There is clearly a relationship between $0$-forms and $3$-forms, given by relating $f$ and $f\,dV$. We can therefore define a map from $0$-forms to $3$-forms, which we write as \begin{equation} *f = f\,dV \end{equation} Similarly, relating $\FF\cdot d\rr$ and $\FF\cdot d\AA$ yields a relationship between $1$-forms and $2$-forms, which we write as \begin{equation} *(\FF\cdot d\rr) = \FF\cdot d\AA \end{equation} In three (Euclidean) dimensions, we also write $*$ for the inverse maps, so that \begin{equation} *(f\,dV) = f \end{equation} and \begin{equation} *(\FF\cdot d\AA) = \FF\cdot d\rr \end{equation}

We can work out the action of the Hodge dual map $*$ on a basis at each rank. The standard basis of $1$-forms in $\RR^3$ is the set $\{dx,dy,dz\}$; any 1-form can be expressed as a linear combination of these basis $1$-forms, with coefficients that are functions. Similarly, the standard basis of $2$-forms is $\{dy\wedge dz,dz\wedge dx,dx\wedge dy\}$, and the standard basis of $3$-forms is $\{dx\wedge dy\wedge dz\}$. Finally, the standard basis of $0$-forms is just $\{1\}$, as any function is a multiple of the constant function $1$ (with a coefficient that is a function). The action of $*$ is therefore given by \begin{align} *1 &= dx\wedge dy\wedge dz \\ *dx &= dy\wedge dz \\ *dy &= dz\wedge dx \\ *dz &= dx\wedge dy \\ *(dy\wedge dz) &= dx \\ *(dz\wedge dx) &= dy \\ *(dx\wedge dy) &= dz \\ *(dx\wedge dy\wedge dz) &= 1 \end{align} along with the linearity property \begin{equation} *(f\,\alpha+\beta) = f\,{*}\alpha + {*}\beta \end{equation} for any function $f$ and any $p$-forms $\alpha$, $\beta$.