This is a small question. Rudin's law for differential forms is given on page 254 as
$$ \omega = \sum \alpha_{i_1 \cdots i_k}(x) \, dx_{i_1} \wedge \cdots \wedge dx_{i_k}$$
according to the rule
$$ \int \limits_\Phi \omega = \int \limits_D \sum \alpha_{i_1 \cdots i_k}(\Phi (u)) \, {\partial(x_{i_1} , \dots , x_{i_k}) \over \partial(u_{1} , \dots , u_{k}) } du$$
His first example (pages 254-255):
Let $ [0,1] \xrightarrow{\quad s \quad} \mathbb R^2$ be a 1--surface that is $\mathscr C^1$. Set $(x_1, x_2) := (x, y) = X$
$$ \omega = y \, dx + x \, dy$$
Then $$ \int \limits_ s \omega = \int \limits_{[0,1]} \left( s_1(t) \, s_2'(t) + s_1'(t) \, s_2(t) \right)\;dt = \cdots$$
The rest of his solution is trivial and not germane, but I'm tripping up on what is likely an easy problem. I'm having some problems understanding what he means by his notation. It seems our $\alpha$ functions are rather easy to define. Have
$$ \alpha_x (X) = x_2 = y$$
that is, retrieve the second element from whatever vector we feed in. Similarly,
$$ \alpha_y (X) = x_1 = x$$
Then we can write out our form as being
$$ \omega = \alpha_x (X) \, dx + \alpha_y (X) \, dy$$
The major point of concern is how Rudin is writing out his Jacobian determinant. I don't see clearly how he is indexing his denominator. We know that $s(t) := \left( s_1 (t) , s_2 (t) \right)$. I'm naively interpreting the entire series inside the integral as being
$$ \int \limits_{[0,1]} \alpha_x (s) {dx \over ds_1} ds_1 + \alpha_y (s) {dy \over ds_2} ds_2 = \int \limits_{[0,1]} s_2(t) {dx \over ds_1} ds_1 + s_1(t) {dy \over ds_2} ds_2 $$
but I feel this is wrong, for instance,
$$ {dx \over ds_1} ds_1 \neq s_1' $$
I assume It's an issue with how I'm interpreting the Jacobian determinant, so I would appreciate anyone who shows the correct computation.
