# 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!

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 $$\int_S \cos \theta\, dS = \frac1{|F|}\int_S \vec{F}\cdot \hat{n} \, dS = \frac1{|F|}\int_V \nabla\cdot \vec{F}\,dV = \frac1{|F|} 0=0$$

• What's the meaning of $\int_S cos \theta dS$? Is that just algebra? – mavavilj Sep 23 '18 at 23:51

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