Parametrising a Sphere If parametrising a sphere $r=(\cos{u}\cos{v},\cos{u}\sin{v},\sin{u})$ is it true that limits for $u$ and $v$ will be $-\pi/2$ and $\pi/2$ and $0$ and $2\pi$ respectively? Why is this different to when parametrising a sphere, with parametrization $\rho\sin{\phi}\cos{\theta}, \rho\sin{\phi}\sin{\theta}, \rho\cos{\phi}$ where the limits of $\phi$ are between and $0$ and $\pi$?
 A: It is almost the same with $\rho=1$ (unit sphere) and $\phi=\pi/2-u$ $v=\theta$
A: Simply make a co-ordinate change via substitution:
substitute $u = \pi/2 - \phi$ and then adjust the range of the $u$ value accordingly. I.e:
$-\pi/2 \leq u \leq \pi/2$ becomes $0 \leq \phi \leq \pi$.
Note: $cos(\pi/2 - \phi) = sin(\phi)$ and $sin(\pi/2 - \phi) = cos(\phi)$.
This is simply a rotation of the co-ordinate system , but since spheres are symmetric it is invariant under rotation.
A: Let me explain my intuition using diagrams , I just see it as difference of angle. As sphere is symmetric both of them gives same result. Moreover, here we just defined $\phi$ to be a different angle but in the end $z$ coordinate is same 

Sorry for the bad diagrams, i am on laptop.
Here as you can see $cos \phi$ moves along only one direction clockwise or anticlockwise whereas $sin \phi$ has to move toward up as well as down, moreover think of their respective domain.
A: A useful exercise might be to consider these six points in $(x,y,z)$ coordinates:
$$(A,0,0)_{x,y,z}$$
$$(0,A,0)_{x,y,z}$$
$$(0,0,A)_{x,y,z}$$
$$(-A,0,0)_{x,y,z}$$
$$(0,-A,0)_{x,y,z}$$
$$(0,0,-A)_{x,y,z}$$
Write each of these in standard $(\rho,\phi,\theta)$ spherical coordinates.
Then write the same point in $(r,u,v)$ coordinates,
where $x=r\cos u \cos v,$ $y = r \cos u \sin v,$ and $z = r \sin u.$
Then you'll see how the coordinate systems relate to one another
and why the range of $u$ is different from the range of $\phi$.
The six points have been chosen so that it will be easy to guess the correct
values of the sines and cosines in every case.
This is not meant to be a difficult exercise.
