Let $f:\Bbb R^{n-1}\to\Bbb R$ be an everywhere positive $C^\infty$ function and let $U$ be a bounded open subset of $\Bbb R^{n-1}.$ Verify directly that Stokes' theorem is true if $D$ is the domain defined by $$0<x_n<f(x_1,\ldots,x_{n-1}),\quad (x_1,\ldots,x_{n-1})\in U$$ and $\mu$ an $(n-1)$-form $$\varphi(x_1,\ldots,x_n)dx_1\wedge\ldots\wedge dx_{n-1},$$ where $\varphi\in C^\infty(\Bbb R^n;\Bbb R).$
Source: Differential Forms, page $132$
My attempt:
First,
$\begin{aligned}d\mu(x_1,\ldots,x_n)&=d\varphi(x_1,\ldots,x_n) dx_1\wedge\ldots\wedge dx_{n-1}\\&=\partial_n\varphi(x_1,\ldots,x_n)dx_n\wedge dx_1\wedge\ldots\wedge dx_{n-1}.\end{aligned}$
I need to verify that $$\int_Dd\mu=\int_{\partial D}\mu.$$
Since $\partial D$ is of Jordan-measure $0,$ I integrated over the closure $\overline D$ and applied Fubini's theorem: $$\begin{aligned}\int_Dd\mu&=\int_{\overline D}d\mu\\&=\int_{\overline U}\int_0^{f(x_1,\ldots,x_{n-1})}\partial_n\varphi(x_1\ldots,x_n)dx_ndx_1\ldots dx_{n-1}\\&=\int_{\overline U}(\varphi(x_1,\ldots,x_{n-1},f(x_1,\ldots,x_{n-1}))-\varphi(x_1,\ldots,x_{n-1},0))dx_1\ldots dx_{n-1}.\boldsymbol{(1)}\end{aligned}$$
On the other hand, (I believe) $$\begin{aligned}&\color{white}=\partial D\\&=\partial\{(x,y)\in\Bbb R^{n-1}\times\Bbb R\mid x\in U, 0<y<f(x)\}\\&=\overline U\times f(\overline U)\cup\overline U\times\{0\}\cup V,\end{aligned}$$ where $$V=\partial U\times\{y\in\Bbb R\mid 0\le y\le f(x),x\in\partial U\},$$ but, I think we should be able to parametrize $V\subset\Bbb R^n$ in some coordinates with $n-1$ conditions so that one of the differentials $dx_1,\ldots,dx_{n-1}$ turns into $0$ when pulled-back in the integral of the differential $(n-1)$-form $\mu,$ as in the proof of Stokes' theorem for the unit cube, where one coordinate on each edge is constant. However, I got a bit confused.
Question:
How should I proceed?
If $\Phi_{\overline U\times\{0\}}$ and $\Phi_{\overline U\times f(\overline U)}$ are parametrizations of $\overline U\times\{0\}$ and $\overline U\times f(\overline U)$ respectively s. t. the normal to $\overline U\times\{0\}$ has a negative $x_n$ component, and the normal to $\overline U\times f(\overline U)$ has a positive $x_n$ component, I'm going to get $\boldsymbol{(1)}.$
Part of the issue is your notation here. The set $\overline U\times f(\overline U)$ is not what you need. You need $\{(x,y): x\in\overline U, y=f(x)\}$. You're right that the boundary consists of three parts:
(a) $\{(x,y): x\in\partial U, 0\le y\le f(x)\}$
(b) $\{(x,y): x\in\overline U: y=0\}$
(c) $\{(x,y): x\in\overline U: y=f(x)\}$
The integral of $\mu$ over set (a) will be $0$ because no $dx_n$ appears in $\mu$, as you indicated.
You have to look at boundary orientation (think about the boundary orientation on a rectangle, for example). The outward normal to (b) is $(-1)^{n-1}e_n$ and the outward normal to (c) has positive $e_n$ component. But remember that if $\Bbb H^n$ is the upper half-space in $\Bbb R^n$, $\partial\Bbb H^n$ is $(-1)^{n-1}\Bbb R^{n-1}$, so the signs work out exactly right. You should have been through this carefully in the proof of Stokes's Theorem.