Theorem: Let $\omega$ be a smooth $(n-1)$-form with compact support on the upper half-space $\mathbb{H}^n = \{x \in \mathbb{R}^n : x_n \geq 0\}$. Then $\int_{\partial \mathbb{H}^n} \omega = \int_{\mathbb{H}^n} d\omega$ where $\partial \mathbb{H}^n = \{x_n = 0\} \cong \mathbb{R}^{n-1}$.

Step 1: Write $\omega$ in coordinates.

A general $(n-1)$-form on $\mathbb{R}^n$ is $\omega = \sum_{i=1}^n f_i \, dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n$ where $\widehat{dx_i}$ means $dx_i$ is omitted. By linearity, it suffices to prove the theorem for a single term $\omega = f \, dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n$ where $f$ has compact support in $\mathbb{H}^n$.

Step 2: Compute $d\omega$.

$d\omega = df \wedge dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n$ $= \sum_{j=1}^n \frac{\partial f}{\partial x_j} dx_j \wedge dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n$

The only nonzero term is $j = i$ (all others have a repeated $dx_j$): $d\omega = (-1)^{i-1} \frac{\partial f}{\partial x_i} \, dx_1 \wedge \cdots \wedge dx_n$

Step 3: Compute $\int_{\mathbb{H}^n} d\omega$.

$\int_{\mathbb{H}^n} d\omega = (-1)^{i-1} \int_{\mathbb{R}^{n-1}} \int_0^\infty \frac{\partial f}{\partial x_i} \, dx_i \, dx_1 \cdots \widehat{dx_i} \cdots dx_n$

Case $i \neq n$: The inner integral $\int_{-\infty}^\infty \frac{\partial f}{\partial x_i}\,dx_i = f|_{x_i = -\infty}^{x_i = \infty} = 0$ by compact support. Also, $\omega|_{\partial \mathbb{H}^n}$ contains $dx_n$ (since $i \neq n$, the form restricted to $\{x_n = 0\}$ involves $dx_n = 0$)... more precisely, the pullback of $\omega$ to $\{x_n = 0\}$ is zero because it contains $dx_n$ as a factor when $i \neq n$. Wait — when $i \neq n$, the form $\omega = f\,dx_1 \wedge \cdots \wedge \widehat{dx_i} \wedge \cdots \wedge dx_n$ contains $dx_n$, so its pullback to $\partial \mathbb{H}^n = \{x_n = 0\}$ (where $dx_n = 0$) vanishes. Thus both sides are $0$. $\checkmark$

Case $i = n$: We have $\omega = f\,dx_1 \wedge \cdots \wedge dx_{n-1}$ and $d\omega = (-1)^{n-1} \frac{\partial f}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n$

$\int_{\mathbb{H}^n} d\omega = (-1)^{n-1} \int_{\mathbb{R}^{n-1}} \left[\int_0^\infty \frac{\partial f}{\partial x_n}\,dx_n\right] dx_1 \cdots dx_{n-1}$

By the fundamental theorem of calculus: $\int_0^\infty \frac{\partial f}{\partial x_n}\,dx_n = \lim_{x_n \to \infty} f - f|_{x_n = 0} = 0 - f(x_1, \ldots, x_{n-1}, 0) = -f|_{x_n=0}$

So $\int_{\mathbb{H}^n} d\omega = (-1)^n \int_{\mathbb{R}^{n-1}} f(x_1, \ldots, x_{n-1}, 0)\,dx_1 \cdots dx_{n-1}$.

For the boundary integral: $\partial \mathbb{H}^n = \{x_n = 0\}$ with the induced orientation, which is $(-1)^n$ times the standard orientation. The pullback of $\omega$ to $\partial \mathbb{H}^n$ is $f|_{x_n=0}\,dx_1 \wedge \cdots \wedge dx_{n-1}$, so: $\int_{\partial \mathbb{H}^n} \omega = (-1)^n \int_{\mathbb{R}^{n-1}} f|_{x_n=0}\,dx_1 \cdots dx_{n-1}$

Both sides agree. $\checkmark$

Step 4: Extension to manifolds.

For a general manifold $M$, use a partition of unity $\{\rho_\alpha\}$ subordinate to coordinate charts to write $\omega = \sum \rho_\alpha \omega$. Each term is supported in a coordinate chart, where $M$ looks like $\mathbb{R}^n$ or $\mathbb{H}^n$. Apply the above calculation to each term and sum. $\square$