$\text{Calculate the area of } S = \{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=4z, z\geq x^2+y^2\}$
As far as I know, what I should use here are the spherical coordinates of $x,y,z$: $$\begin{cases}x=r\sin\theta\cos\phi \\ y=r\sin\theta\sin\phi \\ z=r\cos\theta\end{cases}$$ where $r$ is a constant I will need to determine, $\theta\in[0,\pi],\phi\in[0,2\pi]$. $$x^2+y^2+z^2=4z \iff r^2=4r\cos\theta \iff r =4\cos\theta$$ and $$z\geq x^2+y^2 \iff 4cos^2\theta\geq 16\cos^2\theta\sin^2\theta\iff \sin\theta\leq\frac{1}{2}\iff \theta\in\left[0,\frac{\pi}{6}\right]$$ The area I'm trying to find should be $$\int\int_D||T_\theta \times T_\phi||\text{ d}\phi\text{d}\theta$$ where $$T_\theta=r\cos\theta\cos\phi\vec{i}+r\cos\theta\sin\phi\vec{j}+-r\sin\theta\vec{k} \\ T_\phi = -r\sin\theta\sin\phi\vec{i} + r\sin\theta\cos\phi\vec{j}+0\vec{k}\\ \Rightarrow ||T_\theta\times T_\phi||=r^2\sqrt{\sin^4\theta\cos^2\phi+\sin^4\theta\sin^2\phi+\sin^2\theta\cos^2\theta}=r^2\sin\theta=16\cos^2\theta\sin\theta$$ So, my integral becomes $$\int_{0}^{\frac{\pi}{6}}\int_0^{2\pi}16\cos^2\theta\sin\theta\text{ d}\phi\text{d}\theta$$ According to Desmos, the result is $11.74...$ (my calculations say $32\pi\left(\frac{1}{3}-\frac{\sqrt{3}^3}{24}\right)$), however my textbook says the answer is $4\pi$, which is $12.5...$ That being said, I do not know where I am mistaken, unless it is a pesky calculation error that I am not seeing, but I think it must be in my method. Any help is much appreciated!
As noticed the big mistake is in the assumption that $r$ is constant.
To simplify the derivation and the calculation note that
$$x^2+y^2+z^2=4z \iff x^2+y^2+z^2-4z=0\iff x^2+y^2+(z-2)^2=4$$
which is a sphere centered at $C=(0,0,2)$ with radius $R=2$.
This observation confirms that the result should be $4\pi$ (by $S=2\pi R h$) and, if we want proceed by integration, suggests to change axes (by $Z=z-2$) making the sphere centered at the origin which finally allows to have $r$ constant.