Q:

Why isn't $d\mathbf{A}$ normalized in Stokes' theorem? 

For a nice curve $C$ which is a boundary of a smooth surface $D$, Stokes' theorem says that
$$\begin{align*}
\oint_C \mathbf{F}\cdot d \mathbf{s} = \iint_D (\nabla \times \mathbf{F} )\cdot d\mathbf{A}
\end{align*}$$
However, I don't think I understand exactly what is going on with the $d\mathbf{A}$ term.  I would think that it is the vector of magnitude differential area $dA$ and direction $\hat{n}$, where $\hat{n}$ is the unit vector in the direction perpendicular to the orientation of $dA$. However, by reading practice problems online, I see that this is not the case.  What exactly is the definition of this $d\mathbf{A}$? 
In addition, many sources online show that after evaluating the dot product $(\nabla \times \mathbf{F} )\cdot \mathbf{A}$ (no $dA$ term), the surface integral is magically done only in the $xy$ plane, rather than any other randomly defined plane.  Is this legal? 

A:

The $d\vec S$ in the surface integral of Stokes' Theorem can be written $$d\vec S=\hat n dS$$
where $\hat n$ is the unit normal to the surface and oriented to conform with the right-hand rule. 
The scalar $dS$ is precisely a differential element of surface area.   So, $$\text{Surface Area of S}=\int_S dS$$ 
To compute $$\int_S V\cdot \hat n dS$$
often one projects the surface onto on of the coordinate planes (e.g., the $x-y$ plane).   However, $dS\ne dxdy$!   Rather, we need to account for the surface orientation upon projecting.   So, in this example $dS =\frac{dxdy}{\cos \gamma}=\frac{dxdy}{\hat z \cdot \hat n}$, where $\gamma$ is the angle between the surface normal and the axis of projection (here, the $z$-axis). 
If $\nabla \times \vec F$ is constant on $S$ and the surface normal $\hat n$ is constant (i.,e.,  a planar surface), then we have 
$$\int_S \nabla \times \vec F\cdot \hat n dS=(\nabla \times \vec F\cdot \hat n)\int_S dS=(\nabla \times \vec F\cdot \hat n)S=\nabla \times \vec F\cdot \vec S$$

