We began our study of differential forms by claiming that differential forms are integrands. Which ones?

Using the results of the last section, line integrals are easy. We have \begin{equation} \int_C \FF\cdot d\rr = \int_C F \end{equation} where of course \begin{equation} F = \FF\cdot d\rr \end{equation} is the 1-form corresponding to $\FF$.

What about surface integrals? Consider first some simple examples. What is the flux of $\FF$ upward through the $xy$-plane? We have \begin{equation} \int_S \FF\cdot d\AA = \int_S \FF\cdot \zhat\,dx\,dy = \int_S F_z\,dx\,dy \end{equation} where $F_z$ is the $z$-component of $\FF$. The integrand on the right-hand side can be reinterpreted as $F_z\,dx\wedge dy$, which is one of the components of \begin{equation} {*}F = F_x \,dy\wedge dz + F_y \,dz\wedge dx + F_z \,dx\wedge dy \end{equation} But since $z=0$ on $S$, $dz=0$ there, and we have \begin{equation} \int_S \FF\cdot d\AA = \int_S {*}F \end{equation}

A similar argument in spherical coordinates for the flux of $\FF$ out of a sphere centered at the origin results in \begin{equation} \int_S \FF\cdot d\AA = \int_S F_r \,r^2\sin\theta\,d\theta\,d\phi \end{equation} and again the integrand is just ${*}F$ restricted to the sphere, since $r=\hbox{constant}$ implies that $dr=0$. We are thus led to conjecture that \begin{equation} {*}F = \FF\cdot d\AA \end{equation} One more example should suffice.

Figure 1: Chopping up a surface in rectangular coordinates.

Consider the graph of a function of the form $z=f(x,y)$. What is the surface element on such a surface? As shown in Figure 1, we chop the surface along curves with $y=\hbox{constant}$ and $x=\hbox{constant}$, and consider $d\rr$ along such curves, obtaining \begin{align} d\rr_1 &= \left( \xhat + \Partial{z}{x}\,\zhat \right) dx \\ d\rr_2 &= \left( \yhat + \Partial{z}{y}\,\zhat \right) dy \end{align} so that \begin{equation} d\AA = d\rr_1 \times d\rr_2 = \left( -\Partial{z}{x}\,\xhat - \Partial{z}{y}\,\yhat + \zhat \right) dx\,dy \end{equation} and \begin{equation} \FF\cdot d\AA = \left( -F_x \Partial{z}{x} - F_y\Partial{z}{y} + F_z \right) dx\,dy \end{equation} But \begin{equation} dz = \Partial{z}{x}\,dx + \Partial{z}{y}\,dy \end{equation} so that \begin{align} {*}F &= F_x \,dy\wedge dz + F_y \,dz\wedge dx + F_z \,dx\wedge dy \nonumber\\ &= \left( -F_x \Partial{z}{x} - F_y\Partial{z}{y} + F_z \right) dx\wedge dy \end{align} when restricted to the surface.

The above examples were all 3-dimensional, but the argument is the same in $n$ dimensions (and arbitrary signature, although the orientation must be checked). Line integrals are just as you'd expect, but surface integrals now correspond to the flux through an ($n-1$)-dimensional surface — exactly what you would expect, since ${*}F$ is an ($n-1$)-form.