What kind of surface is this? Is there a way to plot this? I am given the surface:
$$S=\{ \vec{x} \in \mathbb R^3: {\|\vec{x} \|}_2^2=4, x^2+y^2 \le 1, z >0 \}$$
and I want to calculate the Mass of $S$ given a density $\rho$. It sort of looks like the upper half of a sphere. The problem I have is that the first equation ${\| \vec{x} \|}_2^2=4$ means that the radius of this sphere is $R=2$. 
However, the condition $x^2+y^2 \le1$ would mean that it is some kind of half sphere with a smaller "base". I tried to plot this in Wolfram Alpha but I couldn't get it to work.
Is there any way I can parameterize/transform this surface in spherical coordinates?
 A: The surface is a “spherical cap”.  Like you said, the equation $x^2+y^2 + z^2 = 4$ defines the sphere of radius $2$ centered at the origin.  The surface $x^2+y^2 =1$ defines an infinite cylinder of radius $1$ centered along the $z$-axis.  If you think of this cylinder as a straw and run it through the sphere, it punches out two portions of the sphere at the top and bottom.  The third condition $z>0$ keeps the top one of these cutouts.
The sphere and cylinder intersect when $x^2+y^2+z^2 = 4$ and $x^2+y^2 = 1$, so $z^2 = 3$.  This circle can be described in spherical coordinates as
$$
    \cos \phi = \frac{z}{\rho} = \frac{\sqrt{3}}{2} \implies \phi = \frac{\pi}{6}
$$
Therefore you can parametrize the entire surface as 
$
   (\rho,\theta,\phi)
$, where $\rho = 2$, $0 \leq \theta \leq 2\pi$, $0 \leq \phi \leq \frac{\pi}{6}$.  
You could also use cylindrical coordinates, because the shadow this surface makes on the $xy$-plane is the disk of radius $1$.  The surface is represented as $(r,\theta,z)$, where $0 \leq r \leq 1$, $0 \leq \theta \leq 2\pi$, and $z=\sqrt{4-r^2}$.  
A: In spherical coordinates (warning: notations may differ)
$$ \eqalign{x &= r \sin (\theta) \cos (\phi)\cr
            y &= r \sin (\theta) \sin(\phi)\cr
            z &= r \cos(\theta)}$$
you have $\|\vec{x}\| = r$ and $x^2 + y^2 = r^2 \sin^2(\theta)$, so in this case
you want $r = 2$ and $0 \le \theta \le \arcsin(1/2) = \pi/6$.
A: The function $\varphi:B(0,1)\to\mathbb{R}^3$ given by 
$$\varphi(x,y)=\left(x,y,\sqrt{4-x^2-y^ 2}\right)$$ 
will parametrize your surface, where $B(0,1)=\{(x,y)\in\mathbb{R}^2|x^2+y^2\leq 1\}$.
