What kind of surface is this? Is there a way to plot this?

Question

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?

Answer 1

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 $z=\sqrt{4-r^2}$.

Answer 2

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 $0 \le \theta \le \arcsin(1/2) = \pi/6$.

Answer 3

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 $B(0,1)=\{(x,y)\in\mathbb{R}^2|x^2+y^2\leq 1\}$.