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

Question

In the book, Introduction to Electrodynamics by Griffiths (4th edition) in question 1.62 part c, we are asked to prove that the vector area is the same for all surfaces sharing the same boundary. The vector area for a surface is defined as:

\begin{equation} \mathbf{a} = \int_S d \mathbf{a} \end{equation}

I have solved this question using the hints given by Griffiths however I found another solution and I am having difficult time understanding it. The solution is given here:

Suppose that two surfaces sharing a boundary are parameterized by vectors $\mathbf{x}(u, v), \mathbf{x}(a, b)$ respectively. The area integral with the first parameterization is $$\begin{aligned} \mathbf{a} &=\int \frac{\partial \mathbf{x}}{\partial u} \times \frac{\partial \mathbf{x}}{\partial v} d u d v \\ &=\epsilon_{i j k} \mathbf{e}_{i} \int \frac{\partial x_{j}}{\partial u} \frac{\partial x_{k}}{\partial v} d u d v \\ &=\epsilon_{i j k} \mathbf{e}_{i} \int\left(\frac{\partial x_{j}}{\partial a} \frac{\partial a}{\partial u}+\frac{\partial x_{j}}{\partial b} \frac{\partial b}{\partial u}\right)\left(\frac{\partial x_{k}}{\partial a} \frac{\partial a}{\partial v}+\frac{\partial x_{k}}{\partial b} \frac{\partial b}{\partial v}\right) d u d v \\ &=\epsilon_{i j k} \mathbf{e}_{i} \int d u d v\left(\frac{\partial x_{j}}{\partial a} \frac{\partial a}{\partial u} \frac{\partial x_{k}}{\partial a} \frac{\partial a}{\partial v}+\frac{\partial x_{j}}{\partial b} \frac{\partial b}{\partial u} \frac{\partial x_{k}}{\partial b} \frac{\partial b}{\partial v}+\frac{\partial x_{j}}{\partial b} \frac{\partial b}{\partial u} \frac{\partial x_{k}}{\partial a} \frac{\partial a}{\partial v}+\frac{\partial x_{j}}{\partial a} \frac{\partial a}{\partial u} \frac{\partial x_{k}}{\partial b} \frac{\partial b}{\partial v}\right) \\ &=\epsilon_{i j k} \mathbf{e}_{i} \int d u d v\left(\frac{\partial x_{j}}{\partial a} \frac{\partial x_{k}}{\partial a} \frac{\partial a}{\partial u} \frac{\partial a}{\partial v}+\frac{\partial x_{j}}{\partial b} \frac{\partial x_{k}}{\partial b} \frac{\partial b}{\partial u} \frac{\partial b}{\partial v}\right)+\epsilon_{i j k} \mathbf{e}_{i} \int d u d v\left(\frac{\partial x_{j}}{\partial b} \frac{\partial x_{k}}{\partial a} \frac{\partial b}{\partial u} \frac{\partial a}{\partial v}-\frac{\partial x_{k}}{\partial a} \frac{\partial x_{j}}{\partial b} \frac{\partial a}{\partial u} \frac{\partial b}{\partial v}\right) \end{aligned}\tag{6}$$ In the last step a $j, k$ index swap was performed for the last term of the second integral. The first integral is zero, since the integrand is symmetric in $j, k$. This leaves $$\begin{aligned} \mathbf{a} &=\epsilon_{i j k} \mathbf{e}_{i} \int d u d v\left(\frac{\partial x_{j}}{\partial b} \frac{\partial x_{k}}{\partial a} \frac{\partial b}{\partial u} \frac{\partial a}{\partial v}-\frac{\partial x_{k}}{\partial a} \frac{\partial x_{j}}{\partial b} \frac{\partial a}{\partial u} \frac{\partial b}{\partial v}\right) \\ &=\epsilon_{i j k} \mathbf{e}_{i} \int \frac{\partial x_{j}}{\partial b} \frac{\partial x_{k}}{\partial a}\left(\frac{\partial b}{\partial u} \frac{\partial a}{\partial v}-\frac{\partial a}{\partial u} \frac{\partial b}{\partial v}\right) d u d v \\ &=\epsilon_{i j k} \mathbf{e}_{i} \int \frac{\partial x_{j}}{\partial b} \frac{\partial x_{k}}{\partial a} \frac{\partial(b, a)}{\partial(u, v)} d u d v \\ &=-\int \frac{\partial \mathbf{x}}{\partial b} \times \frac{\partial \mathbf{x}}{\partial a} d a d b \\ &=\int \frac{\partial \mathbf{x}}{\partial a} \times \frac{\partial \mathbf{x}}{\partial b} d a d b \end{aligned}\tag{7}$$ However, this is the area integral with the second parameterization, proving that the area-integral for any given boundary is independant of the surface.

My questions are:

1. In the 5th line (the last line) of equation 6, the author argues that the first integral is zero since the integrand is symmetric in j, k. I didn't understand why this is true. Can you please show step-by-step why the integrand being symmetric in indices makes this integral zero?

2. Where in the equations did the author use the fact that the two surfaces are sharing a boundary?

Answer 1

Ah, I think I got what they did now. Let's say we have two surfaces $S_1$ and $S_2$ parameterized by $(u,v)$ coordinates and $(a,b)$ coordinates respectively. To get the second equation from the first one , begin with the first equation:

$$ \int \frac{\partial x}{\partial u} \times \frac{\partial x}{\partial v} du dv$$

Then have a look at the Einstein expression for the cross product, we can write:

$$ \frac{\partial x}{\partial u} \times \frac{\partial x}{\partial v} = \frac{\partial x_j}{\partial u} \frac{\partial x_k}{\partial u} \epsilon_{ijk} e_i \tag{1} $$

Now, here is the critical step, the author has written the components of the $x_j$ as functions of the other parameterization: $x_j(u,v) = x_j( a(u,v) , b(u,v) )$ then used the multivariable chain rule. This looks like a hack, but recall that we are talking about a physical surface area and we know such regions of space can be reparametrized. Here is an explicit computation of the above mentioned point:

$$ \frac{\partial x_j \big( a(u,v) , b(u,v) \big)}{\partial u} = \frac{\partial x_j}{\partial a} \frac{\partial a}{\partial u} + \frac{\partial x_j}{\partial b} \frac{\partial b}{\partial u}$$

Similarly they have done for $x_k$.

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For line 5 simplification, the deal can be understood from the fisrt term:

$$e_i \epsilon_{ijk} \frac{\partial x_j}{\partial b} \frac{\partial x_k}{\partial b} \frac{\partial a}{\partial v} \frac{\partial a}{\partial u} $$

This expression is just the definition of cross product is just: $ \frac{\partial x}{\partial b} \times \frac{\partial x}{\partial b}$ in (1) which is always zero.

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The idea in the very last step is that even when you reparametrize the integral, you must change the bounds such that you are integrating over the same physical surface area. In sense that, the real life surface area corresponds to some domain $D$ in the $(u,v)$ plane and some other domain $D'$ in the $(a,b)$ plane. When we convert from $(u,v)$ to $(a,b)$ our $D$ gets mapped to $D'$ but still it represents the same physical space. [When domains get mapped, so you can imagine their boundaries as well]

So, the change of variables captured the 'same boundary' idea.

Hope this helps.

Edit: For the last step, it's a bit tricky. This proof is actually not general at all. See this post I had made on mathstackexchange.