Surface area of $x^2+y^2+z^2=9$, where $1\leq x^2+y^2\leq4$ and $z\geq0$ Let $S$ be the portion of the sphere $x^2+y^2+z^2=9$, where $1\leq x^2+y^2\leq4$ and $z\geq0$. Calculate the surface area of $S$
Ok i'm really confused with this one. I know i have to apply the surface area formula but and possibly spherical coordinates but i can't seem how to get the integral out.
The shape. I thought of using spherical system but after doing i ended up with a 3 coordinate system. I'm not even sure how to begin with this one.
 A: The surface $S$ in question is a spherical zone. Its area can be found by elementary means: If $R$ is the radius of the sphere and $h$ is the $z$-height of the zone then the area $\omega(S)$ is given by $$\omega(S)=2\pi R h\ .$$
As $R=3$ and $h$ is easily computed as $h=\sqrt{9-1}-\sqrt{9-4}$ we obtain $$\omega(S)=6\pi(\sqrt{8}-\sqrt{5})\ .$$
Now for the integral: Use polar coordinates in the $(x,y)$-plane as parameter variables. Then $S$ is produced by
$$S:\quad(r,\phi)\mapsto{\bf x}(r,\phi):=(r\cos\phi, \>r\sin\phi, \>\sqrt{9-r^2})\qquad(0\leq r\leq 2, \ 0\leq\phi\leq 2\pi)\ .$$
Then
$${\bf x}_r=\bigl(\cos\phi,\sin\phi,-{r\over\sqrt{9-r^2}}\bigr),\quad{\bf x}_\phi=(-r\sin\phi,r\cos\phi,0)$$
and
$${\bf x}_r\times{\bf x}_\phi=\left({r^2\cos\phi\over\sqrt{9-r^2}}, \ {r^2\sin\phi\over\sqrt{9-r^2}}, \ r\right)\ .$$
Therefore the area element becomes
$${\rm d}\omega=|{\bf x}_r\times{\bf x}_\phi|\>{\rm d}(r,\phi)={3r\over\sqrt{9-r^2}}\>{\rm d}(r,\phi)\ ,$$
and we obtain
$$\omega(S)=\int_0^{2\pi} \int_1^2{3r\over\sqrt{9-r^2}}\>dr\>d\phi=2\pi\ \left(-3\sqrt{9-r^2}\right)\biggr|_1^2=6\pi(\sqrt{8}-\sqrt{5})\ ,$$
as before.
A: Cylindrical coordinates are the way to go! 
Recall that if you have a surface S the surface integral is equal to 
\begin{equation}
\int\int f(x,y,z) dS
\end{equation}
Well, you can represent z as a function of x and y. 
Also recall that dS stands for the "arc-length" at that point.
Therefore, you can rewrite the integral as
\begin{equation}
\int\int f(x,y,z(x,y))\sqrt{(\frac{\partial{z}}{\partial{x}})^{2} + (\frac{\partial{z}}{\partial{y}})^{2} + 1} dA
\end{equation}
Now, convert to cylindrical coordinates. You know that $\theta$ ranges from 0 to 2$\pi$ and $r$ ranges from 1 to 4.
I believe you can go from there.
A: I shall outline symbolically at first:
Due to possibility two axial positions and independence of $ \theta$ ( full $ 0 to 2 \pi) you can use either spherical coordinate system or cylindrical. I used the latter.
$$ dA= 2 \pi r ds  = 2 \pi \frac {dz}{\cos \phi} =2 \pi  R  dz  $$
$$ (\because \cos \phi = \frac {r}{R}) $$
Now integrating the above,
$$ A = 2 \pi R ( z_2-z_1); \,  z_1= \sqrt{R^2-r_1^2}, z_2= \sqrt{R^2-r_2^2}  \tag{1}$$
where generally we have $ (r_1<r_2< R) $.
The result shows that area depends on product of $2 \pi R$ and spherical segment height, well known formula. 
However there are two cases with different segment heights:
Case1: $ (z_1<0 < z_2); $ and \, $ (z_1< z_2<0); $
A rough sketch, z-axis horizontal.

So the areas in the two cases are: 
$$ 2 \pi R (z_2-z_1), 2  \pi R (z_2+z_1)  \tag{2}$$
where z' values are taken from (1), i.e., depending upon whether the end sections lie on  same side or on opposite sides  of the sphere equator/ max. radius. 
In this particular case we have $ R=3, r_1= 1 , r_2 =2. $ and  $ z_1= 2 \sqrt2, z_2= \sqrt 5$ 
$$ A_{1,2}= 6 \pi ( 2 \sqrt2 \pm  \sqrt 5 ) $$
