[[Noteworthy Uses]]:: *[[Noteworthy Uses]]* [[Proved By]]:: *[[Proved By|Crucial Dependencies]]* Intuition:: *[[Intuition]]* Specializations:: *[[Specializations]]* Generalizations:: [[Generalized Stokes' Theorem]] ---- - Let $M$ be a $n$-[[differentiable Euclidean submanifold (with or without boundary)|manifold]] in $\rr ^{n}$; - Let $\omega = P \ dy \wedge dz - Q \ dx \wedge dz + R \ dx \wedge dy$ be an arbitrary $2$-form defined on an [[open set]] containing $M$. - (To relate to [[vector field]]s, recall [[div grad diagram]] and [[div grad curl diagram]] (musical isomorphisms)). > [!theorem] Theorem. ([[divergence theorem]]) > Then $\int _{M} \text{div } F \, \d V = \int _{\partial{M}} (F \cdot N) \ \d S. $ > > [!specialization] > > Any open subset $\Omega \subset \mathbb{R}^{n}$ is an [[embedded submanifold]], the [[differentiable Euclidean submanifold (with or without boundary)|manifold boundary of which]] agrees with the [[boundary|topological boundary]]. Thus [[closure|for]] $F \in C^{1}(\overline{\Omega}, \mathbb{R}^{n})$ we have $\int _{\Omega} \text{div }F \, d \lambda= \int _{\partial M} (F \cdot N) \, \d S, $ > (Might want to reconcile the surface measure $\d S$ with the [[Riemannian volume form]] $\d V$.) It is enough to assume [[Lipschitz boundary]] on $\Omega$. ([[Rademacher's Theorem]] will ensure the normal field is still well-defined a.e.) > > In particular, if $f \in C_{c}^{1}(\Omega, \mathbb{R}^{n})$), then $F(\partial_{}M)=0$, thus giving $\int_{\Omega}^{} \operatorname{div}F\,d \lambda =0.$ > > [!proof]- Proof. ([[divergence theorem]]) > Compute $d\omega=(\frac{ \partial P }{ \partial x }+ \frac{ \partial Q }{ \partial y }+\frac{ \partial R }{ \partial z }) dx \wedge dy \wedge dz$, and note that this equals $(\text{div }F) \ dx \wedge dy \wedge dz$ for $F=(P,Q,R)$. Hence by [[Generalized Stokes' Theorem]] we get $\int _{M} (\text{div }F) \, dx \wedge dy \wedge dz = \int _{\partial{M}} P \ dy \wedge dz - Q \ dx \wedge dz + R \ dx \wedge dy. $ Using [[integral of k-form over parameterized k-manifold]] the LHS is just $\int _{M} \text{div }F$. Meanwhile, using [[flux integral]] the RHS equals $\int _{\partial{M}} (F \cdot N) \ \d V$. This is the proof. $\qedin$ ---- #### > [!intuition] > So, [[divergence]] associates with 'flux'... to be made more precise elsewhere. ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```