Is $\vec{F}\cdot \hat{n}$ zero on the boundary? I'm trying to do the following exercise

Let $\Omega\subset\mathbb R^3$ be a connected bounded subset with differentiable boundary $\partial\Omega$. If the divergence of $F:\mathbb R^3 \rightarrow \mathbb R^3$ is zero in $\Omega$, then
$$ \int_\Omega F^TJ^TFdx=0, $$
where $J$ means the jacobian of $F$ and $J^T$ the transpose matrix of $J$.

I have asked a solution for it here and @MarkViola has proved that the integral is equal to
$$\int_{\partial\Omega} (F\cdot F)(F\cdot n) \, dS , $$
where $n$ is the normal vector to the surface defined by the boundary. Now, why is this integral also zero??? I can see that if $\operatorname{div}F=0$ on $\Omega$ then
$$\int_{\partial \Omega} F\cdot n \, dS = 0. $$
But, does it imply that $F\cdot n =0$ on the boundary? If not, why the integral
$$\int_{\partial\Omega} (F\cdot F)(F\cdot n) \, dS $$
is equal to zero?
Remark I should remark that @Mark also gave an example for which $F\cdot n$ is not zero on the boundary and the integral is. The only problem I see in that case is that the boundary is $C^1$ a.e. but not actually $C^1$.
 A: It is not necessary that
$$
F\cdot n=0.
$$
For example, you may consider


*

*$F(x,y,z)=\left(y,x,0\right)$, and

*$\Omega=\left\{x\in\mathbb{R}^3:\left\|x\right\|\le 1\right\}$.


In this case, $\partial\Omega$ is the unit sphere, with its outward unit normal being
$$
n=\left(x,y,z\right).
$$
Then on $\partial\Omega$, we have
$$
F\cdot n=\left(y,x,0\right)\cdot\left(x,y,z\right)=2xy\not\equiv 0.
$$

In general, it is not necessary that
$$
\int_{\Omega}FJ^{\top}F{\rm d}V=0,
$$
even if $F$ is divergence-free. For example, consider


*

*$F(x,y,z)=\left(x,y,-2z\right)$, so that $F$ is divergence-free, and

*$\Omega=\left\{x\in\mathbb{R}^3:\left\|x\right\|\le 1\right\}$.


In this case, $\partial\Omega$ is still the unit sphere, with its outward unit normal being
$$
n=\left(x,y,z\right).
$$
Then $\partial\Omega$, we have
\begin{align}
F\cdot F&=x^2+y^2+4z^2=1+3z^2,\\
F\cdot n&=x^2+y^2-2z^2=1-3z^2.
\end{align}
We will make use of this standard spherical coordinate transform
\begin{align}
x&=r\sin\theta\cos\varphi,\\
y&=r\sin\theta\sin\varphi,\\
z&=r\cos\theta
\end{align}
with $r=1$ on $\partial\Omega$, $\theta\in\left[0,\pi\right)$, and $\varphi\in\left[0,2\pi\right)$. Then
\begin{align}
\int_{\Omega}FJ^{\top}F{\rm d}V&=\int_{\partial\Omega}\left(F\cdot F\right)\left(F\cdot n\right){\rm d}S\\
&=\int_{\partial\Omega}\left(1+3z^2\right)\left(1-3z^2\right){\rm d}S\\
&=\int_{\partial\Omega}\left(1-9z^4\right){\rm d}S\\
&=\int_0^{\pi}\sin\theta{\rm d}\theta\int_0^{2\pi}\left(1-9z^4\right){\rm d}\varphi\\
&=\int_0^{\pi}\sin\theta{\rm d}\theta\int_0^{2\pi}\left(1-9\cos^4\theta\right){\rm d}\varphi\\
&=\int_0^{\pi}\sin\theta\left(1-9\cos^4\theta\right){\rm d}\theta\int_0^{2\pi}{\rm d}\varphi\\
&=2\pi\int_0^{\pi}\left(9\cos^4\theta-1\right){\rm d}\cos\theta\\
&=-\frac{16\pi}{5}\\
&\ne 0.
\end{align}
