any rotation matrix is one. Thus, we have
$$V(X) = \iiint_X dx dy dz = k^{-3} \iiint_{Y} dx' dy' dz' = \frac{V(Y)}{k^3} \iff V(Y) = k^3 \cdot V(X)$$
Proof of the "Square" Part:
As for the surface area, note that
$$A(X) = \iint_{\partial X} dA = \iint_{T} \left\|\frac{\partial \vec{x}(s, t)}{\partial s} \times \frac{\partial \vec{x}(s, t)}{\partial t}\right\| ds dt$$
where $\vec{x}: (s, t) \to (x, y, z)$ is parameterization of the surface of $X$ and $T$ is the appropriate region over which $(s, t)$ varies. Consider a similar parameterization $\vec{x}' : (s, t) \to U(S(x, y, z))$ where $(s, t)$ again varies in the region $T$. Clearly, this parameterization traces out the surface $Y$, namely
$$A(Y) = \iint_{T} \left\|\frac{\partial \vec{x}'(s, t)}{\partial s} \times \frac{\partial \vec{x}'(s, t)}{\partial t}\right\| ds dt$$
As $U$ and $S$ are just constant linear transforms, we may take them out of the derivative:
$$\iint_{T} \left\|\frac{\partial \vec{x}'(s, t)}{\partial s} \times \frac{\partial \vec{x}'(s, t)}{\partial t}\right\| ds dt = \iint_{T} \left\|U\left(S\left(\frac{\partial \vec{x}(s, t)}{\partial s}\right)\right) \times U\left(S\left(\frac{\partial \vec{x}(s, t)}{\partial t}\right)\right) \right\| ds dt $$
Recalling that all orthogonal transforms preserve lengths and angles and thus the magnitude of the cross-product, it follows that we can just ignore $U$ in the above expression. Additionally, $S$ merely scales its argument by $k$, so ultimately we have
\begin{align*}
\iint_{T} \left\|U\left(S\left(\frac{\partial \vec{x}(s, t)}{\partial s}\right)\right) \times U\left(S\left(\frac{\partial \vec{x}(s, t)}{\partial t}\right)\right) \right\| ds dt = \iint_{T} \left\|k \frac{\partial \vec{x}'(s, t)}{\partial s} \times  k \frac{\partial \vec{x}'(s, t)}{\partial t}\right\| ds dt \\ \\
= k^2 \iint_{T} \left\|\frac{\partial \vec{x}(s, t)}{\partial s} \times \frac{\partial \vec{x}(s, t)}{\partial t}\right\| ds dt = k^2 A(X)
\end{align*}
as desired. $\square$
