why $\iint \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\:dx\:dy$ means surface area? Why the following integral means the area of surface $f(x,y)=z$?
$$\iint \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}\:dx\:dy$$
 A: $\frac{\partial f}{\partial x}$ - Think of this as saying "If x increases by $\partial x$, how much does f increase by?".  Now above every little square patch $dx dy$, you are computing the area of a small plane.  The plane can be represented by the four corners $$(x, y, f(x,y))$$ $$ (x + \partial x, y, f(x,y) + \frac{\partial f}{\partial x} \partial x)$$ $$(x, y + \partial y, f(x,y) + \frac{\partial f}{\partial y} \partial y)$$ $$ (x + \partial x, y + \partial y, f(x,y) + \frac{\partial f}{\partial x}\partial x + \frac{\partial f}{\partial y} \partial y)$$
If you compute the area of this parallelogram, you get $$\sqrt{\partial x^2 \partial y^2 + \left(\frac{\partial f}{\partial x}\right)^2 (\partial x)^2 (\partial y)^2  + \left(\frac{\partial f}{\partial y}\right)^2(\partial y)^2 (\partial x)^2}$$
Taking out the $(\partial x)^2(\partial y)^2$ gets you the result.
A: There is a fact that $||\vec{v}\times\vec{w}||$ is the area of the parallelogram spanned by $\vec{v}$ and $\vec{w}$.  So, a good approximation of the area around a point $x$ on a surface is $||\vec{T_1}\times \vec{T_2}||$, where $\vec{T_1}$ and $\vec{T_2}$ are linearly independent tangent vectors at $x$.  Your surface is given by $z=f(x,y)$.  Think of this as all the points $(x,y,f(x,y))$.  Two tangent vectors will then be $(1,0,f_x(x,y))$ and $(0,1,f_y(x,y))$.  These are just the coordinate-wise partial derivatives of $(x,y,f(x,y))$.  Therefore your good approximation to the area around any point on the surface is given by the magnitude of the cross product, which works out to be
$$
\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}
$$
Now just integrate over your surface.
A: There's a pretty good summary here, but it uses some elementary differential geometry concepts.  The truth is that surface area is a notoriously tricky concept to define precisely.  In any case, the square root factor results from the definition of surface area in terms of a cross-product of tangent vectors in directions defined by a parametrization $(u,v)$.  
For such a general parametrization, the surface area is defined in terms of the first fundamental form
$$\iint_S du\,dv \sqrt{E\,G-F^2}$$
where
$$E=\left(\frac{\partial x}{\partial u} \right)^2+\left(\frac{\partial y}{\partial u} \right)^2+\left(\frac{\partial z}{\partial u} \right)^2$$
$$F=\left(\frac{\partial x}{\partial u} \right)\left(\frac{\partial x}{\partial v} \right) + \left(\frac{\partial y}{\partial u} \right)\left(\frac{\partial y}{\partial v} \right) + \left(\frac{\partial z}{\partial u} \right)\left(\frac{\partial z}{\partial v} \right) $$
$$G=\left(\frac{\partial x}{\partial v} \right)^2+\left(\frac{\partial y}{\partial v} \right)^2+\left(\frac{\partial z}{\partial v} \right)^2$$
When $z=f(x,y)$, $x=u$, $y=v$.  Then
$$E\,G-F^2 = \left[1+\left(\frac{\partial z}{\partial x} \right)^2\right]
 \left[1+\left(\frac{\partial z}{\partial y} \right)^2\right] - \left(\frac{\partial z}{\partial x} \right)^2\left(\frac{\partial z}{\partial y} \right)^2$$
The result follows.
A: The following is redundant as an answer, but since I had already typed it and it's too long for a comment, I decided to post it anyway.
Let $\vec {r}\colon \operatorname{dom}(f)\longrightarrow \Bbb R^3$, be defined by $\vec {r}(x,y)=(x,y,f(x,y))$, for all $(x,y)\in \operatorname{dom}(f)$.
Let $1_\varphi\colon \Bbb R^3\longrightarrow \Bbb R$ be constantly $1$.
If $\vec{r},f$ and $\operatorname{dom}(f)$ respect certain conditions then the area is given by 
$$\displaystyle \iint \limits_{\operatorname{dom}(f)} (1_\varphi \circ \vec{r})(x,y)\left\Vert\frac{\partial \vec{r}}{\partial x}(x,y) \times \frac{\partial \vec{r}}{\partial y}(x,y) \right\Vert\mathrm dx\mathrm dy$$
Since  for all $(x,y)\in \operatorname{dom}(f)$ it is true that $(1_\varphi \circ \vec{r})(x,y)=1\\$, $$\displaystyle \frac{\partial \vec{r}}{\partial x}(x,y)=\left(1,0,\frac{\partial f}{\partial x}(x,y)\right),$$  $$\displaystyle  \frac{\partial \vec{r}}{\partial y}(x,y)=\left(0,1,\frac{\partial f}{\partial y}(x,y)\right),$$ $$\displaystyle  \frac{\partial \vec{r}}{\partial x}(x,y) \times \frac{\partial \vec{r}}{\partial y}(x,y)=\left(\displaystyle -\frac{\partial f}{\partial x}(x,y),-\frac{\partial f}{\partial x}(x,y),1\right)$$ and $$\displaystyle \left\Vert\left(\displaystyle -\frac{\partial f}{\partial x}(x,y),-\frac{\partial f}{\partial x}(x,y),1\right)\right\Vert=\sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial y}(x,y)\right)^2,}$$ it follows that
$$\displaystyle \iint \limits_{\operatorname{dom}(f)} (1_\varphi \circ \vec{r})(x,y)\left\Vert\frac{\partial \vec{r}}{\partial x}(x,y) \times \frac{\partial \vec{r}}{\partial y}(x,y) \right\Vert\mathrm dx\mathrm dy=\\=\iint \limits_{\operatorname{dom}(f)}\sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2+\left(\frac{\partial f}{\partial y}(x,y)\right)^2}\mathrm dx\mathrm dy.$$
