I am trying to prove that the integral in three dimensions of dot product (F, curl G) is equal to the integral of dot product $(G, \operatorname{curl} F)$ on a domain $D$ in $\mathbb{R}^3$ that is bounded by a closed surface $S$. The surface normal $n$ of $S$ is a linear combination of $F$ and $G$ (two continuously differentiable vector fields).
Thus I think I should first rewrite for the $(S,n)$ with the divergence theorem and use the distributive property of the dot product – using the linear combination fact for the surface normal – and write:
$$
\iiint \limits_D (F, \operatorname{curl} G) \text{d}V = \iint \limits_S (aF,\operatorname{curl} G) \text{d}A + \iint \limits_S (bG,\operatorname{curl}G) \text{d}A \\
\iiint \limits_D (G, \operatorname{curl} F) \text{d}V = \iint \limits_S (aF,\operatorname{curl} F) \text{d}A + \iint \limits_S (bG,\operatorname{curl} F) \text{d}A
$$
I feel like I should be able to set some things equal to each other here, but I get stuck.
Also, I am wondering if I am switching in the right way between the domain and the surface. Any help would be greatly appreciated!
Best Answer
The vector identity you need is this: For $C^1$ vector fields $\vec F$ and $\vec G$ on $\Bbb R^3$, we have $$\text{div}(\vec F\times\vec G) = (\text{curl}\,\vec F,\vec G) - (\vec F,\text{curl}\,G).$$ Then the Divergence Theorem tells us that $$\iiint_V \text{div}(\vec F\times\vec G)\,dV = \iint_S (\vec F\times\vec G,\vec n)\,dS,$$ where $\vec n$ is the unit outward-pointing normal to $S=\partial V$. Now, if you assume that at each point of $S$, $\vec n$ is a linear combination of $\vec F$ and $\vec G$, then the integrand on the right-hand side vanishes. This gives what I believe is the result you were trying to establish.