Specializations:: [[the fundamental theorem of calculus]], [[Green's Theorem]], [[classical Stokes' Theorem]], [[divergence theorem]] ---- - Let $k \in \nn$. - If $k>1$: - Let $M$ be a [[compact]], [[orientation of a Euclidean submanifold|oriented]] $k$-[[differentiable Euclidean submanifold (with or without boundary)|manifold]] in $\rrn$; - Let $\omega$ be a $(k-1)$-[[k-form on Euclidean space|form]] on an [[open set]] in $\rrn$ containing $M$; - Give $\partial{M}$ [[the induced orientation on a boundary manifold|induced orientation]]. - If $k=1:$ - Let $M$ be a [[smooth arc]] in $\rrn$ with initial point $p$ and final point $q$ parametrized; - Let $f$ be a $0$-[[k-form on Euclidean space|form]] on an [[open set]] in $\rrn$ containing $M$. > [!theorem] Theorem. ([[Generalized Stokes' Theorem]]) > If $k>1$, $\int _{M} d \omega = \int _{\partial{M}} \omega, $ > where we define $\int_{\partial{M}} \omega=0$ in the case $\partial{M}=\emptyset.$ \ If $k=1$, $\int _{M} df = f(q) - f(p). $ - [ ] for smooth manifolds without boundary — just gives zero > [!proof] Proof for smooth manifolds without boundary. > Let $d \eta$ be an [[exact form]], compactly supported, say, $\text{supp } \eta=\bigcup_{i=1}^{N}U_{i}$. Let $\{ \rho_{i} \}$ be a [[partition of unity]] subordinate $U_{1},\dots,U_{N}$ and $U_{0}=\{ \text{everything else} \}$. Now, $\eta=\sum_{i=1}^{N} \rho_{i} \eta,$ > hence $d \eta=\sum_{i=1}^{N} d(\rho_{i} \eta)$. We will show each $\int _{M} \, d(\rho_{i} \eta)$, a single-patch integral, is zero. $\rho_{i}\eta=\sum_{j} (-1)^{j-1}\omega_{j} dx^{1} \wedge\dots \wedge \hat{dx^{j}} \wedge \dots \wedge dx^{n}$; WLOG enough to show for the $j=1$ term $h \ dx^{2} \wedge \dots \wedge dx^{n}$ ($h=\omega_{1})$. Then $d(\rho_{i}\eta)= \frac{ \partial h }{ \partial x^{1} } dx^{1} \wedge \dots \wedge dx^{n}.$ > So $\begin{align} > \int _{M} d(\rho_{i} \eta)&=\int _{M} \frac{ \partial h }{ \partial x^{1} } dx^{1} \wedge \dots \wedge dx^{n} \\ > &= \int _{\mathbb{R}^{n}} \frac{ \partial h }{ \partial x^{1} } dx_{1}\dots dx_{n} \\ > &= \int _{\mathbb{R}^{n-1}} ( \int _{-R}^{R} \frac{ \partial h }{ \partial x^{1} } dx_{1} ) dx_{2} \dots dx_{n} > \end{align}$ > where we have used [[Fubini's Theorem]] to split the integral and finite support assumption to bound the inner integral with $[-R,R]$ enveloping the support. But then the inner integral vanishes by [[the fundamental theorem of calculus]]. > [!proof]- Proof. ([[Generalized Stokes' Theorem]]) > First assume $k>1$. > **Step 1.** We carefully choose a particular set of [[coordinate patch]]es covering $M$. For $p \in M \cut \partial{M}$, there exists $\alpha : \mathcal{W} := \itr I ^{k} \to M$ s.t. $p \in \alpha(\mathcal{W})$ and $\alpha$ belongs to the [[orientation of a Euclidean submanifold|orientation]]. ![[CleanShot 2023-03-17 at [email protected]]] For $p \in \partial{M}$, there exists $\alpha: \mathcal{W}:=(\itr I ^{k}) \cup (\itr I ^{k-1} \times \{ 0 \}) \to M$ s.t. $p \in \alpha(\itr I ^{k-1} \times \{ 0 \})$ and $\alpha$ belongs to the [[orientation of a Euclidean submanifold|orientation]]. ![[CleanShot 2023-03-17 at [email protected]]] \ **Step 2**. $\int_{M} d \omega:= \sum_{i=1}^{\ell} \int _{M} \phi_{i}d \omega$, where $\phi_{1},\dots,\phi_{\ell}$ is a [[Partition of Unity on a manifold (deprecated)|partition of unity on]] $M$ [[partition of unity (deprecated)|dominated by]] the collection of [[coordinate patch]]es constructed in **step 1**. (So, $\fa i$, $\text{supp } \phi_{i} \cap M$ is contained within one [[coordinate patch]].) \ We claim that if $\int _{M} d(\phi_{i} \omega)= \int _{\partial{M}} \phi_{i} \omega$, then the theorem follows. Let us verify this: we would get $\begin{align} \textcolor{Skyblue}{\sum_{i=1}^{ \ell} \int _{M} d(\phi _{i} \omega)} = & \sum_{i=1}^{\ell} \int _{\partial{M}} \phi _{i} \omega = \textcolor{LimeGreen}{\int _{\partial{M}} \omega } \end{align},$\ where the LHS is really $\begin{align} \textcolor{Skyblue}{\sum_{i=1}^{\ell} \int _{M} d(\phi _{i} \omega)} = & \sum_{i=1}^{\ell} \int _{M} d \phi_{i} \wedge \omega + (-1) ^{0} \phi _{i} d\omega \\ = & \int _{M} \overbrace{d\left( \sum_{i=1}^{\ell} \phi_{i} \right)}^{= d(1)=0} \wedge \omega + \sum_{i=1}^{\ell} \int _{M} \phi _{i} \ d\omega \\ =: & \textcolor{LimeGreen}{\int _{M} d \omega. } \end{align} $\ (We used the [[exterior derivative wedge product rule|wedge product rule]] and [[linear map|lots of linearity]].) Thus, we indeed want to show that $\int _{M} d(\phi_{i} \omega)= \int _{\partial{M}} \phi_{i} \omega$, as doing so would complete the proof. \ **Step 3**. By **2**, it is enough to prove the theorem for $\omega$ such that $\text{supp }\omega \cap M$ is contained in one [[coordinate patch]] $\alpha: \mathcal{W} \to M$ constructed in **step 1**. \ **(a)** Suppose $\mathcal{W}=\itr I ^{k}$. ![[CleanShot 2023-03-17 at 14.05.07.jpg]] We get $\begin{align} \int _{M} d \omega = & \int _{\mathcal{W}:=\itr I ^{k}} \alpha^{*}(d\omega) \\ = & (-1) ^{k} \int _{\itr I ^{k-1}} b ^{*} (\alpha^{*} (\omega) ) \\ = & 0 \ \ \ \big(\text{since supp $\alpha^{*}(\omega) \subset \itr I^k$}\big) \end{align},$ where the second line follows from [[the cube lemma]]. Similarly, $\int _{\partial{M}} \omega = 0. $ Now the last case. Suppose $\mathcal{W} = (\itr I ^{k}) \cup (\itr I ^{k} \times \{ 0 \}$. Then, $\begin{align} \int _{M} d\omega = & \int _{\mathcal{W}} \alpha^{*}(d\omega) \\ = & \int _{\itr I ^{k}} \alpha^{*}(d\omega) \\ = & (-1)^{k} \int _{\itr I ^{k-1}} b ^{*}\alpha^ * (\omega). \end{align}$ Put $\beta= \alpha \circ b: \itr I ^{k-1} \to \partial{M}$. Then $\beta$ is a [[coordinate patch]] given by a [[the induced orientation on a boundary manifold|restriction of]] $\alpha$. Therefore $\beta$ belongs to the [[orientation of a Euclidean submanifold|orientation]] of $\partial{M}$ of $k$ is even. If $k$ is odd then we choose the [[reverse orientation]] of $\beta$. Thus, $\int _{\partial{M}} \omega = (-1) ^{k} \int _{\itr I ^{k-1}} \beta^{*}\omega =(-1)^{k} \int _{\itr I ^{k-1}} b ^{*} \alpha^{*}(\omega) . $ \ Next assume $k=1$. Suppose $\alpha:[a,b] \to M$ is as in the [[smooth arc]] definition. Then $\alpha:(a,b) \to M - p - q$ is a [[coordinate patch]] covering $M$ except a set of [[measure zero]]; by [[computation of integrals of k-forms on parameterized k-manifolds|this result]] we proceed to write $\begin{align} \int _{M} df = & \int_{a}^b \alpha ^*(df) \\ = & \int _{a}^b d \alpha^{*}(f) \\ = & \int_{a}^b d(f \circ \alpha ) \\ = & \int _{a} ^b D (f \circ \alpha) \ dt \\ = & f(\alpha(a)) - f(\alpha(b)) \\ =f(p) - f(q) \end{align}$ by [[the fundamental theorem of calculus]]. $\qedin$ > [!basicexample] > Take $M=\{ (x,y,9-x^{2}-y^{2}) : x^{2}+y^{2} \leq 9\}$ to be a 2-[[differentiable Euclidean submanifold (with or without boundary)|manifold]] in $\rr ^{3}$. ![[CleanShot 2023-03-17 at [email protected]]] [[orientation of a Euclidean submanifold|Orient]] $M$ so that $N(x,y,z)=(*,*,\text{positive})$. \ Define $\omega := (2z-y) dx + (x+z) dy + (3x-2y) dz$. \ **Step 1.** Observe that $M$ can be naturally covered by a [[coordinate patch]] $\alpha:\UU:=\mathbb{S}^{2}(3) \to M$ given by $\alpha(u,v)=(u, v, 9-u^{2}-v^{2})$. So, $\begin{align} \int _{M} d\omega = & \int _{M} 2 dx \wedge dy + 1dx \wedge dz - 3 dy \wedge dz \\ = & \int_{\UU} \alpha^{*}(2 dx \wedge dy + 1dx \wedge dz - 3 dy \wedge dz) \\ = & \int _{\UU} (2-2v-6u) \ du \wedge dv \\ = & \int _{\UU} 2 - 2v - 6u \\ = & 18\pi. \end{align}$ **Step 2.** $\int _{\partial{M}} \omega = ?$. First need $\partial{M}$. Since $M$ is a $2$-[[differentiable Euclidean submanifold (with or without boundary)|manifold]], [[the induced orientation on a boundary manifold|induced orientation]] is just the restriction orientation. Cover $\partial{M}$ with the [[coordinate patch]] $\beta:(0,2\pi) \to M$ given by $\beta(t)=(\cos t, \sin t, t)$. ![[CleanShot 2023-03-17 at 17.34.40.jpg]] So, $\begin{align} \int _{\partial{M}} \omega = & \int _{(0,2\pi)} (2z-y) dx + (x+z)dy + (3x-2y)dz = & 18\pi. \end{align}$ ![[CleanShot 2023-03-17 at 17.38.01.jpg]] > > **2.** Let $M:=\{ (t,t^{2}) : -1 \leq t \leq 1\}$ with initial point $(-1,1)$ and final point $(1,1)$. Compute $\int _{M} y ^{2} \ dx + 2xy \ dy$. > > By inspection, $d(xy ^{2})=y^{2} \ dx + 2xy \ dy$. Thus the result is obtained via GST in the $k=1$ case as $xy ^{2} |_{(1,1)}-xy^{2} \vert_{(-1,1)}=2$. > > **3.** Let $F(x,y,z)=\frac{-(x,y,z)}{(x^{2}+y^{2}+z^{2})^{3/2}}$, $(x,y,z) \neq \v 0$. ![[CleanShot 2023-03-18 at 16.39.53.jpg]] > Note that we can rewrite $F$ as $F(x)=-\frac{x}{\|x\|^{3}}^{}=-\frac{1}{\|x\|^{2}} \frac{x}{\|x\|}=-\frac{1}{\|x\|^{2}} \hat{x}$, where $\hat{x}$ notes normalized $x$. *What's the total work done by the force field $F$ moving a particle from $(1,0,1)$ to $(1,0,0)$ along the depicted arc?* > > We recall the characterization of total work using a [[circulation integral]]: that if $M$ a $1$-manifold then $\int _{M} F_{1}\ dx_{1} + \dots + F_{n} \ dx_{n}=\int _{M} (F \cdot T) \ \d V$. The RHS is the total work definition, the LHS is how we'll compute it using GST. This requires us to rewrite the $1$-form $F_{1}dx_{1} + \dots + F_{n} dx_{n}$ as $df$ (called the 'potential'), where the $0$-form $f$ is given by $f(x,y,z)= \frac{1}{(x^2 + y ^{2} + z ^{2}) ^{ 1/2}}$ (check and in hw). Now, GST immediately gives us the answer $f(1,0,1) - f(1,0,0)$. We see something special: the work done depends not on the trajectory at all other than the start/end points! (is this a more general property of [[exact form]]s?) ---- #### ----- #### 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 > ```