How to interpret the integrand in this surface integral? Let Ω be the region in $ℝ^3$ defined by
$$ Ω={(x_1,x_2,x_3):max(∣∣x_1∣∣,∣∣x_2∣∣,∣∣x_3∣∣)≤1}$$
Let ∂Ω denote the boundary of Ω.
Calculate
$$∫_{∂Ω}ϕF⋅ndσ$$
where n is the unit normal vector, dσ denotes integration over ∂Ω,
$F_i=\large \frac{x_i}{(x_1^2+x_2^2+x_3^2)^{\frac{3}{2}}}^=\frac{x_i}{r^3}$
and $ϕ(y_1,y_2,y_3)$ is a continuously differentiable function of $\large y_i=\frac{x_i}{r}$. Assume that ϕ has unit mean over the unit sphere.
I just started on this problem so I don't want solutions.  
My question is:  how should I interpret $\phi F$?  Is $\phi$ another vector field and that I should take the inner product of $\phi$ with $F$?  
This wouldn't make much sense since; I'd end up with a scalar, and then scalar.$\vec n$ wouldn't really make sense either.
Any hints or suggestions are welcome.
Thanks,
EDIT: I'd welcome solutions at this point.  I am getting weird computations -- such as an integral that is equal to zero.  I tried using the "product rule" that I found on Wolfram Alpha to compute the divergence of $\phi F$.  I notice first that divF=0, so F alone is divergence-free.  But I honestly do not know whether I have the correct vector field after multiplication with $\phi$.  So, when computing the divergence of $\phi F$, I might be using an incorrect vector field.
 A: First of all, the expression is 
$$\int_{\partial \Omega} \phi F \cdot \vec n d\sigma.$$
To use divergence theorem, you better need to calculate 
$$\text{div} (\phi F) = \nabla \phi \cdot F + \phi \text{div} F.$$
Now there are two terms on the right hand side. Note that the second term is zero as $\text{div} F = 0$. The first term is zero too, as $F = \frac{1}{r^3} (x_1, x_2, x_3)$ and $\phi$ is constant along this direction. 
Thus you actually have $\text{div}(\phi F) = 0$. Now because you know almost nothing about $\phi$, you only know that it's average on the unit sphere is $1$. Thus you want to change $\partial \Omega$ to the unit sphere. Now let $B$ be the unit ball in $\mathbb R^3$ centered at $0$. So $\partial B$ is the unit sphere. Note that $B \subset \Omega$. Let $M = \Omega\setminus B$. Then $\partial M = \partial \Omega - \partial B$. By the divergence theorem, 
$$\int_{\partial M} \phi F \cdot \vec n d\sigma = \pm \int_M \text{div} (\phi F) dx = 0. $$
This imply 
$$\int_{\partial \Omega} \phi F \cdot \vec n d\sigma = \int_{\partial B} \phi F \cdot \vec n d\sigma$$
Now on the sphere $\partial B$, the normal vector is $(x_1, x_2, x_3)$, thus $F\cdot \vec n  =1$ and so 
$$\int_{\partial B} \phi F \cdot \vec n d\sigma = \int_{\partial B} \phi d\sigma = \text{Area of the unit sphere} = 4\pi.$$
