Compute the surface area of the unit sphere $x^2+y^2+z^2=1$ Compute the surface area of the unit sphere $x^2+y^2+z^2=1$
The following is a solution the book suggests:  

The upper hemisphere is the graph of the function $\varphi(x,y)=\sqrt{1-x^2-y^2}.$
  A little calculation yields $$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}=\frac 1{\sqrt{1-x^2-y^2}} \tag {1}$$
Note that $d A=\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}d x d y \tag {*} $
  So, by $(*)$, the area of the hemisphere is obtained by integrating the function in $(1)$ over the unit disc. Switching to polar cordinates yields $$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi \tag{2}$$

Firstly, in $(1)$   how does this "$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}$ " become equal to $\sqrt{1-x^2-y^2}$? Also how do we get $(2)$ , aren't we supposed to compute $d A=\sin\varphi d\theta d\varphi $?

Note that
$$(\partial_x\varphi)^2=\frac{x^2}{\varphi^2}$$
$$(\partial_y\varphi)^2=\frac{y^2}{\varphi^2}$$
and then 
$$\sqrt{1+(\partial_x\varphi)^2+(\partial_y\varphi)^2}=\frac1{\sqrt{1-x^2-y^2}}$$
and thus
$$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi $$
