Proof that the vector area is the same for all surfaces sharing the same boundary

Question

A proof of the title was posted on physics stack exchange asking for explanation of the steps, I had written an explanation of the steps as an answer. However, after writing this answer there has been something which has been bothering me about the proof.

In the last step, I had assumed that for two surfaces $S_1$ and $S_2$ attached to the same boundary $C$ parameterized by $(a,b)$ and $(u,v)$ that reparametrizing $(a,b) \to (u,v)$ for $S_1$ that the boundary of $S_1$'s parameterization in the $(u,v)$ coordinates is same that of $S_2$

For naive surfaces, I found this to be true. Consider a circle cantered at origin of the x-y plane and consider the following surfaces attached to it: A cone, a portion of paraboloid , a hemisphere. It is clear that for all these surfaces that the interior of the circle maps entirely onto the surface.

For explicit surfaces which don't work: Suppose I took the hemisphere and portion of paraboloid as the surfaces involved in the proof, if the hemisphere's equation was initially expressed in spherical and I turned the coordinates into cartesian, I'd find that the domain of the circle and the paraboloid as same.

Now, consider these two surfaces:

Clearly they share same boundary but I can't imagine that both of their parameterization domains are same (if they have any that is)[ Eg: Suppose I parameterized both in (x,y)]. Hence, my question: Is the statement in the question general? If so, how does one extend the proof from the physics stack exchange post for an arbitrary surface.

Answer 1

The proof only applies to globally parameterized surfaces, i.e. those surfaces which can be realized as the image of an embedding $\varphi:U\to\mathbb{R}^3$ where $U\subseteq\mathbb{R}^2$ is a bounded regular domain. If the boundary of the surface is a single closed curve, the same is true of $U$, and thus, by the smooth Jordan-Schoenflies theorem we can always choose the parameterization so that $U$ is the unit disc. This is true of the surface you've drawn, though it may be difficult to write down an explicit parameterization. It seems Griffiths relies on the domains being the same for his argument.

There are, however, surfaces which cannot be globally parameterized, such as a torus with a disc removed. If you want a proof which applies to these surfaces, it would require a different approach than Griffiths uses. Here's one proof using Stokes' theorem for surfaces:

Let $\mathbf{e}_i$, $i\in\{1,2,3\}$ be the standard constant unit vector fields on $\mathbb{R}^3$. We can choose choose vector fields $\mathbf{v}_i$ such that $\operatorname{curl}(\mathbf{v}_i)=\mathbf{e}_i$, one possible choice being $\mathbf{v}_i=\frac{1}{2}\epsilon_{ijk}x_j\mathbf{e}_k$. These will be useful later.

Let $S\subset\mathbb{R}^3$ be a compact, smooth, oriented surfaces whose boundary is a closed, oriented curve $C\subset\mathbb{R}^3$ (with the induced orientation). Let $\mathbf{n}:S\to\mathbb{R}^3$ be the unit normal vector chosen consistently with the orientation of $S$. Writing components as $\mathbf{n}=\langle n_1,n_2,n_3\rangle$, note that $n_i=\mathbf{n}\cdot\mathbf{e}_i$. Now, using $\mathbf{v}_i$, we can apply Stokes' theorem: $$ \int_{S}n_i\ dA=\int_S\mathbf{e}_i\cdot\mathbf{n}\ dA=\int_S\operatorname{curl}(\mathbf{v}_i)\cdot\ \mathbf{dA}=\int_C\mathbf{v}_i\cdot\mathbf{dL} $$ This means that the integral $\int_S n_i\ dA=\left(\int_S\mathbf{dA}\right)_i$ depends only on the oriented boundary of $S$. Thus, given compact, oriented, smooth surfaces $S_1,S_2$ with identical (and identically oriented) boundaries, we have $\int_{S_1}\mathbf{dA}=\int_{S_2}\mathbf{dA}$.