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.


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