## On integrating cross products of cross products

[Updated on 20-6-2011 by adding a lemma, in the hope that this will make things more transparent. ]
Inspired by Wonging’s result on integrating products of dot products, we have another Wongting-type theorem:

Theorem 1 (Wongting’s theorem, 2011) For ${v_1, v_{2}\in \mathbb{R}^3 }$,

$\displaystyle \int_{\mathbb{S}^{2}} (v_1\times x)\times (v_2\times x)dS(x)= \omega_3 (v_1\times v_{2})\in \mathbb{R}^3$

where ${\omega_3=\frac{4\pi}{3}}$ is the volume of the unit ball in ${\mathbb{R}^3}$.

I intend to generalize it further, but since $u\times x$ is not defined in $\mathbb{R}^n$, $n>3$, it is inappropriate to integrate on $\mathbb{S}^{n-1}$, so we have to find other space to integrate the corresponding cross product of cross product. I plan to do it later.

To prove the theorem, we use the following lemma.

Lemma 2 If ${u, v,x\in \mathbb{R}^3}$, then

$\displaystyle (u\times x)\times (v\times x)= \det(u,v,x)x.$

Proof: We can w.l.o.g. assume $|x|=1$. Let ${e_3=x}$ and let $\{e_i\}_{i=1}^3$ forms a positive orthnormal basis. Note that any component of $u$ which is parallel to $x$ will not contribute to both sides of the equation. So we can assume $\displaystyle {u=u^1 e_1+ u^2 e_2}$, thus ${u\times x= u^2 e_1 - u^1e_2}$. Similarly ${v\times x= v^2 e_1 -v^1e_2}$, so ${(u\times x)\times (v\times x)=\left |\begin{matrix} u^1 & u^2\\ v^1 & v^2 \end{matrix}\right|e_3=\det(u,v,x)x}$ as ${x=e_3}$. $\Box$
Note that this lemma can actually be seen geometrically.

Proof of Wongting’s theorem:
By the linearity of both sides of the expression of Wongting’s theorem and the anti-symmetric property of ${(v_1, v_{2})\mapsto v_1\times v_{2}}$, we can assume that

$\displaystyle v_1=e_1, v_2=e_{2}\quad \text{(WHY??)}$

By the lemma, the integrand on LHS of the statement in the theorem is $x_3x\in \mathbb{R}^3$.

By the antisymmetry of the funcions ${x_1 x_3}$ and ${x_2 x_3}$ upon reflection of the ${x_3}$-plane, the first two components integrate to be zero over the sphere. For the third component,

$\displaystyle \int_{\mathbb{S}^2} x_3^2 dS(x)=\frac{1}{3} \int_{\mathbb{S}^2} (x_1^2+x_2^2+x_3^2)dS(x)= \frac{4\pi}{3}=\omega_3.$

Thus ${LHS= \omega_3 e_3}$. Clearly ${v_1\times v_2=e_3}$, so we get the result.

$\Box$