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 $?
 A: 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
$$\int_0^1\int_0^{2\pi}\frac {r}{\sqrt{1-r^2}}d\theta dr=2\pi $$
A: This is an alternative solution in case one's interested in:  
By the spherical coordinates we get $x=\cos\theta\sin\varphi, \; \;y=\sin\theta\sin\varphi, \; \;z=\cos\varphi$
And, since  $$d A= \sqrt{\left[ \frac {\partial y \partial z}{\partial \theta \partial \varphi}\right]^2 +\left[ \frac {\partial z \partial x}{\partial \theta \partial \varphi}\right]^2+\left[ \frac {\partial x \partial y}{\partial \theta \partial \varphi}\right]^2 }d \theta d \varphi$$
we get, $$d A=\sqrt{(\sin^2\varphi\cos\theta)^2+(\sin^2\varphi\sin\theta)^2+(\cos\varphi\sin\varphi)^2}d \theta d \varphi=\sin\varphi d \theta d \varphi$$
Thus, the area of the sphere is $$\int_0^{\pi}\int_0^{2\pi}\sin\varphi d \theta d \varphi=-2\pi\cos\varphi|_0^\pi=4\pi$$
