Caculate surface of vivianis window I want to calculate the surface of
$$S=\{(x,y,z)\in\mathbb{R}^3 | x^2+y^2+z^2=4 \;\wedge\; (x-1)^2 + y^2 \leq 1 \}$$
My attempt: try to solve with a substitution of polar coordinates, but I have not been able to parametrize $x^2+y^2+z^2=4 \;\wedge\; (x-1)^2 + y^2 \leq 1$.
Edit: Per @Math Lover's suggestion, I tried his parametrization:
$z\in[-2,2]$ implies $z=2\cos(\phi)$ (then $\phi\in [0,\pi]$ or $\phi\in[0,2\pi]$?). Choose $y=2\sin(\theta)\sin(\phi),\;x=2\cos(\theta)\sin(\phi)$.
$\implies 4\cos^2(\theta)\sin^2(\phi)+4\sin^2(\theta)\sin^2(\phi)+4\cos^2(\phi)=4$, which satisfies the first equation.
To intersect the two equations, I wrote $$(2\cos(\theta)\sin(\phi)-1)^2+4\sin^2(\theta)\sin^2(\phi) \overset{!}{=} 4$$
The solutions I got from this were $$\theta=\pm \cos^{-1}\left(\sin(\phi) - \frac{1}{\sin(\phi)}\right)$$
So I have $0\leq\phi\leq\pi$ (?) and $-\cos^{-1}\left(\sin(\phi) - \frac{1}{\sin(\phi)}\right)\leq\theta\leq +\cos^{-1}\left(\sin(\phi) - \frac{1}{\sin(\phi)}\right)$.
I am now unsure what to integrate over to get the area. $z^2$? If yes, then
$\implies \int\limits_Sz^2\;do = \int\limits_{\phi=0}^\pi\int\limits_{\theta=-\cos^{-1}\left(\sin(\phi) - \frac{1}{\sin(\phi)}\right)}^{\cos^{-1}\left(\sin(\phi) - \frac{1}{\sin(\phi)}\right)}
4\cos^2(\phi) \sin(\phi) \;d\theta d\phi$
Is $\sin(\phi)$ the correct factor for this substitution? Do I need another factor like $\sqrt{2}$? Thanks a lot
 A: The question asks you to find surface area of the sphere $x^2 + y^2 + z^2 = 4~$ for $~(x-1)^2 + y^2 \leq 1$.
The parametrization for the sphere is correct. You wrote the intersection of surfaces incorrectly. The RHS should be $1$ and not $4$.
$~(x-1)^2 + y^2 \leq 1$ can be simplified as $x^2 + y^2 \leq 2x$. As you can see from the equation, $x^2 + y^2 \leq 2x$ only forms for $x \geq 0$ and hence $-\pi/2 \leq \theta \leq \pi/2$.
At the intersection with the surface of the sphere, $x^2 + y^2 = 4 \sin^2\phi = 2 \cdot 2 \cos\theta \sin\phi \implies \cos\theta = \sin\phi$
So if we only evaluate the surface area for $z \geq 0$ and then using symmetry, multiply by $2$ to include surface below $z \leq 0$,
$- (\pi/2 - \phi) \leq \theta \leq (\pi/2 - \phi), 0 \leq \phi \leq \pi/2$
As we know the surface area element of sphere is $4 \sin \phi ~ d\theta ~ d\phi$ so the integral becomes,
$ \displaystyle S = 2 \int_0^{\pi/2} \int_{- (\pi/2 - \phi)}^{(\pi/2 - \phi)} 4 \sin\phi ~ d\theta ~ d\phi$
or,
$ \displaystyle S = 4 \int_0^{\pi/2} \int_{0}^{(\pi/2 - \theta)} 4 \sin\phi ~ d\phi ~ d\theta$
