[[Noteworthy Uses]]:: *[[Noteworthy Uses]]*
[[Proved By]]:: *[[Proved By|Crucial Dependencies]]*
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- Let $\Omega$ be [[open set|open in]] $\rr^2$;
- Let $g: \rr ^{2} \to \rr$ be [[smooth]];
- Let $M := \{ \big(x,y,g(x,y)\big) : (x,y) \in \Omega \}$ be the [[graph]] of $g$.
- **Remark.** This result actually generalizes to $M$ being *any* oriented $(n-1)$-[[differentiable Euclidean submanifold (with or without boundary)|manifold]] in $\rrn$, but showing that more general result is less elucidating.
- Naturally cover $M$ with the [[coordinate patch]] $\alpha: \rr ^{2} \to \rr ^{3}$ given by $\alpha(u,v)=\big(u,v,g(u,v)\big)$;
- $M$ is [[connected]] and hence has exactly two orientations. [[orientation of a Euclidean submanifold|Orient]] it using whichever one contains $\alpha$;
- Define a [[vector field]] $F(x,y,z)$ on $M$.
> [!theorem] Theorem. ([[flux integral]])
> We have that $\int_M F_{1} \ dy \wedge dz - F_{2} \ dx \wedge dz + F_{3} \ dx \wedge dy = \int _M (F \cdot N) \, \d V , $
where $N$ is the [[orientation of a Euclidean submanifold|unit normal field]] corresponding the the [[orientable manifold|orientable|orientation]] of $M$. Note that the LHS is the [[integral of a form over a compact oriented Euclidean submanifold]], while the the RHS is the [[integral of a scalar function over a parameterized manifold]].
[[circulation integral]]
> [!proof]+ Derivation. ([[flux integral]])
>
> We've chosen the [[orientation of a Euclidean submanifold|orientation]] of $M$ containing $\alpha$ — is it comprised of upward or downward pointing arrows? We find out by getting a [[unit normal vector to a parameterized curve]] via $\frac{ \partial \alpha }{ \partial u } \times \frac{ \partial \alpha }{ \partial v } / \|\frac{ \partial \alpha }{ \partial u } \times \frac{ \partial \alpha }{ \partial v }\|$. This is $\begin{align}
\frac{\frac{ \partial \alpha }{ \partial u } \times \frac{ \partial \alpha }{ \partial v }
}{ \|\frac{ \partial \alpha }{ \partial u } \times \frac{ \partial \alpha }{ \partial v }\|}= & \frac{\left( 1,0,\frac{ \partial g }{ \partial u } \right) \times \left( 0,1,\frac{ \partial g }{ \partial v } \right)}{ \|\frac{ \partial \alpha }{ \partial u } \times \frac{ \partial \alpha }{ \partial v }\|} \\
= & \frac{\left( -\frac{ \partial g }{ \partial u } , -\frac{ \partial g }{ \partial v }, 1 \right)}{ \|\frac{ \partial \alpha }{ \partial u } \times \frac{ \partial \alpha }{ \partial v }\|},
\end{align}$
and so we get that $N\big(\alpha(u,v)\big)=\frac{\left( -\frac{ \partial g }{ \partial u }, -\frac{ \partial g }{ \partial v }, 1 \right)}{\sqrt{ \left( \frac{ \partial g }{ \partial u } \right)^{2} + \left(\frac{ \partial g }{ \partial v } \right)^{2}+1}}.$
Now with the $\beta_{2}$ [[linear isomorphism|isomorphism]] from [[div grad curl diagram]] in mind, we compute [^1] $\begin{align}
&\int _{M} (F_{1} \ dy \wedge dz - F_{2} \ dx \wedge dz + F_{3} \ dx \wedge dy )& \\
= & \int _{\Omega} \alpha^{*}(F_{1} \ dy \wedge dz - F_{2} \ dx \wedge dz + F_{3} \ dx \wedge dy ) \\
= & \int _{\Omega} (F_{1} \circ \alpha) \ dv \wedge\left( \frac{ \partial g }{ \partial u } du + \frac{ \partial g }{ \partial v } dv \right) - (F_{2} \circ \alpha) du \wedge \left( \frac{ \partial g }{ \partial u } du + \frac{ \partial g }{ \partial v } dv \right) + (F_{3} \circ \alpha) du \wedge dv \\
= & \int _{\Omega} \big(-(F_{1} \circ \alpha) \frac{ \partial g }{ \partial u } - (F_{2} \circ \alpha) \frac{ \partial g }{ \partial v } + (F_{3} \circ \alpha) \big) \ du \wedge dv \\
= & \int _{\Omega}\big(-(F_{1} \circ \alpha) \frac{ \partial g }{ \partial u } - (F_{2} \circ \alpha) \frac{ \partial g }{ \partial v } + (F_{3} \circ \alpha) \big) \\
= & \int _{\Omega } (F_{1} \circ \alpha, F_{2} \circ \alpha, F_{3} \circ \alpha) \cdot (-\frac{ \partial g }{ \partial u } ,- \frac{ \partial g }{ \partial v } , 1) \ \ (**).
\end{align}$
**Recall**: For $h: M \to \rr$, [[MOC integral of scalar field over parameterized manifold|we have]] $\int _{M} h \, \d V := \int _{\Omega} (h \circ \alpha )\, V(D\alpha) \ (*)$. For our case, $V(D\alpha)=\sqrt{ \det (D\alpha^{\top}D \alpha) }= \det \sqrt{ \begin{bmatrix}
1 & 0 & \frac{ \partial g }{ \partial u } \\
0 & 1 & \frac{ \partial g }{ \partial v }
\end{bmatrix} \begin{bmatrix}
1 & 0 \\
0 & 1 \\
\frac{ \partial g }{ \partial u } & \frac{ \partial g }{ \partial v }
\end{bmatrix}} = \sqrt{ \frac{ \partial g }{ \partial u } ^{2} + \frac{ \partial g }{ \partial v } ^{2} +1 }.$
Now, from $(*)$, $\begin{align}
\int _{M} (h \circ \alpha) V(D\alpha) = & \int _{M} (h \circ \alpha) \sqrt{ \frac{ \partial g }{ \partial u } ^{2} + \frac{ \partial g }{ \partial v } ^{2} +1} := \int _{M} h \, \d V.
\end{align}$
Consider $(**)$, multiply it by '$1