Showing that the unit sphere is a surface I was going through Andrew Pressley's book and on the place where they have discussed surfaces, there is one example which deals with the unit sphere. I have understood to the point where they have taken the surface patch and the fact that it will not be able to cover the whole sphere and will leave out a semicircle. But after that they are saying they want to rotate the given surface patch by π radians about the z axis and π/2 radians about the x axis to obtain another surface patch which would cover the region not covered before. Here my doubt is, wouldn't rotating it just by π radians about the z axis solve my problem? I'm confused here. 
Ps. I wanted to upload a photo but the site wouldn't let me because I don't have enough reputation. It's on page 72 of Elementary differential geometry by Andrew Pressley

Lots of ways to do this. One way would be to consider the function $f: \mathbb R^3\rightarrow\mathbb R$ defined by $x \mapsto ||x||^2 $. Check that $1$ is a regular value of this smooth map and $S^2=f^{-1}(1)$ (Use Implicit Function Theorem). The other way would be to look at the stereographic projection $S^2-N\rightarrow\mathbb R^2$ where $N $ is the north pole. and similarly $S^2-S\rightarrow\mathbb R^2$ where $S$ is the south pole. The way you mention it the parametrization is given by $(-\pi/2,\pi/2)\times (0,2\pi)\rightarrow S^2$ given by $(\theta,\phi)\mapsto(cos\ \theta\ cos\ \phi,cos\ \theta\ sin\ \phi, sin\ \theta) $. This covers $S^2-\{x\in S^2:x_1>0,x_2=0\}$ i.e. the sphere minus a semicircle. It is easy to see if you rotate the surface patch with axis of rotation as the $z-axis$ followed by the $x-axis$ then the image is still in $S^2$ and it is still a surface patch since $rotation$ is an isometry. Finally the two surface patches cover $S^2$ and this shows $S^2$ is a regular surface in $\mathbb R^3$ 
