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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.