Q:

Parametrization of truncated sphere not centered on the origin.

The sphere is $x^2+y^2+(z-1)^2=25$ with $0 \le z$ and i wanted to parametrize it with spherical coordinates.
How does it change with the fact that it is not centered on the origin? 
From what value to what value are the variables defined? And what whould happen if the sphere is $x^2+y^2+(z-5)^2=25$?
Thanks for your time!

A:

A parametrization, as I assume you're referring to, is the description of the points on a surface in three dimensions as:
$$x = f(s, t)\\y=g(s, t)\\z = h(s, y).$$
If this is the case, to translate a curve like a sphere centered at the origin, all you have to do is add the "new center's" coordinates to the parametrization centered at the origin.
For example, the parametrization of a sphere with radius $\rho$ centered at the origin is:
$$x=\rho\sin(\phi)\cos(\theta)\\
y = \rho\sin(\phi)\sin(\theta)\\
z = \rho\cos(\phi)$$
While the same sphere centered at $\langle2, \pi, e\rangle$ would be parametrized:
$$x=2+\rho\sin(\phi)\cos(\theta)\\
y = \pi+\rho\sin(\phi)\sin(\theta)\\
z = e+\rho\cos(\phi)$$
In these cases, your bounds on $\theta$ and $\phi$ don't change.

For such a limit as $0\le z$, you would have to change the bounds of $\phi$ and $\theta$. Simply take the bounds used previously, which in the parametrization I gave were $0<\phi\le\pi$ and $0 < \theta \le 2\pi$, and take their intersection with the bounds of $\phi$ and $\theta$ for which the restriction holds.
For your example given, we have the parametrization:
$$x=5\sin(\phi)\cos(\theta)\\
y = 5\sin(\phi)\sin(\theta)\\
z = 1+5\cos(\phi)\\
0<\phi\le\pi\\
0<\theta\le2\pi$$
So asserting that $0 \le z$ implies $0\le 1+5\cos(\phi)$, which in turn implies that $-\frac15\le\cos(\phi)$. The intersection of this and the previous bound on $\phi$ gives that the bound is $0 < \phi \le \arccos(-\frac15).$
The bounds on $\theta$ are unchanged by restrictions on $z$ in this specific case.
Does this help?

