Let \(D_{1}\) and \(D_{2}\) be two compact, convex polyhedra in the oriented \(k\)-dimensional space \(R^{k}\) and \(f: D_{1} \rightarrow D_{2}\) a differentiable map which is an orientation-preserving diffeomorphism \({ }^{55}\) of the interior of \(D_{1}\) onto the interior of \(D_{2}\). Then, for any differential \(k\)-form \(\omega^{k}\) on \(D_{2}\). $$ \int_{D_{1}} f^{*} \omega^{k}=\int_{D_{2}} \omega^{k} $$

Question: Prove that for two compact, convex polyhedra \(D_1\) and \(D_2\) in \(k\)-dimensional space and an orientation-preserving diffeomorphism \(f:D_1 \rightarrow D_2\), the integral of the pullback by \(f\) of a differential \(k\)-form \(\omega^k\) on \(D_2\) equals the integral of \(\omega^k\) on \(D_2\). That is, \(\int_{D_1} f^*\omega^k = \int_{D_2} \omega^k\). Answer: The proof relies on understanding the pullback of a differential form by a differentiable map, and utilizing the change of variables formula for integration along with the fact that the diffeomorphism \(f\) preserves orientation. By substituting the relationship between volume forms, we find that \(\int_{D_1} f^*\omega^k = \int_{D_2} \omega^k\), thus proving the desired equality.