Differential forms in three (Euclidean) dimensions are just the integrands discussed in the previous section. 1) A 1-form $F$ is the integrand of a line integral, that is \begin{equation} F = \FF\cdot d\rr \end{equation} where we deliberately use the same base letter $F$ for the vector field and the corresponding 1-form. A 2-form $\alpha$ is the integrand of a surface integral, that is \begin{equation} \alpha = \FF\cdot d\AA \end{equation} and a 3-form $\beta$ is the integrand of a volume integral, that is \begin{equation} \beta = f\,dV \end{equation} Finally, a 0-form $f$ is the “integrand” of the 0-dimensional integral defined in the previous section, namely $f$ itself.

Examples of 0-forms are easy: Let $f$ be any (reasonable) function. Each 0-form (function) $f$ also defines a 3-form, which in rectangular coordinates takes the form \begin{equation} \beta = f\,dx\,dy\,dz \end{equation} Each vector field $\FF$ defines a 1-form, which in rectangular coordinates takes the form \begin{equation} F = F_x\,dx + F_y\,dy + F_z\,dz \end{equation} But what are 2-forms?

Suppose the surface $S$ is the $xy$-plane, given by $z=0$. Then clearly \begin{equation} d\AA = dx\,\xhat \times dy\,\yhat = dx\,dy\,\zhat \end{equation} so that \begin{equation} \FF\cdot d\AA = F_z\,dx\,dy \end{equation} Repeating this argument for the other two coordinate planes correctly suggests that the 2-form defined by $\FF$ is given by \begin{equation} \alpha = F_x\,dy\,dz + F_y\,dz\,dx + F_z\,dx\,dy \end{equation}