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let $\alpha, \beta \in C^2(\Omega)$ be zero forms

Where $\Omega$ is a regular surface with boundary $\partial{\Omega}$. I have to write the following formula using differential forms

\begin{equation} \int_{\Omega} \nabla\alpha \times \nabla \beta \ \cdot d\Omega = \int_{\partial{\Omega}} \alpha \nabla \beta \ \cdot dr \end{equation}

What I have done so far:

$\nabla \alpha = \frac{\partial \alpha}{\partial x }\hat{x}+\frac{\partial \alpha}{\partial y }\hat{y}+\frac{\partial \alpha}{\partial z }\hat{z}$

$\nabla \beta = \frac{\partial \beta}{\partial x }\hat{x}+\frac{\partial \beta}{\partial y }\hat{y}+\frac{\partial \beta}{\partial z }\hat{z}$

\begin{equation} \nabla \alpha \times \nabla \beta = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial \alpha}{\partial x } & \frac{\partial \alpha}{\partial y } & \frac{\partial \alpha}{\partial z } \\ \frac{\partial \beta}{\partial x } & \frac{\partial \beta}{\partial y } & \frac{\partial \beta}{\partial z } \end{vmatrix} = ( \frac{\partial \beta}{\partial z } \frac{\partial \alpha}{\partial y } - \frac{\partial \beta}{\partial y } \frac{\partial \alpha}{\partial z } ) \hat{x} -( \frac{\partial \beta}{\partial z } \frac{\partial \alpha}{\partial x } - \frac{\partial \alpha}{\partial z } \frac{\partial \beta}{\partial x } ) \hat{y} + (\frac{\partial \alpha}{\partial x }\frac{\partial \beta}{\partial y } - \frac{\partial \beta}{\partial x }\frac{\partial \alpha}{\partial y } )\hat{z} \end{equation}

But I'm stuck, I don't know how to continue. Maybe $d\Omega = (dydz,dzdx,dxdy)$, $dr = (dx,dy,dz)$ and then I have to compute the dot product but I'm not sure.

Any hint? Thank you.

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  • $\begingroup$ Your right-hand side makes no sense, as you can't take the cross product of a scalar and a vector. It should just be $\int_{\partial\Omega}\alpha\nabla\beta\cdot dr$. $\endgroup$ – Ted Shifrin Apr 5 at 18:19
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This is one step with Stokes's Theorem: $$\int_\Omega d\alpha\wedge d\beta = \int_{\partial\Omega} \alpha\,d\beta,$$ since $d(\alpha\,d\beta) = d\alpha\wedge d\beta$. Note that if $\vec F = (F_1,F_2,F_3)$, the flux integral $\displaystyle\int \vec F\cdot d\vec S$ is precisely given by integrating the $2$-form $F_1\,dy\wedge dz + F_2\,dz\wedge dx + F_3\,dx\wedge dy$, as you suggested.

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