Question Number 220228 by SdC355 last updated on 09/May/25
$$\mathrm{1}-\mathrm{form} \\ $$$$\alpha={A}_{{x}} \mathrm{d}{x}+{A}_{{y}} \mathrm{d}{y}+{A}_{{z}} \mathrm{d}{z} \\ $$$$\mathrm{d}\alpha=\mathrm{d}{A}_{{x}} \wedge\mathrm{d}{x}+\mathrm{d}{A}_{{y}} \wedge\mathrm{d}{y}+\mathrm{d}{A}_{{z}} \wedge\mathrm{d}{z} \\ $$$$\mathrm{d}\alpha=\left(\frac{\partial{A}_{{z}} }{\partial{y}}−\frac{\partial{A}_{{x}} }{\partial{z}}\right)\mathrm{d}{y}\wedge\mathrm{d}{z}+\left(\frac{\partial{A}_{{x}} }{\partial{z}}−\frac{\partial{A}_{{z}} }{\partial{x}}\right)\mathrm{d}{z}\wedge\mathrm{d}{x}+\left(\frac{\partial{A}_{{y}} }{\partial{x}}−\frac{\partial{A}_{{x}} }{\partial{y}}\right)\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\mathrm{2}-\mathrm{form} \\ $$$$\omega={B}_{{x}} \mathrm{d}{y}\wedge\mathrm{d}{z}+{B}_{{y}} \mathrm{d}{z}\wedge\mathrm{d}{x}+{B}_{{z}} \mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\mathrm{d}\omega=\mathrm{d}{B}_{{x}} \mathrm{d}{y}\wedge\mathrm{d}{z}+\mathrm{d}{B}_{{y}} \mathrm{d}{z}\wedge\mathrm{d}{x}+\mathrm{d}{B}_{{z}} \mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\mathrm{d}\omega=\left(\frac{\partial{B}_{{x}} }{\partial{x}}+\frac{\partial{B}_{{y}} }{\partial{y}}+\frac{\partial{B}_{{z}} }{\partial{z}}\right)\mathrm{d}{x}\wedge\mathrm{d}{y}\wedge\mathrm{d}{z} \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{process} \\ $$