Surface Integral for the plane The formula for surface integral is given by:
$$\iint_S \overrightarrow F \cdot \hat n \space  dS $$
I want to know about the $dS$ part.
For example:
If the plane lies in the first octant then it is:
$$dS = \frac{dx \space dy}{|\hat k \cdot \hat n|}$$
There are also other case such as:
$$dS = \frac{dz \space dy}{|\hat i \cdot \hat n|}$$
and 
$$dS = \frac{dx \space dz}{|\hat j \cdot \hat n|}$$
How do I exactly decide which $dS$ formula to choose i.e. if the plane lies in fourth or second octant. How will I choose?
 A: The plane (which appears only in a comment of yours) is of course infinite. I guess you are told to integrate over the triangle $S$ cut out of this plane by the first octant. This triangle projects down to the $(x,y)$-plane to the triangle
$$S':=\left\{(x,y)\>\biggm|\>0\leq x\leq 6, \  \>0\leq y\leq4-{2\over3}x\right\}\ .$$
Now use the parametrization
$$\phi:\quad S'\to S,\qquad (x,y)\mapsto \left(x,y,\ 2-{1\over3}x-{1\over2}y\right)$$
to compute the requested flow. 
In order to proceed further we need the exact description of the (supposedly nonconstant) vector field $\vec F$ and the intended orientation of $S$.
A: To calculate a surface integral over a vector field you need parametrization of that oriented surface. Without too much of rigour:
$$\begin{align}\iint_S {\mathbf v}\cdot\mathrm d{\mathbf {S}} &= \iint_S \left({\mathbf v}\cdot {\mathbf n}\right)\,\mathrm dS\\&{}= \iint_T \left({\mathbf v}(\mathbf{x}(s, t)) \cdot {\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \over \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\|}\right) \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\| \mathrm ds\, \mathrm dt\\&{}=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \mathrm ds\, \mathrm dt.\end{align}$$
$d{\mathbf {S}}$ means that you are integrating vector field over an oriented surface. When you "extract" a unit normal vector from it you are left with a scalar function and $dS$ which correspond to a "normal" (scalar) surface integral. In this notation $$dS= \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\| \mathrm ds\, \mathrm dt$$
where $\mathbf{x}(s,t): T\rightarrow S$ is a parametrization of your surface.
Also please note how $\left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\| $ cancel each other and simplify calculations.
A: Those dot products come about by projecting a small area $dS$ on a surface onto a small area $dA$ in the plane. Specifically, we get that $dS\cos{\theta}=dA$ where $dA=dxdy$, for example, and $\theta$ is the angle between $dA$ and $dS$. Note that the angle between two rectangles is the same as the angle between their normal vectors.
If $dS$ were being projected onto the $xz$ plane then we would have $dA=dxdz$.
We know how to integrate over $dxdy$ but not how to integrate over $dS$ so we want to convert $\int dS$ into an integral over $dA=dxdy$. Now, $$dS\cos{\theta}=dxdy\implies dS=\frac{dxdy}{\cos{\theta}}$$
We use the dot product to compute $\cos{\theta}$.
Let $\hat n$ be the normal to the surface at the point around which we consider $dS$. We get $$\hat n\cdot\hat k=\lvert\hat n\rvert\lvert\hat k\rvert\cos{\theta}=\cos{\theta}$$
Therefore, we get that $$dS=\frac{dxdy}{\hat n\cdot\hat k}$$
For the case when we project upon another plane like $xz$, for example, we get that $$dS=\frac{dxdz}{\cos{\theta}}=\frac{dxdz}{\hat n\cdot\hat j}$$ because $\hat j$ is the normal vector to the $xz$ plane.
