> [!theorem] > > Let $X$ be an [[Orientation on Manifold|orientated]] [[Riemannian Manifold|Riemannian n-manifold]], $\xi \in \vf(X)$ be a [[Compactly Supported|compactly supported]] [[Vector Field|vector field]], and $N$ be the outward-pointing normal vector field on $\partial X$, then > $ > \iota^*_{\partial X}(dV_g(\xi, \cdots)) = \angles{\xi, N}_g dV_{g^*} > $ > and > $ > \int_X \div{\xi}dV_g = \int_{\partial X} \angles{\xi, N}_g dV_{g^*} > $ > > where $\partial X$ is given the induced Stokes orientation, and $g^*$ is the induced Riemannian metric. > > *Proof*. Let > $ > \xi = \xi_\perp + \xi_\partial \quad \xi_\perp = \angles{\xi, N}_g N > $ > From here, > $ > \begin{align*} > \iota_{\partial X}^*(dV_g(\xi_\perp, \cdots)) &= \angles{\xi, N}_g dV_{g}(N, \cdots) \\ > &= \angles{\xi, N}_g dV_{g^*}(\cdots) > \end{align*} > $ > On the other hand, $\xi_\partial$ takes values in $T\partial X$, so $dV_g(\xi_\partial, \cdots) = 0$. > > Using [[Stokes' Theorem]], > $ > \begin{align*} > \int_X \div{\xi}dV_g &= \int_X d[dV_g(\xi, \cdots)] = \int_{\partial X}\iota_{\partial X}^*dV_g(\xi, \cdots) \\ > &= \int_{\partial X}\iota_{\partial X}^*\angles{\xi, N}_gdV_{g^*} > \end{align*} > $