We have seen that (in $\RR^3$) \begin{align} F &= \FF\cdot d\rr \\ {*}F &= \FF\cdot d\AA \end{align} when we integrate both sides. This restriction leads us to a slightly different interpretation of integrands than you may be used to. Integrands by themselves can be thought of as being defined everywhere; it is only after you decide where to integrate them that they are restricted to the domain of integration. If two integrands are equal after being integrated over any domain, we say that the integrands themselves are equal. It is in this sense that equality holds in the expression above.

So much for line and surface integrals. What about volume integrals? The volume element $dV$ in $n$ dimensions is just the orientation $\omega={*}1$, so we have \begin{equation} f\,dV = f\,\omega = f\,{*}1 = {*}f \end{equation} or simply \begin{equation} {*}f = f\,dV \end{equation}

Finally, since we now know how to integrate $F$, ${*}F$, and ${*}f$, it seems natural to ask whether we can integrate $f$. Be careful! This is $f$ the 0-form, not $f\,dx$ the 1-form! A 0-form must be integrated over a 0-dimensional “surface”, that is, over one or more points. So we define \begin{equation} \int_p f = f(p) \end{equation}