[Math] Surface integral via differential forms

differential-formsdifferential-geometryintegration

I am trying to figure out why the change of variables formulas for surface integrals (for example) is what it is. I was wondering if there is a justification in terms of differential forms.

So, let $g : V \to \mathbb{R}$ be a smooth function, andlet $S$ be the surface given by $G(V)$, where $G(x, y) = (x, y, g(x, y))$. In my current predicament, $V$ is an open subset of $\mathbb{R}^2$, but supposedly 2 may be replaced with general $n$. Let $F : S \to \mathbb{R}$ be a smooth function. I found the following formula online:
$$
\int_S F\, dS = \int_V (F\circ G)\sqrt{g_x^2 + g_y^2 + 1}\ dA.
$$
Here $dA = dx \wedge dy$ is the standard area form. I have a few questions about this.

Could someone recommend a rigorous proof of this fact, or perhaps a proof sketch? I am familiar with the change of variables formula when we are considering a change of variables by a diffeomorphism between open sets in $\mathbb{R}^n$, but this seems to fall apart here (since $S$ is not open in $\mathbb{R}^3$).

I was trying to justify this with the machinery of differential forms. For instance, we know that if we define $\omega = F \, dS$ to be a 2-form on $S$,
$$
\int_S \omega = \int_V G^*\omega.
$$
This is where I get really confused. As far as I know, $G^*\omega$ is defined as follows:
\begin{align*}
G^*\omega(x, y)(v, w) &= \omega(G(x, y))(dG(x, y)(v), dG(x, y)(w)) \\
&= \omega(G(x, y))(v_1, v_2, g_xv_1 + g_yv_2, w_1, w_2, g_xw_1 + g_yw_2).
\end{align*}
But I have no idea how to proceed from here. How does this end up getting us the square root from above? I see a lot of determinants of transformations, but $dG$ is not a square matrix, so I'm not sure how to get this to appear in this case.

Any help is much appreciated.

Best Answer

The part you are missing is the coordinate expression for the area form $dS$. To understand this, you need to know more about Riemannian volume forms.

So, let $(M,g)$ be an oriented $n$-dimensional Riemannian manifold. The Riemannian volume form of $M$, which we denote by $\mathrm{vol}_g$, is defined geometrically as the unique $n$-form on $M$ satisfying $$\mathrm{vol}_g(e_1,\ldots,e_n)=1$$ whenever $e_1,\ldots,e_n$ is an oriented orthonormal (local) frame. The local coordinate expression of $\mathrm{vol}_g$ is $$\mathrm{vol}_g=\sqrt{\det g_{ij}}\;dx^1\wedge\ldots\wedge dx^n,$$where $g_{ij}$ is the matrix representation of $g$ in the local coordinates $x^1,\ldots,x^n$.

Now, your area form $dS$ is, in fact, the Riemannian volume form of the surface $S$, equipped with the Riemannian metric induced by the embedding in $\mathbb{R}^3$. The map $G$ is a parametrization of $S$, and the matrix representation of the Riemannian metric with respect to $G$ is given by $$g=\left(\begin{array}{cc}g_x^2+1&g_xg_y\\g_xg_y&g_y^2+1\end{array}\right).$$ Hence, $$\det g_{ij}=g_x^2+g_y^2+1,$$and your desired formula follows by putting all the pieces together.

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