Let $S$ be a surface embedded in $\mathbb R^3$. In elementary Calculus, we defined the surface integral
$$\int_S \mathbf F \cdot d\mathbf a,$$
where $\mathbf F=(F_1, F_2, F_3)$ is a vector-valued function. How can I interpret this in terms of modern differential geometry?
I tried as follows. First, since we should integrate over differential forms and also $S$ is a 2-manifold, it is natural to define a 2-form
$$F = F_1 dy\wedge dz + F_2 dz\wedge dx + F_3 dx\wedge dy.$$
Then, I suspect that
$$\int_S \mathbf F \cdot d\mathbf a = \int_S F.$$
Is this true? I also found out that the LHS is
$$\int_S \mathbf F \cdot d\mathbf a = \int_S \langle *F, N \rangle \mathrm{vol}_S,$$
where $*$ denotes the Hodge dual, $N$ denotes the outward normal vector, $\langle \cdot,\cdot\rangle$ denotes the pairing between 1-form and vector, and $\mathrm{vol}_S$ denotes the (induced) volume form on $S$.
However, I don't know how to proceed.
Best Answer
To answer the question directly, yes, in terms of forms, \begin{align} \int_S\mathbf{F}\cdot d\mathbf{a}&=\int_S\iota_S^*\left(F_1\,dy\wedge dz+F_2\,dz\wedge dx+F_3\,dx\wedge dy\right) \end{align} where $\iota_S:S\to\Bbb{R}^3$ simply is the inclusion, and we're pulling back a $2$-form on $\Bbb{R}^3$ to a $2$-form on $S$.
It may be more instructive to look at the situation in generality. So, let $(M,g)$ be an $n$-dimensional Riemannian manifold which is oriented with volume form $\mu_g$, which we may on occasion write as $dV_g$ (it is also common to write this as $\mu_M$ when the metric $g$ is understood), even though it is not the exterior derivative of anything. Recall that $\mu_g$ is characterized by the fact that for any $p\in M$ and any positively-oriented ordered, orthonormal basis $\{\xi_1,\dots, \xi_n\}$ of $T_pM$, we have \begin{align} \mu_g(\xi_1,\dots, \xi_n)&=1. \end{align}
Let $S\subset M$ be a smooth $(n-1)$-dimensional embedded submanifold (i.e a smooth hypersurface), and equip it with the induced Riemannian metric $\iota^*g$. Suppose also that there is a smooth unit normal $N$ defined on $S$ (i.e $N:S\to TM$ is smooth and for each $p\in S$, $N(p)\in (T_pS)^{\perp}$). With this, one typically defines an orientation on $S$ by saying that an ordered basis $\{\xi_1,\dots, \xi_{n-1}\}$ of $T_pS$ is positively-oriented if and only if $\{N(p), \xi_1,\dots, \xi_{n-1}\}$ is a positively-oriented ordered basis for $T_pM$ (i.e $\mu_g$ evaluated on the $n$ vectors in this order is strictly positive).
Since $(S,\iota^*g)$ is an oriented Riemannian manifold in its own right, it has its own volume form, $\mu_{\iota^*g}$ (other common notations include $\mu_S$ or $dS$ or $dA$ or $da$ or $d^{n-1}V$ or $dV_{\iota^*g}$ etc). Now, we can investigate the relationship between the induced volume form on $S$ and the volume form on $M$. For this, let $F:S\to TM$ be a smooth map such that for each $p\in S$, $F(p)\in T_pM$ (i.e a field of vectors defined on $S$, but tangent to $M$; we're not requiring it to be tangent to $S$). Let us also define the normal and tangential components of $F$ in the usual way: \begin{align} F_{\perp}&:= g(F,N)\,N\equiv \langle F,N\rangle N\quad \text{and}\quad F_{\parallel}:= F-F_{\perp} \end{align} Let $p\in M$ and let $\{\xi_1,\dots, \xi_{n-1}\}$ be any basis for $T_pM$. We now look at the interior product: \begin{align} (F_{\parallel}\,\lrcorner \,\mu_g) (\xi_1,\dots, \xi_{n-1})&:= \mu_g(F_{\parallel}(p), \xi_1,\dots, \xi_{n-1})=0, \end{align} where the last equality is because $F_{\parallel}(p)\in T_pS$ is tangent to $S$, meaning it can be written as a linear combination of the $\xi_i$'s and hence each summand vanishes (because the same $\xi$ appears twice in the evaluation of $\mu_g$). Therefore, the interior product of $F_{\parallel}$ with $\mu_g$ vanishes, or rearrranging this equation, we have \begin{align} F\lrcorner\,\, \mu_g &= \langle F,N\rangle\, (N\lrcorner\,\, \mu_g) =\langle F,N\rangle\,\, \mu_{\iota^*g},\tag{$*$} \end{align} where the last equality can be easily verified by the characterization of Riemannian volume forms I mentioned above: take any $p\in S$ and any positively-oriented, ordered, orthonormal basis $\{\xi_1,\dots, \xi_{n-1}\}$ of $T_pS$. Then, \begin{align} (N\,\lrcorner\,\,\mu_{g})(\xi_1,\dots, \xi_{n-1})&= \mu_g(N(p),\xi_1,\dots, \xi_{n-1})=+1, \end{align} where the final equality is because $\{N(p),\xi_1,\dots, \xi_{n-1}\}$ is an ordered orthonormal basis for $T_pM$, which is also positively-oriented by definition of the orientation on $S$. Note also, that $\mu_{\iota^*g}$ is the unique such form. Hence, $\mu_{\iota^*g}= N\,\lrcorner \, \mu_g$, as claimed above in $(*)$.
Now, let us connect this all back to your question. We have by $(*)$ that \begin{align} \mathbf{F}\cdot d\mathbf{a}&\equiv\langle F,N\rangle\, \mu_{\iota^*g}=F\,\lrcorner\,\mu_g\equiv F\,\lrcorner \,\, dV_g. \tag{$**$} \end{align} Here, the first and last equal signs are just notation, only the middle one is significant. Hence, \begin{align} \int_S\,\mathbf{F}\cdot d\mathbf{a}&= \int_S F\,\lrcorner\,dV_g. \end{align} If we want, we can now extract the formula in terms of local coordinates. Let $(U,x= (x^1,\dots, x^n))$ be a positively-oriented coordinate chart. Then, we can write \begin{align} \mu_g\equiv dV_g&= \sqrt{|\det (g_{ij})|}\,dx^1\wedge \cdots \wedge dx^n\quad \text{and}\quad F=\sum_{i=1}^nF^i\frac{\partial}{\partial x^i} \end{align} Here, $F^i:S\cap U\to\Bbb{R}$ are smooth functions. So, the interior product is given by \begin{align} F\,\lrcorner\,dV_g&=\left(\sum_{i=1}^nF^i\frac{\partial}{\partial x^i}\right)\,\lrcorner \sqrt{|\det g|}\,dx^1\wedge \cdots \wedge dx^n\\ &=\sum_{i=1}^n \sqrt{|\det g|}F^i \left(\frac{\partial}{\partial x^i}\,\lrcorner\,dx^1\wedge \cdots\wedge dx^n\right)\\ &=\sum_{i=1}^n\sqrt{|\det g|}F^i\, (-1)^{i-1}\,dx^1\wedge \cdots \wedge \widehat{dx^i}\wedge \cdots\wedge dx^n \end{align} (ok here I've been a little lazy with the notation regarding restrictions/pullbacks to $S\cap U$). Note btw that in the very special case of $\Bbb{R}^3$ with its standard Riemannian metric, and in Cartesian coordinates $(x^1,x^2,x^3)=(x,y,z)$, we have $\sqrt{|\det g_{ij}|}=1$, and the summation fully expanded out is \begin{align} F^1\,dx^2\wedge dx^3-F^2\,dx^1\wedge dx^3+F_3\,dx^1\wedge dx^2&\equiv F^1\,dy\wedge dz+F^2\,dz\wedge dx+F^3\,dx\wedge dy, \end{align} which is exactly what you found.