Vector by integral, notation convention or mistake? A part of a formula in my engineering book states $$\int_S\overrightarrow D\,\mathrm d\!\overrightarrow S$$
What I'm wondering about is bounds of integrals being a scalar, and the differential being a vector. The books is generally very non-rigorous math-wise , so I'm wondering if this is another "overlook" in the book, or it is a math convention that we don't have to write $$ \int_{\overrightarrow {S}}\overrightarrow D\,\mathrm d\!\overrightarrow S \qquad\text{or maybe}\qquad \int_{\overrightarrow {S}}\overrightarrow D\,\mathrm dS , $$ or is maybe the book correct?
Also a few words about differentials having vectors? In this case, the scalar product will take care of it, but would something like $\int\overrightarrow a\times\mathrm d\!\!\overrightarrow b$ even make sense? What could be possible values for a definite version of that integral?
 A: The fact that $dS$ and $S$ use the same letter is somewhat of a coincidence; in this notation, $S$ is usually the surface over which you're integrating, and $\vec{dS}$ the directed element of area. They're not supposed to be the same. For example, if I was integrating over a surface $\Sigma$, I would write the integral as $\int_\Sigma \vec{D} \cdot \vec{dS}$.
Regarding your second question: you have to remember that in physics/engineering, when a notation makes sense we don't usually need a rigorous definition for it if we can just use it. In this case, the physical meaning is pretty clear: you have to divide your surface into small elements, take the normal vector to each one, do the cross product with your vector field and then sum all the vectors you get.
Let's use a simpler example. The magnetic force on a wire loop can be writen as $\oint i\ \vec{dl}\times \vec{B}$. If you just plug in everything, you'll find that this makes perfect sense: if you have a parametrization $\vec{r}(t)$ for the loop, then $\vec{dl}$ = $\frac{d\vec{r}}{dt}dt$, exactly as if we were doing the more usual $\oint \vec{B}\cdot\vec{dl}$. Next, $\vec{B}$ will be a function of position, so we plug in the parametrization to get $\vec{B}(\vec{r}(t))$. Now we have two vectors the depend on $t$, so we just take the cross product. One of those vectors has a differential inside, but who cares? The result is an integral over $t$ of $\frac{d\vec{r}}{dt}\times\vec{B}$, which makes perfect sense.
And we did all of this without needing a definition, because in any real world scenario, blindly applying the formula just works.
