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If $X$ is an oriented $n$-manifold with boundary, $F:\partial X\hookrightarrow X$ is the inclusion, and $\omega$ is a compactly supported $(n-1)$-form, thenwhere the boundary has the outward-normal-first boundary orientation. A subordinate partition of unity reduces the theorem to the fundamental theorem of calculus in oriented coordinate half-spaces; local finiteness and compact support ensure that only finitely many terms contribute.

Real line bundles over a paracompact space $X$ are classified byEuclidean space ${\mathbb{R}}^n$ is a contractible space, so its first cohomology vanishes and every real line bundle on it is trivial. Apply this to the defining bundle from part (c). In a global trivialization, $s$ is a smooth real function with $Y=s^{-1}(0)$ and $ds{|}_Y\ne 0$. Thus $ds$, or a metric-dual normal vector field, gives a global orientation of the normal line. Combining this with the standard orientation of ${\mathbb{R}}^n$ gives an orientation of $Y$ using the same normal-first convention as the outward-normal-first boundary orientation. Hence every properly embedded hypersurface in ${\mathbb{R}}^n$ is orientable.

Give $\partial X$ the outward-normal-first boundary orientation. The Generalized Stokes theorem states that, for every compactly supported $(n-1)$-form $\omega$,

Choose an oriented coordinate cover by charts into ${\mathbb{R}}^n$ or the half-space ${\mathbb{H}}^n=\{x^1\ge 0\}$, and choose a partition of unity $({\rho }_i)$ subordinate to it. Since the family is locally finite and $\omega$ has compact support, only finitely many ${\rho }_i\omega$ are nonzero. It is therefore legitimate to write both integrals as finite sums and prove the identity for a form supported in one chart.

In an interior chart the integral of an exact compactly supported top form is zero by the fundamental theorem of calculus. In a boundary chart writeIntegrating $d\omega =({\sum }_j{\partial }_j{a}_j)d{x}^1\wedge \cdots \wedge d{x}^n$ coordinate by coordinate kills every tangential derivative. The normal derivative leaves precisely the restriction to $x^1=0$, with the sign selected by the outward-normal-first convention. This is ${\int }_{\partial X}{F}^{\ast }\omega$, proving the theorem.

Let $({\theta }^1,\dots ,{\theta }^n)$ be the dual basis of the positively oriented orthonormal basis $(v_1,\dots ,v_n)$. By the definition of the Riemannian volume form,The interior product of a differential form with the outward unit normal isThe vectors $(v_2,\dots ,v_n)$ form a positive orthonormal frame of $T\partial X$ by the outward-normal-first boundary orientation. Consequently the pullback of the last display is the positive unit boundary volume form: