surface integral of angle between fixed vector and the unit normal so I came across this problem while studying and I just dont understand it...S is the smooth solid boundary of an object in R3, and v is its outward unit normal. l is fixed vector in R3. The angle between l and v is $\theta$. Prove:
$$\iint_S \cos(\theta) \;\mathrm{dS}=0$$ 
Now, I just dont understand this at all. The reason I dont understand is because $\cos \theta$ is not a vector or a scalar, it is just a value and is different for each unit normal. How can we take a surface integral of that?
I was thinking maybe it could be tied into the divergence theorem, which would make sense but I dont know how to make the leap from the divergence theorem and the 2 vectors there to the angle between them. Help!
 A: Let $\vec{F}$ be some fixed vector (I hate to use the symbol l which is easily confused with $1$).
Then since $F$ is constant, $\nabla\cdot \vec{F} = 0$ everywhere. And in particular, if $V$ is the volume enclosed by the surface in question,
$$\int_V \nabla\cdot \vec{F}\,dV = 0$$
Now look at $\cos \theta$.  Since at each point on the surface, with normal $\hat{n}$,
$$\vec{F}\cdot \hat{n} = |F|\cos \theta$$ we can say that 
$$\cos\theta  = \frac{\vec{F}\cdot \hat{n}}{|F|}$$
Therefore
$$\int_S \cos \theta\, dS = \int_S \frac{\vec{F}\cdot \hat{n}}{|F|}\, dS 
= \frac1{|F|}\int_S \vec{F}\cdot \hat{n} \, dS 
$$
Now use Gauss's law (another name for that is the divergence theorem) which says 
$$\int_S \vec{F}\cdot \hat{n} \, dS =\int_V \nabla\cdot \vec{F}\,dV $$
to get 
$$\int_S \cos \theta\, dS  = \frac1{|F|}\int_S \vec{F}\cdot \hat{n} \, dS   = \frac1{|F|}\int_V \nabla\cdot \vec{F}\,dV = \frac1{|F|} 0=0$$
A: Since $I$ is constant, we may assume it extends to a vector field defined on all of $\Bbb R^n$ (we don't need to restrict ourselves to $n = 3$).  Also, since $I$ is constant, we have
$\text{div}(I) = \nabla \cdot I = 0; \tag 1$
furthermore, if $I \ne 0$, 
$v \cdot I = \Vert I \Vert \cos \theta. \tag 2$
We now apply the divergence theorem to $I$ over the "object" $\mathscr O$ bounded by $S$:
$\displaystyle \Vert I \Vert \int_S \cos \theta \; dS =  \int_S I \cdot v dS = \int_{\mathscr O} \nabla \cdot I dV = \int_{\mathscr O} 0 dV = 0, \tag 3$
the desired result, provided $I \ne 0$.
Divergence Theorem: https://en.m.wikipedia.org/wiki/Divergence_theorem
