Prove that under an affine transformation of space the ratio…
Prove that under an affine transformation of space the ratio of the volumes of any two bodies is invariant.
Explanation of Method
Under an affine transformation of space the ratio of the volumes of two bodies is invariant.
Step 1: Parallelepipeds (lemma)
A parallelepiped with edge vectors $\vec u,\vec v,\vec w$ has volume $V=|\det(\vec u\,|\,\vec v\,|\,\vec w)|$. Its image has edges $M\vec u, M\vec v, M\vec w$ (Problem 171, step 1), so
\[ V'=|\det\big(M(\vec u\,|\,\vec v\,|\,\vec w)\big)|=|\det M|\cdot V . \]
Step 2: Arbitrary bodies
Via the change of variables. The Jacobian of $\vec r\,'=M\vec r+\vec b$ is the constant $\det M$, so
\[ V(\Omega')=\iiint_{\Omega'}dx'\,dy'\,dz'=\iiint_{\Omega}|\det M|\,dx\,dy\,dz=|\det M|\,V(\Omega). \]
Elementary version. Approximate $\Omega$ by small cubes. Each cube's volume is multiplied by the same factor $|\det M|$, by the lemma. Passing to the limit gives the same formula.
\[V(\Omega')=|\det M|\cdot V(\Omega)\quad\text{for every body }\Omega\]
Step 3: The ratio
\[ \frac{V(\Omega_1')}{V(\Omega_2')}=\frac{|\det M|\,V(\Omega_1)}{|\det M|\,V(\Omega_2)}=\frac{V(\Omega_1)}{V(\Omega_2)} . \]
\[\frac{V_1'}{V_2'}=\frac{V_1}{V_2}\qquad\blacksquare\]
The key point is that the scaling factor is the same constant for all bodies. For a non-affine map the Jacobian varies from point to point, and ratios of volumes change. The plane analogue is that affine maps preserve ratios of areas, with factor $|\det M|$ for $2\times2$ matrices.
Result
\[\boxed{\,V(\Omega')=|\det M|\cdot V(\Omega)\quad\text{for every body }\Omega\,}\]
\[\boxed{\,\frac{V_1'}{V_2'}=\frac{V_1}{V_2}\qquad\blacksquare\,}\]
Remark
Example: in any tetrahedron the medial tetrahedron and similar constructions keep their volume ratios under affine maps. This is why such ratios can be computed on a convenient (e.g. regular or right-angled) tetrahedron.