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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.

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    $\begingroup$ As you said, $\phi$ is actually a function, so $\phi F$ is a function times a vector fields, thus is still a vector field. Then $\phi F \cdot n$ is the inner product between these two vector fields. $\endgroup$ – user99914 Oct 7 '15 at 5:24
  • $\begingroup$ Hi @JohnMa is $\phi$ a scalar-valued function? Is $\phi F$ a composition of a scalar function with a vector field? I'm a little confused with what this object is. Can you elaborate just a bit more? Thanks, $\endgroup$ – User001 Oct 7 '15 at 5:27
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    $\begingroup$ For example, if $F = (F_1, F_2, F_3)$ then $\phi F$ is another vector fields given by $(\phi F_1, \phi F_2, \phi F_3)$. $\endgroup$ – user99914 Oct 7 '15 at 5:28
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    $\begingroup$ $\phi$ is a function with three variable, so $\phi(y_1, y_2, y_3)$ is a number. The sentence "Assume that $\phi$ has unit mean over the unit sphere" is a hint that my interpretation is correct (What does this mean if $\phi$ is a vector fields?) $\endgroup$ – user99914 Oct 7 '15 at 5:36
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    $\begingroup$ It's not composition, just multiplication. $\endgroup$ – mrf Oct 7 '15 at 6:17
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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

$$\int_{\partial B} \phi F \cdot \vec n d\sigma = \int_{\partial B} \phi d\sigma = \text{Area of the unit sphere} = 4\pi.$$

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    $\begingroup$ Note that $\phi (x) = \phi (\frac{x_1}{r}, \frac{x_2}{r}, \frac{x_3}{r})$, so $\phi(tx) = \phi (\frac{tx_1}{tr}, \frac{tx_2}{tr}, \frac{tx_3}{tr}) = \phi (\frac{x_1}{r}, \frac{x_2}{r}, \frac{x_3}{r}) = \phi(x)$. Thus the value of $\phi$ is the same along each ray $\{tx : t>0\}$. @LebronJames $\endgroup$ – user99914 Oct 8 '15 at 11:11
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    $\begingroup$ No! @LebronJames it is not known if $\nabla \phi = 0$, I only know that along the $r$ direction, $\nabla_r \phi = 0$, but it is sufficient to show that $\nabla \phi \cdot F = 0$ as $F$ is parallel to the $r$-direction. $\endgroup$ – user99914 Oct 9 '15 at 0:33
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    $\begingroup$ $\nabla \phi$ is a vector, and it has no $F$ component as $\phi$ is constant along that direction. Basically we are using the formula $\nabla_V \phi = \nabla \phi \cdot V$ for all fixed vector $V$ ($\nabla_V \phi$ is the directional derivative of $\phi$ along $V$. @LebronJames $\endgroup$ – user99914 Oct 9 '15 at 0:47
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    $\begingroup$ Ahhhh...that's so pretty. Yes, of course. That is an awesome argument. Thanks a ton for your time and patience @JohnMa. I learned a lot from you on this problem. Have a great night :-) $\endgroup$ – User001 Oct 9 '15 at 0:51
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    $\begingroup$ In Euclidean space, if you have a function $\phi$ defined on an open set in $\mathbb R^n$, then as a vector $\nabla \phi$ is given by $\nabla \phi = (\partial_1 \phi, \partial_2 \phi, \cdots, \nabla_n \phi)$. It will also satisfies $\nabla \phi \cdot V = \partial_V \phi$. @LebronJames $\endgroup$ – user99914 Oct 9 '15 at 1:39

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