Line and Surface Integrals

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.

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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.

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