parametrization of surface element in surface integrals I don't understand this

How $ dS = \sqrt{ \left ( \partial g \over \partial x\right )^2 + \left ( \partial g \over \partial y\right )^2 + 1 } \; dA \; \; $ ?? Is $ dA = dx\times dy$??
 A: The surface integral is given by:
$$
\int_{S} f \,dS 
= \iint_{D} f(\mathbf{\sigma}(u, v)) \left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\| \,du\,dv
$$
Where $\mathbf{\sigma}(u, v)$ is a parametrization of the surface $S$.
Now, if $S$ is given by the function $z = g(x, y)$, we have:
$$
\sigma(u, v) = \sigma(x, y) = (x, y, g(x, y))
$$
Therefore:
$$
\left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\| = \left\|\left(1, 0, {\partial g \over \partial x}\right)\times \left(0, 1, {\partial g \over \partial y}\right)\right\|
$$
Calculate the cross product on the RHS to find that:
$$
\left(1, 0, {\partial g \over \partial x}\right)\times \left(0, 1, {\partial g \over \partial y}\right) = \left(-{\partial g \over \partial x}, -{\partial g \over \partial y}, 1\right)
$$
And its norm is equal to:
$$
\sqrt{\left({\partial g \over \partial x}\right)^2+\left({\partial g \over \partial y}\right)^2+1}
$$
Therefore, the surface integral for $S$ given by $z = f(x, y)$ is:
$$
\iint_{D} f(x, y, g(x, y)) \sqrt{\left({\partial g \over \partial x}\right)^2+\left({\partial g \over \partial y}\right)^2+1} \,dx\,dy
$$
A: The surface in question is given by $z=g(x,y)$.
The vector in the surface that comes as a small change, $\mathrm{d}x$, to $x$ is
$$
\left(1,0,\frac{\partial g}{\partial x}\right)\mathrm{d}x\tag{1}
$$
The vector in the surface that comes as a small change, $\mathrm{d}y$, to $y$ is
$$
\left(0,1,\frac{\partial g}{\partial y}\right)\mathrm{d}y\tag{2}
$$
If we take the cross product of $(1)$ and $(2)$, we get
$$
\left(1,0,\frac{\partial g}{\partial x}\right)\times\left(0,1,\frac{\partial g}{\partial y}\right)\,\mathrm{d}x\,\mathrm{d}y = \left(-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1\right)\,\mathrm{d}x\,\mathrm{d}y\tag{3}
$$
The area represented is the absolute value of $(3)$:
$$
\sqrt{\left(\frac{\partial g}{\partial x}\right)^2+\left(\frac{\partial g}{\partial y}\right)^2+1}\quad\mathrm{d}x\,\mathrm{d}y\tag{4}
$$
A: For a surface integral (or line integral, or path integral, or change of variables...), the computation breaks into four steps.  Notation: let $f(x,y,z)$ denote the integrand and let $S$ be the surface parametrized by $(x(u,v),y(u,v),z(u,v))$ so that $u$ and $v$ are the parameters.


*

*Convert integrand to $u$'s and $v$'s by plugging in the parametrizations for $(x,y,z)$,

*Convert differential by $dS = ||T_u \times T_v||dudv$, where 
$$ T_u = \left( \frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u} \right), $$
$$ T_v = \left( \frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v} \right), $$

*Convert integral bounds so that they are in terms of $u$ and $v$

*Put steps 1-3 together and compute!


As applies to your situation, your surface is defined by the portion of the graph of $z = g(x,y)$ which lies over the region $D$ in the $xy$-plane.  Thus, we can parametrize the surface using the $x$ and $y$ as the parameters.  Pedantically, we have $x=u$, $y=v$, and $z=g(u,v)$.
Going through the steps now: First, we have the integrand $f(x,y,z)$.  Plugging in $x,y$ and $z$ gives us $f(u,v,g(u,v))$.  But that's just $f(x,y,g(x,y))$.
Step 2: Calculate $T_u$ and $T_v$:
$$ T_u = \left( 1,0,\frac{\partial g}{\partial u} \right) $$
$$ T_v = \left( 0,1,\frac{\partial g}{\partial v} \right) $$
$$ T_u \times T_v = \left( -\frac{\partial g}{\partial u}, -\frac{\partial g}{\partial v}, 1 \right) $$
$$ ||T_u \times T_v|| = \sqrt{ \left( \frac{\partial g}{\partial u} \right)^2+\left( \frac{\partial g}{\partial v} \right)^2+1} $$
$$ dS = ||T_u \times T_v||dudv = \sqrt{ \left( \frac{\partial g}{\partial u} \right)^2+\left( \frac{\partial g}{\partial v} \right)^2+1}dudv $$
Now, $x=u$ and $y=v$, so we may as well write 
$$ dS = \sqrt{ \left( \frac{\partial g}{\partial x} \right)^2+\left( \frac{\partial g}{\partial y} \right)^2+1}dxdy $$
And yes, $dA=dxdy$.
Step 3: We need ranges on $u$ and $v$, which for this problem means ranges on $x$ and $y$.  we know that $(x,y)$ must lie in $D$.
That's how all the pieces come together.  It's a standard surface integral with the wrinkle that we get to use $x$ and $y$ as the parameters.
A: It is simple 
$$dS = \sqrt{(dxdy)^2+(dydz)^2+(dxdz)^2} = \sqrt{1+\left(\dfrac{dz}{dx}\right)^2+\left(\dfrac{dz}{dy}\right)^2}\hspace{1mm}‌​dxdy$$
