Q:

Why this does not have a dot product on the right: $\int_\Omega \nabla \phi dx =\int_{\partial \Omega}\phi \nu ds$

My main concern is because, on the right side of the expression, there is a vector $\nu$, but there is not any dot product.
$$\int_\Omega \nabla \phi dx =\int_{\partial \Omega}\phi \nu ds$$ 
Where $\phi $ is a scalar function, and $\nu$ is the normal vector.
Where can I find solved exercises of applications of this theorem? I would like to get an intuition about the performed operations

A:

The equation is correct if interpreted as $n$ scalar equations where $n$ is the dimension. I.e. writing $ \nu=(\nu_1,\dots, \nu_n)$, its true that (for nice enough functions $\phi$ and sets $\Omega$)
$$ \frac{\partial \phi}{\partial x_i} dx = \int_{\partial \Omega} \phi \nu_i dS, \quad i=1,\dots, n$$
where $dS$ is the surface measure. In other words, what your equation is saying is:
$$ \begin{pmatrix}\int_\Omega \frac{\partial \phi}{\partial x_1}dx \\ \vdots \\ \int_\Omega \frac{\partial \phi}{\partial x_n}dx\end{pmatrix}  =\begin{pmatrix} \int_{\partial \Omega} \phi \nu_1 dS\\ \vdots \\ \int_{\partial \Omega} \phi \nu_n dS\end{pmatrix}$$
The proof is simple: just apply the Divergence theorem $n$ times, once for each function $F$ defined below, where $i=1,\dots n$:
$$ F=(F_1,\dots,F_n):\Omega\to\mathbb R^n, F_j(x) = \begin{cases}0 & i\neq j \\ \phi & i=j\end{cases}$$
then $\nabla \cdot F = \partial\phi/\partial x_i$ and $F\cdot \nu = \phi \nu_i$.
I don't know where to find exercises. 

