Is this problem ill-defined?

Question

I am confused by a problem from Schey's Div, Grad, Curl and All That:

Use the divergence theorem to show that $$\iint_S\hat{\mathbf{n}}dS=0,$$ where S is a closed surface and $\hat{\mathbf{n}}$ the unit vector normal to the surface $S$.

The whole discussion of surface integrals in this text so far has dealt with scalar-valued integrands, like:

$$\iint_S\mathbf{F}\cdot\hat{\mathbf{n}}dS.$$

So it would seem the author is either introducing vector-valued integrands for the first time, or suggesting $\mathbf{F}=1$, which I assume is a typo.

Any ideas about how to resolve this issue? I guess I could just say that, over a closed surface, for every surface normal there will be a normal elsewhere on the surface pointing in the opposite direction so they all cancel. But this approach doesn't use the divergence theorem.

Answer 1

$\DeclareMathOperator{div}{div}\DeclareMathOperator{grad}{grad}$ We have $$ \int\limits_S n \,dS = \int\limits_S \sum_i n_i e_i \, dS = \sum_i e_i \int\limits_S (e_i \cdot n) dS = \sum_i e_i \int\limits_V \div e_i \, dV = \sum_i e_i \int\limits_V 0 \, dV = 0 $$ where $e_i$ is the canonical base vector in $i$-direction: $(e_i)_j = \delta_{ij}$.

Answer 2

The approach you suggest would not work. One-by-one cancellation works with sums but in the case of integrals you'd also need to show that the contributions with opposite signs are represented with the same weight (size of a differential patch corresponding to that $\vec n$ or something).

I think they mean to show that

$$\iint_S n_j\,\mathrm{d}S = 0, \quad \forall j \in \{x,y,z\}$$

because integral of a vector, unless defined more concretely, is understood as a vector of integrals of its components, and that is zero if and only if each of the integrals is. (The other possible understanding

$$\iint_S \vec n \cdot \,\vec{\mathrm{d}S} = 0$$

could not be asked because it is not true: the value of the left-hand side is the surface area of $S$.)

In the above sense, that's not hard, because $n_j = \vec e_j \cdot \vec n$, where $\vec e_j$ is a unit vector in the $j$-th coordinate. I'm sure you can take it from there.