Baby Rudin 10.11 and 10.12

Solution 1:

On RHS of $$ \omega = \sum_{i_1,\dots, i_k = 1}^n a_{i_1\dots i_k}(x) dx_{i_1} \wedge \dots \wedge dx_{i_k} \tag{34} $$ the sum is meant to be over all indices $i_1\dots i_k$, each of which runs from $1$ to $n$. Later in the text, he restricts them to be such that $1 \le i_1 < \dots < i_k \le n $ with the standard presentation. The set $\{a_{i_1\dots i_k}\}$ is a set of functions, indexed as above.

So for your $\gamma: [0,1] \to E \subset \mathbb R^3$, and we have $\gamma(t) = (\gamma_1(t),\gamma_2(t),\gamma_3(t)) = (x(t),y(t),z(t))$ and the jacobians are $\{\frac{\partial \gamma_1}{\partial t},\frac{\partial \gamma_2}{\partial t},\frac{\partial \gamma_3}{\partial t}\}$ or $\{x'(t), y'(t), z'(t)\}$ , you can compute them as described below. Now with $\omega = xdy + ydx$, we have $a_1(x,y,z) = y, a_2(x,y,z) = x$ and $a_3(x,y,z) = 0$ and with $(35)$ we get $$ \omega(\gamma) \equiv \int_\gamma \omega = \int_{[0,1]}\sum_{j = 1}^3a_j(\gamma(t))\gamma_j'(t)dt = \int_{[0,1]} (\gamma_1(t)\gamma_2'(t) + \gamma_2(t)\gamma_1'(t))dt\\ = \int_{[0,1]} (x(t)y'(t) + y(t)x'(t))dt $$

Now for the intergal you know, I hope, the product rule $(f(t) g(t))' = f'(t) g(t) + f(t) g'(t)$ that shouold explain how you get the equality. Then $$ \int_a^b f'(t) g(t) + f(t) g'(t) dt = f(b) g(b) - f(a) g(a) $$

Appendix:

The differential form (in Rudin's book) is a map $\omega : \mathscr{C}'(D,E) \to \mathbb R$ such that $$ \omega : \Phi \mapsto \omega(\Phi) \equiv \int_\Phi \omega := \int_D \sum a_{i_1\dots i_k}(\Phi(u)) \frac{\partial (x_{i_1},\dots,x_{i_k})}{\partial (u_1,\dots, u_k)} d u \tag{35} $$ (where $\mathscr{C}'(D,E)$ is a set not a vector spaec!). The symbols $a_{i_1\dots i_k}$ are indexed functions $a_{i_1\dots i_k}:\mathbb R^n \supset E \to \mathbb R$, they can be precomposed with $\Phi:D \to E$ to get $a_{i_1\dots i_k} \circ \Phi:D \to \mathbb R$.

The Notation $$ \frac{\partial (x_{i_1},\dots,x_{i_k})}{\partial (u_1,\dots, u_k)} $$ denotes certain determinants from the jacobian of the $k$-surface $\Phi$, not the absolute value of them since it is a signed integral. More explicitly we have $$ \Phi: D \ni (u_1,\dots, u_k) \mapsto (x_1,\dots,x_n) \in E. $$ and its jacobian is the $(n\times k)$-matrix $$ D\Phi = \begin{pmatrix} \partial x_1 / \partial u_1 & \dots & \partial x_1 / \partial u_k \\ \vdots & \ddots & \vdots \\ \partial x_n / \partial u_1 & \dots & \partial x_n / \partial u_k \\ \end{pmatrix}. $$ By taking only the $k$ rows $(i_1,\dots,i_k)$ we have a restriction (or projection $P_I$, with $I = (i_1,\dots, i_k)$ on) to a $k$-dimensional subspace of $\mathbb R^n$, and the jacobian $D(P_I \circ \Phi)$ of the composition $P_I \circ \Phi : D \to P_I(E) \subset \mathbb R^k$ is then a (square) $(k\times k)$-matrix, of which one can take the determinant, that is $$ \frac{\partial (x_{i_1},\dots,x_{i_k})}{\partial (u_1,\dots, u_k)} = \det \begin{pmatrix} \partial x_{i_1} / \partial u_1 & \dots & \partial x_{i_1} / \partial u_k \\ \vdots & \ddots & \vdots \\ \partial x_{i_k} / \partial u_1 & \dots & \partial x_{i_k} / \partial u_k \\ \end{pmatrix} $$