Calculating the surface area of a sphere of radius $a$ with spherical coordinates is quite easy ($4\pi a^2$).
I'm trying to do the same with cylindrical coordinates, $\rho$, $\theta$, $z$, (just for fun) and I can't seem to find the right area...
I put the center of the sphere of radius $a$ at $\rho=z=0$, I integrate over $z\in[-a,a]$ and $\theta\in[0,2\pi]$ with $\rho (z)d\theta dz$, the surface element.
$$\int_{-a}^{a}\int_{0}^{2\pi}\rho(z)d\theta dz,$$ where $\rho(z)=\sqrt{a^2-z^2}$ is the parameterization of the sphere's surface in cylindrical coordinates (independent of $\theta$)
$$ 2\pi\int_{-a}^{a}\sqrt{a^2-z^2} dz=2\pi\left[\frac{1}{2}z\sqrt{a^2-z^2}+\frac{1}{2}a^2\arctan\left(\frac{x}{\sqrt{a^2-z^2}}\right)\right]_{-a}^{a}=\pi^2a^2 $$
That is not equal $4\pi a^2$. I'm a $\frac{4}{\pi}$ factor away, that's basically 1, right? :P
Parameterization of the radius is fine... At this point I'm stumped.
To compare it to something that does work: finding the sphere's volume in cylindrical coordinates is fine. $$ \begin{align} V=\int_{-a}^{a}\int_{0}^{\rho(z)}\int_{0}^{2\pi}\rho(z)d\theta d\rho dz & = 2\pi\int_{-a}^{a}\int_{0}^{\sqrt{a^2-z^2}}\rho(z)d\rho dz\\ & =2\pi\int_{-a}^{a}\left[\frac{\left(\rho(z)\right)^2}{2}\right]_{0}^{\sqrt{a^2-z^2}}dz\\ &=\pi\int_{-a}^{a}\left(a^2-z^2\right)dz\\ &=\pi\left[a^2z-\frac{z^3}{3}\right]_{-a}^{a}\\ V&=\frac{4}{3}\pi a^3 \end{align} $$
I would parametrize the top hemi sphere in cylindrical coordinates as follows: $$ x=r\cos(\theta),\quad y=r\sin(\theta), \quad z=\sqrt{a^2-r^2}, $$
with $(r,\theta)\in D=\{(r,\theta)\;|\;0\le r\le a, 0\le \theta\le 2\pi\}$. It follows that
$$ A=2\iint_D \|r_r\times r_{\theta}\| dA = 2 \int_0^{2\pi}\int_0^a\frac{ar}{\sqrt{a^2-r^2}}drd\theta = 4\pi a^2. $$