# Write Formula using differential forms

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

$$$$\int_{\Omega} \nabla\alpha \times \nabla \beta \ \cdot d\Omega = \int_{\partial{\Omega}} \alpha \nabla \beta \ \cdot dr$$$$

What I have done so far:

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

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

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

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.

• 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 $\int_{\partial\Omega}\alpha\nabla\beta\cdot dr$. – 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.